Why does sin(x) - sin(y) equal this? The Next CEO of Stack OverflowProve that...

How does a dynamic QR code work?

Avoiding the "not like other girls" trope?

Direct Implications Between USA and UK in Event of No-Deal Brexit

Why was Sir Cadogan fired?

Calculating discount not working

What steps are necessary to read a Modern SSD in Medieval Europe?

Could a dragon use hot air to help it take off?

pgfplots: How to draw a tangent graph below two others?

What is a typical Mizrachi Seder like?

Free fall ellipse or parabola?

Does Germany produce more waste than the US?

Can you teleport closer to a creature you are Frightened of?

Identify and count spells (Distinctive events within each group)

Read/write a pipe-delimited file line by line with some simple text manipulation

Is it reasonable to ask other researchers to send me their previous grant applications?

Car headlights in a world without electricity

Can this transistor (2N2222) take 6 V on emitter-base? Am I reading the datasheet incorrectly?

Prodigo = pro + ago?

Does the Idaho Potato Commission associate potato skins with healthy eating?

Why do we say “un seul M” and not “une seule M” even though M is a “consonne”?

How badly should I try to prevent a user from XSSing themselves?

Gauss' Posthumous Publications?

Arrows in tikz Markov chain diagram overlap

Is it OK to decorate a log book cover?



Why does sin(x) - sin(y) equal this?



The Next CEO of Stack OverflowProve that $sin(2A)+sin(2B)+sin(2C)=4sin(A)sin(B)sin(C)$ when $A,B,C$ are angles of a triangleWhy $sin(pi)$ sometimes equal to $0$?Understanding expanding trig identitiesWhy does this always equal $1$?When does this equation $cos(alpha + beta) = cos(alpha) + cos(beta)$ hold?Solve $ cos 2x - sin x +1=0$Writing equation in terms of sin and cosSolve Trigonometric Equality, Multiple Angle TrigonometryFinding relationships between angles, a, b and c when $sin a - sin b - sin c = 0$Does $sin^2x-cos^2x$ equal $cos(2x)$












1












$begingroup$


Why does this equality hold?



$sin x - sin y = 2 cos(frac{x+y}{2}) sin(frac{x-y}{2})$.



My professor was saying that since



(i) $sin(A+B)=sin A cos B+ sin B cos A$



and



(ii) $sin(A-B) = sin A cos B - sin B cos A$



we just let $A=frac{x+y}{2}$ and $B=frac{x-y}{2}$. But I tried to write this out and could not figure it out. Any help would be appreciated






real-analysis analysis trigonometry







share|cite|improve this question







New contributor




Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    Let A and B be as you defined. Then $sin(A+B)=sin(frac{x+y}{2}+frac{x-y}{2})$. Evaluate this and use the given identities.
    $endgroup$
    – Newman
    2 hours ago










  • $begingroup$
    After substituting for A and B in the equations (i) and (ii) you have to calculate (i) - (ii)
    $endgroup$
    – R_D
    2 hours ago
















1












$begingroup$


Why does this equality hold?



$sin x - sin y = 2 cos(frac{x+y}{2}) sin(frac{x-y}{2})$.



My professor was saying that since



(i) $sin(A+B)=sin A cos B+ sin B cos A$



and



(ii) $sin(A-B) = sin A cos B - sin B cos A$



we just let $A=frac{x+y}{2}$ and $B=frac{x-y}{2}$. But I tried to write this out and could not figure it out. Any help would be appreciated






real-analysis analysis trigonometry







share|cite|improve this question







New contributor




Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    Let A and B be as you defined. Then $sin(A+B)=sin(frac{x+y}{2}+frac{x-y}{2})$. Evaluate this and use the given identities.
    $endgroup$
    – Newman
    2 hours ago










  • $begingroup$
    After substituting for A and B in the equations (i) and (ii) you have to calculate (i) - (ii)
    $endgroup$
    – R_D
    2 hours ago














1












1








1





$begingroup$


Why does this equality hold?



