This is question 3 from the APICS Mathematics Competition paper of 1999:

Prove that $$\sin^2(x+\alpha)+\sin^2(x+\beta)-2\cos(\alpha-\beta)\sin(x+\alpha)\sin(x+\beta)$$ is a constant function of $x$.

Expanding it seems rather daunting, in particular the last term, and nothing I've tried has been useful towards cancelling terms out. It was assigned in a pre-calculus course, so it should be possible to solve without using derivatives. However, showing that $f'(x)=0$ would obviously be a valid solution.

Any ideas are greatly welcome.

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    $\begingroup$ The natural idea would be to rewrite it in terms of complex exponentials and simplify. $\endgroup$ – Carl Mummert Jul 24 '16 at 14:30
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    $\begingroup$ My gut feeling says think in terms of law of cosines. $\endgroup$ – Siong Thye Goh Jul 24 '16 at 14:34
  • $\begingroup$ Differentiating, you get a trig identity whose proof is really no simpler than the original. $\endgroup$ – DanielWainfleet Jul 24 '16 at 21:31
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    $\begingroup$ Rewriting everything in terms of $\sin(x)=\frac{1}{2i}(e^{ix}-e^{-ix})$ and $\cos(x)=\frac{1}2(e^{ix}+e^{-ix})$ is a trick that always works for this sort of problem. Not so elegant and maybe less comfortable for "pre-calculus", but if someone puts this problem in front of me, that's what's going to happen. $\endgroup$ – Milo Brandt Jul 25 '16 at 3:47

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Taking @SiongthyeGoh's comment, in the above picture, $AE=1$, $\angle BAE=x$, $\angle BAC=\beta$, and $\angle BAD=\alpha$.

It follows that $DE=\sin(x+\alpha)$ and $CE=\sin(x+\beta)$. Applying the cosine theorem to $\triangle DCE$, the given expression is equal to $DC^2$, hence is a constant.

  • $\begingroup$ Quick question: I can see how the given expression is equal to $DC^2$. However, how do we know that $DC$ is constant as the angle x varies? $\endgroup$ – user200783 Jul 25 '16 at 11:02
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    $\begingroup$ $DC$ spans an arc of $2(\alpha-\beta)$ of the circle with diameter $AB=1$. $\endgroup$ – Quang Hoang Jul 25 '16 at 15:13

Let $$E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\color {blue}{2\cos(\alpha-\beta)\sin(x+\alpha)}\sin(x+\beta)$$

Now use $2\cos A \sin B = \sin (A+B) - \sin (A-B)$.

$$E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\left[\sin\left(\alpha-\beta +x+\alpha\right)-\sin\left(\alpha-\beta -x-\alpha\right)\right]\sin(x+\beta)$$

$$E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\left[\sin\left(\alpha-\beta +x+\alpha\right)-\sin\left(-\beta -x\right)\right]\sin(x+\beta)$$

$$E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\left[\sin\left(\alpha-\beta +x+\alpha\right)+\sin\left(x+\beta \right)\right]\sin(x+\beta)$$

$$E=\sin^2(x+\alpha)-\color {red}{\sin\left(\alpha-\beta +x+\alpha\right)\sin(x+\beta)}$$

Now use $2\sin A \sin B = \cos (A-B) - \cos (A+B)$.








Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$ and Werner Formula$(2\sin A\sin B=\cdots),$ $$D=1-\cos(2x+\alpha+\beta)\cos(\alpha-\beta)-\cos(\alpha-\beta)[\cos(\alpha-\beta)-\cos(2x+\alpha+\beta)]$$

$$=\cdots=1-\cos^2(\alpha-\beta)$$ which is independent of $x$



Use the formulas $\sin^2t=\frac {1-\cos2t} 2$ and $\cos2p+\cos2q=2\cos(p+q)\cos(p-q)$.


$$\text {Let }\quad L=\sin^2 (x+a)+\sin^2(x+b).$$ $$\text {Let }\quad R=-2\cos (a-b)\sin (x+a) \sin (x+b).$$ $$\text {We have }\quad L=\frac {1}{2}(1-\cos (2x+2a) +\frac {1}{2}(1-\cos (2x+2b).$$ Now with $y=2x+a+b$ and $z=a-b$ we have $$\cos (2x+2a)=\cos (y+z)=\cos y \cos z-\sin y \sin z. $$ $$\cos (2x+2b)=\cos (y-z)=\cos y \cos z +\sin y \sin z .$$ $$\text {Thus, }\quad\cos (2x+2a) + \cos (2x+2b )=2 \cos y \cos z.$$ $$\text {Therefore }\quad L=1-\cos y \cos z.$$ The motivation for this comes from examining R : We always have $\sin U \sin V= \frac {1}{2}(\cos (U-V)-\cos (U+V)). $ So we have $$\sin (x+a) \sin (x+b)=\frac {1}{2}(\cos (a-b)-\cos (2x+a+b))=\frac {1}{2}(\cos z-\cos y).$$ $$\text {Therefore }\quad R= (-2\cos z)\cdot \frac {1}{2}(\cos z-\cos y)=-\cos^2 z+\cos z \cos y.$$ $$\text {In conclusion, } \quad L+R=1-\cos^2z=1-\cos^2 (a-b)$$ which is independent of $x.$


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