$sin x - sin y = 2 cos(frac{x+y}{2}) sin(frac{x-y}{2})$.



My professor was saying that since



(i) $sin(A+B)=sin A cos B+ sin B cos A$



and



(ii) $sin(A-B) = sin A cos B - sin B cos A$



we just let $A=frac{x+y}{2}$ and $B=frac{x-y}{2}$. But I tried to write this out and could not figure it out. Any help would be appreciated






real-analysis analysis trigonometry







share|cite|improve this question







New contributor




Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Why does this equality hold?



$sin x - sin y = 2 cos(frac{x+y}{2}) sin(frac{x-y}{2})$.



My professor was saying that since



(i) $sin(A+B)=sin A cos B+ sin B cos A$



and



(ii) $sin(A-B) = sin A cos B - sin B cos A$



we just let $A=frac{x+y}{2}$ and $B=frac{x-y}{2}$. But I tried to write this out and could not figure it out. Any help would be appreciated






real-analysis analysis trigonometry




real-analysis analysis trigonometry






share|cite|improve this question







New contributor




Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 2 hours ago









Ryan DuranRyan Duran

61




61




New contributor




Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    Let A and B be as you defined. Then $sin(A+B)=sin(frac{x+y}{2}+frac{x-y}{2})$. Evaluate this and use the given identities.
    $endgroup$
    – Newman
    2 hours ago










  • $begingroup$
    After substituting for A and B in the equations (i) and (ii) you have to calculate (i) - (ii)
    $endgroup$
    – R_D
    2 hours ago


















  • $begingroup$
    Let A and B be as you defined. Then $sin(A+B)=sin(frac{x+y}{2}+frac{x-y}{2})$. Evaluate this and use the given identities.
    $endgroup$
    – Newman
    2 hours ago










  • $begingroup$
    After substituting for A and B in the equations (i) and (ii) you have to calculate (i) - (ii)
    $endgroup$
    – R_D
    2 hours ago
















$begingroup$
Let A and B be as you defined. Then $sin(A+B)=sin(frac{x+y}{2}+frac{x-y}{2})$. Evaluate this and use the given identities.
$endgroup$
– Newman
2 hours ago




$begingroup$
Let A and B be as you defined. Then $sin(A+B)=sin(frac{x+y}{2}+frac{x-y}{2})$. Evaluate this and use the given identities.
$endgroup$
– Newman
2 hours ago












$begingroup$
After substituting for A and B in the equations (i) and (ii) you have to calculate (i) - (ii)
$endgroup$
– R_D
2 hours ago




$begingroup$
After substituting for A and B in the equations (i) and (ii) you have to calculate (i) - (ii)
$endgroup$
– R_D
2 hours ago










3 Answers
3






active

oldest

votes


















2












$begingroup$

Following your professor's advice, let $A=frac{x+y}{2}$, $B=frac{x-y}{2}$. Then $$x=A+B\y=A-B$$So the LHS of your equation becomes $$sin(A+B)-sin(A-B)$$Now you just use the usual addition/subtraction trigonometric identities (i) and (ii) listed to evaluate this. It should give $2cos Asin B$ as required.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    The main trick is here:



    begin{align}
    color{red} {x = {x+yover2} + {x-yover2}}\[1em]
    color{blue}{y = {x+yover2} - {x-yover2}}
    end{align}



    (You may evaluate the right-hand sides of them to verify that these strange equations are correct.)



    Substituting the right-hand sides for $color{red}x$ and $color{blue}y,,$ you will obtain



    begin{align}
    sin color{red} x - sin color{blue }y = sin left(color{red}{{x+yover2} + {x-yover2} }right) - sin left(color{blue }{{x+yover2} - {x-yover2}} right) \[1em]
    end{align}



    All the rest is then only a routine calculation:



    begin{align}
    require{enclose}
    &= sin left({x+yover2}right) cosleft( {x-yover2} right) +
    sin left({x-yover2}right) cosleft( {x+yover2} right)\[1em]
    &-left[sin left({x+yover2}right) cosleft( {x-yover2} right) -
    sin left({x-yover2}right) cosleft( {x+yover2} right)right]\[3em]
    &= enclose{updiagonalstrike}{sin left({x+yover2}right) cosleft( {x-yover2} right)} +
    sin left({x-yover2}right) cosleft( {x+yover2} right)\[1em]
    &-enclose{updiagonalstrike}{sin left({x+yover2}right) cosleft( {x-yover2} right)} +
    sin left({x-yover2}right) cosleft( {x+yover2} right)
    \[3em]
    &=2sin left({x-yover2}right) cosleft( {x+yover2} right)\
    end{align}






    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      Following your notation, let $A=dfrac{x+y}{2}$ and $B=dfrac{x-y}{2}$.
      Note that $A+B=x$ and $A-B=y$.



      Now, $sin x=sin(A+B)=sin Acos B+cos Asin B$ and $sin y=sin(A-B)=sin Acos B - cos Asin B$ from your professor's advice.



      To get the LHS, $sin x-sin y = 2cos Asin B$. And that's it. Replace $A,B$ in terms of $x$ and $y$.






      share|cite|improve this answer









      $endgroup$














        Your Answer





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "69"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        autoActivateHeartbeat: false,
        convertImagesToLinks: true,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        noCode: true, onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });






        Ryan Duran is a new contributor. Be nice, and check out our Code of Conduct.










        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3171404%2fwhy-does-sinx-siny-equal-this%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        Following your professor's advice, let $A=frac{x+y}{2}$, $B=frac{x-y}{2}$. Then $$x=A+B\y=A-B$$So the LHS of your equation becomes $$sin(A+B)-sin(A-B)$$Now you just use the usual addition/subtraction trigonometric identities (i) and (ii) listed to evaluate this. It should give $2cos Asin B$ as required.






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          Following your professor's advice, let $A=frac{x+y}{2}$, $B=frac{x-y}{2}$. Then $$x=A+B\y=A-B$$So the LHS of your equation becomes $$sin(A+B)-sin(A-B)$$Now you just use the usual addition/subtraction trigonometric identities (i) and (ii) listed to evaluate this. It should give $2cos Asin B$ as required.






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            Following your professor's advice, let $A=frac{x+y}{2}$, $B=frac{x-y}{2}$. Then $$x=A+B\y=A-B$$So the LHS of your equation becomes $$sin(A+B)-sin(A-B)$$Now you just use the usual addition/subtraction trigonometric identities (i) and (ii) listed to evaluate this. It should give $2cos Asin B$ as required.






            share|cite|improve this answer









            $endgroup$



            Following your professor's advice, let $A=frac{x+y}{2}$, $B=frac{x-y}{2}$. Then $$x=A+B\y=A-B$$So the LHS of your equation becomes $$sin(A+B)-sin(A-B)$$Now you just use the usual addition/subtraction trigonometric identities (i) and (ii) listed to evaluate this. It should give $2cos Asin B$ as required.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 hours ago









            John DoeJohn Doe

            11.4k11239




            11.4k11239























                2












                $begingroup$

                The main trick is here:



                begin{align}
                color{red} {x = {x+yover2} + {x-yover2}}\[1em]
                color{blue}{y = {x+yover2} - {x-yover2}}
                end{align}



                (You may evaluate the right-hand sides of them to verify that these strange equations are correct.)



                Substituting the right-hand sides for $color{red}x$ and $color{blue}y,,$ you will obtain



                begin{align}
                sin color{red} x - sin color{blue }y = sin left(color{red}{{x+yover2} + {x-yover2} }right) - sin left(color{blue }{{x+yover2} - {x-yover2}} right) \[1em]
                end{align}



                All the rest is then only a routine calculation:



                begin{align}
                require{enclose}
                &= sin left({x+yover2}right) cosleft( {x-yover2} right) +
                sin left({x-yover2}right) cosleft( {x+yover2} right)\[1em]
                &-left[sin left({x+yover2}right) cosleft( {x-yover2} right) -
                sin left({x-yover2}right) cosleft( {x+yover2} right)right]\[3em]
                &= enclose{updiagonalstrike}{sin left({x+yover2}right) cosleft( {x-yover2} right)} +
                sin left({x-yover2}right) cosleft( {x+yover2} right)\[1em]
                &-enclose{updiagonalstrike}{sin left({x+yover2}right) cosleft( {x-yover2} right)} +
                sin left({x-yover2}right) cosleft( {x+yover2} right)
                \[3em]
                &=2sin left({x-yover2}right) cosleft( {x+yover2} right)\
                end{align}






                share|cite|improve this answer











                $endgroup$


















                  2












                  $begingroup$

                  The main trick is here:



                  begin{align}
                  color{red} {x = {x+yover2} + {x-yover2}}\[1em]
                  color{blue}{y = {x+yover2} - {x-yover2}}
                  end{align}



                  (You may evaluate the right-hand sides of them to verify that these strange equations are correct.)



                  Substituting the right-hand sides for $color{red}x$ and $color{blue}y,,$ you will obtain



                  begin{align}
                  sin color{red} x - sin color{blue }y = sin left(color{red}{{x+yover2} + {x-yover2} }right) - sin left(color{blue }{{x+yover2} - {x-yover2}} right) \[1em]
                  end{align}



                  All the rest is then only a routine calculation:



                  begin{align}
                  require{enclose}
                  &= sin left({x+yover2}right) cosleft( {x-yover2} right) +
                  sin left({x-yover2}right) cosleft( {x+yover2} right)\[1em]
                  &-left[sin left({x+yover2}right) cosleft( {x-yover2} right) -
                  sin left({x-yover2}right) cosleft( {x+yover2} right)right]\[3em]
                  &= enclose{updiagonalstrike}{sin left({x+yover2}right) cosleft( {x-yover2} right)} +
                  sin left({x-yover2}right) cosleft( {x+yover2} right)\[1em]
                  &-enclose{updiagonalstrike}{sin left({x+yover2}right) cosleft( {x-yover2} right)} +
                  sin left({x-yover2}right) cosleft( {x+yover2} right)
                  \[3em]
                  &=2sin left({x-yover2}right) cosleft( {x+yover2} right)\
                  end{align}






                  share|cite|improve this answer











                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    The main trick is here:



                    begin{align}
                    color{red} {x = {x+yover2} + {x-yover2}}\[1em]
                    color{blue}{y = {x+yover2} - {x-yover2}}
                    end{align}



                    (You may evaluate the right-hand sides of them to verify that these strange equations are correct.)



                    Substituting the right-hand sides for $color{red}x$ and $color{blue}y,,$ you will obtain



                    begin{align}
                    sin color{red} x - sin color{blue }y = sin left(color{red}{{x+yover2} + {x-yover2} }right) - sin left(color{blue }{{x+yover2} - {x-yover2}} right) \[1em]
                    end{align}



                    All the rest is then only a routine calculation:



                    begin{align}
                    require{enclose}
                    &= sin left({x+yover2}right) cosleft( {x-yover2} right) +
                    sin left({x-yover2}right) cosleft( {x+yover2} right)\[1em]
                    &-left[sin left({x+yover2}right) cosleft( {x-yover2} right) -
                    sin left({x-yover2}right) cosleft( {x+yover2} right)right]\[3em]
                    &= enclose{updiagonalstrike}{sin left({x+yover2}right) cosleft( {x-yover2} right)} +
                    sin left({x-yover2}right) cosleft( {x+yover2} right)\[1em]
                    &-enclose{updiagonalstrike}{sin left({x+yover2}right) cosleft( {x-yover2} right)} +
                    sin left({x-yover2}right) cosleft( {x+yover2} right)
                    \[3em]
                    &=2sin left({x-yover2}right) cosleft( {x+yover2} right)\
                    end{align}






                    share|cite|improve this answer











                    $endgroup$



                    The main trick is here:



                    begin{align}
                    color{red} {x = {x+yover2} + {x-yover2}}\[1em]
                    color{blue}{y = {x+yover2} - {x-yover2}}
                    end{align}



                    (You may evaluate the right-hand sides of them to verify that these strange equations are correct.)



                    Substituting the right-hand sides for $color{red}x$ and $color{blue}y,,$ you will obtain



                    begin{align}
                    sin color{red} x - sin color{blue }y = sin left(color{red}{{x+yover2} + {x-yover2} }right) - sin left(color{blue }{{x+yover2} - {x-yover2}} right) \[1em]
                    end{align}



                    All the rest is then only a routine calculation:



                    begin{align}
                    require{enclose}
                    &= sin left({x+yover2}right) cosleft( {x-yover2} right) +
                    sin left({x-yover2}right) cosleft( {x+yover2} right)\[1em]
                    &-left[sin left({x+yover2}right) cosleft( {x-yover2} right) -
                    sin left({x-yover2}right) cosleft( {x+yover2} right)right]\[3em]
                    &= enclose{updiagonalstrike}{sin left({x+yover2}right) cosleft( {x-yover2} right)} +
                    sin left({x-yover2}right) cosleft( {x+yover2} right)\[1em]
                    &-enclose{updiagonalstrike}{sin left({x+yover2}right) cosleft( {x-yover2} right)} +
                    sin left({x-yover2}right) cosleft( {x+yover2} right)
                    \[3em]
                    &=2sin left({x-yover2}right) cosleft( {x+yover2} right)\
                    end{align}







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 1 hour ago

























                    answered 1 hour ago









                    MarianDMarianD

                    2,0531617




                    2,0531617























                        1












                        $begingroup$

                        Following your notation, let $A=dfrac{x+y}{2}$ and $B=dfrac{x-y}{2}$.
                        Note that $A+B=x$ and $A-B=y$.



                        Now, $sin x=sin(A+B)=sin Acos B+cos Asin B$ and $sin y=sin(A-B)=sin Acos B - cos Asin B$ from your professor's advice.



                        To get the LHS, $sin x-sin y = 2cos Asin B$. And that's it. Replace $A,B$ in terms of $x$ and $y$.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Following your notation, let $A=dfrac{x+y}{2}$ and $B=dfrac{x-y}{2}$.
                          Note that $A+B=x$ and $A-B=y$.



                          Now, $sin x=sin(A+B)=sin Acos B+cos Asin B$ and $sin y=sin(A-B)=sin Acos B - cos Asin B$ from your professor's advice.



                          To get the LHS, $sin x-sin y = 2cos Asin B$. And that's it. Replace $A,B$ in terms of $x$ and $y$.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Following your notation, let $A=dfrac{x+y}{2}$ and $B=dfrac{x-y}{2}$.
                            Note that $A+B=x$ and $A-B=y$.



                            Now, $sin x=sin(A+B)=sin Acos B+cos Asin B$ and $sin y=sin(A-B)=sin Acos B - cos Asin B$ from your professor's advice.



                            To get the LHS, $sin x-sin y = 2cos Asin B$. And that's it. Replace $A,B$ in terms of $x$ and $y$.






                            share|cite|improve this answer









                            $endgroup$



                            Following your notation, let $A=dfrac{x+y}{2}$ and $B=dfrac{x-y}{2}$.
                            Note that $A+B=x$ and $A-B=y$.



                            Now, $sin x=sin(A+B)=sin Acos B+cos Asin B$ and $sin y=sin(A-B)=sin Acos B - cos Asin B$ from your professor's advice.



                            To get the LHS, $sin x-sin y = 2cos Asin B$. And that's it. Replace $A,B$ in terms of $x$ and $y$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 57 mins ago









                            AdmuthAdmuth

                            585




                            585






















                                Ryan Duran is a new contributor. Be nice, and check out our Code of Conduct.










                                draft saved

                                draft discarded


















                                Ryan Duran is a new contributor. Be nice, and check out our Code of Conduct.













                                Ryan Duran is a new contributor. Be nice, and check out our Code of Conduct.












                                Ryan Duran is a new contributor. Be nice, and check out our Code of Conduct.
















                                Thanks for contributing an answer to Mathematics Stack Exchange!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid …



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3171404%2fwhy-does-sinx-siny-equal-this%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                -

                                Popular posts from this blog

                                What is the “three and three hundred thousand syndrome”?Who wrote the book Arena?What five creatures were...

                                Image
                                What happens when the centripetal force is equal and opposite to the centrifugal force? Why is "la Gestapo" feminine? Nested Dynamic SOQL Query How are passwords stolen from companies if they only store hashes? Exposing a company lying about themselves in a tightly knit industry: Is my career at risk on the long run? Can "few" be used as a subject? If so, what is the rule? Why are there no stars visible in cislunar space? categorizing a variable turns it from insignificant to significant Do I need an EFI partition for each 18.04 ubuntu I have on my HD? What are the consequences of changing the number of hours in ...
                                Read more

                                Gersau Kjelder | Navigasjonsmeny46°59′0″N 8°31′0″E46°59′0″N...

                                Image
                                Tidlegare republikkarSkipingar i 1433Kommunar i SchwyzDistrikt i SchwyzNedleggingar i 1798Stader ved Vierwaldstättersee kommunekantonenSchwyzSveitskm²VierwaldstätterseeLauerz Gersau Kommunevåpenet til Gersau Plassering Styresmakter Land   Sveits Kanton kantonen Schwyz Distrikt Gersau Språk tysk Geografi Flatevidd  - kommune 23,70 km² Innbyggjarar  - kommune (desember 2010)    - folketettleik 2 094   88,4 /km² Koordinatar 46°59′0″N 8°31′0″E  Koordinatar: 46°59′0″N 8°31′0″E  Høgd over havet 435 moh Diverse annan informasjon Postnummer 6442 Nettstad: www.gersau.ch Freistaat und Republik Gersau Fristaten og republikken Gersau By-republikk og alliert protektorat i Det gamle sveitsiske eidsforundet ←   ← 1433–1798 1814–1818 →   → Hovudstad Gersau Styreform Republikk Historisk periode Tidleg moderne tid  - Kjøpte fridom frå ...
                                Read more

                                Hestehale Innhaldsliste Hestehale på kvinner | Hestehale på menn | Galleri | Sjå òg |...

                                Image
                                Frisyrar frisyrehårhårstrikkhårklemmerøyrethesthovudverkeuropeisk motejobbtreningfletterparykkarungdomsopprørmotkulturmusikararmotorsykkelentusiastarpolitietpolitiskforretningverksemdfiskeflette Hestehale sett bakfrå. Hestehale på ein mannleg indianarleiar. Hestehale er ein frisyre der ein samlar det meste eller alt av langt hår [1] i bakhovudet med ein hårstrikk eller liknande. Avhengig av mote kan det òg vere vanleg å sette opp hestehalen med hårklemmer eller å ha hestehalen hengande på den eine sida av hovudet ved å samle håret ved øyret. Hestehale er ein vanleg frisyre for jenter og for gutar med langt hår. Han har fått namnet sitt etter likskapen med halen på ein hest. For stramme hestehaler kan føre til hårtap, [2] i nokre høve også hovudverk. [3] Innhaldsliste 1 Hestehale på kvinner 2 Hestehale på menn 3 Galleri 4 Sjå òg 5 Bakgrunnsstoff 6 Kjelder Hestehale på kvinner | Ungjente med hestehale som held håret unna ansiktet...
                                Read more