If $\sin(\pi \cos\theta) = \cos(\pi\sin\theta)$, then show ........

If $\sin(\pi\cos\theta) = \cos(\pi\sin\theta)$,
then show that $\sin2\theta = \pm 3/4$. I can do it simply by equating $\pi - \pi\cos\theta$ to $\pi\sin\theta$, but that would be technically wrong as those angles could be in different quadrants. So how to solve?

$$\cos(\pi\sin\theta)=\sin(\pi\cos\theta)=\cos\left(\dfrac\pi2-\pi\cos\theta\right)$$

$$\implies\pi\sin\theta=2m\pi\pm\left(\dfrac\pi2-\pi\cos\theta\right)$$ where $m$ is any integer

$$\iff\sin\theta=2m\pm\left(\dfrac12-\cos\theta\right)$$

$$\iff\sin\theta\mp\cos\theta=2m\pm\dfrac12$$

$\sin\theta\mp\cos\theta=\sqrt2\sin\left(\theta\mp\dfrac\pi4\right)$

$\implies-\sqrt2\le\sin\theta\mp\cos\theta\le\sqrt2\implies m=?$

Now square both sides

• I didn't quite understand what to do after finding the min. and max. value of sin(theta) + cos(theta). How do I find 'm'? Jun 21, 2016 at 6:26
• @JamilAhmed, If $m\ge1,$ $$2m+\dfrac12>2m-\dfrac12\ge\dfrac32>\sqrt2$$ What if $m\le-1$ Jun 21, 2016 at 6:35
• I still didn't get you. If m>1 , then 2m+ 1/2 > 3/2 , but how does it have any relation with root over 2? Can you please explain this potion again....? Also, how is it helpful to prove that sin2(theta) = +- 3/4 Jun 21, 2016 at 7:16
• @JamilAhmed, It required $m<1$ and the $-$ will require $m>-1$ $$\implies m$$ has to be zero Jun 21, 2016 at 8:57

We first rewrite $$\cos(\pi/2 - \pi \cos \theta) = \cos(\pi \sin \theta)$$ (cofunction identity).

We notice that $\cos(x)$ is a periodic function with period $2\pi$, so we need a period offset term to be sure that we find all solutions. We also need to account for the fact that $\cos(x)$ is symmetric, so: $$\cos(\pi/2 - \pi \cos \theta) = \cos(\pm \pi \sin \theta + 2 \pi k) \\ \pi/2 - \pi \cos \theta = \pm \pi \sin \theta + 2 \pi k$$ Then we do basic algebra $$1/2 - \cos \theta = \pm \sin \theta + 2k \\ \cos \theta \pm \sin \theta = 1/2 - 2k$$ We'll get back to this.

Trigonometric trick time!

We can do a neat little trick to all functions of the form $a \cos \theta + b \sin \theta$. We rewrite: \begin{align*} a \cos \theta + b \sin \theta &= \sqrt{a^2+b^2} \left(\frac{a}{\sqrt{a^2+b^2}} \cos \theta + \frac{b}{\sqrt{a^2+b^2}} \sin \theta\right) \\ &=: \sqrt{a^2+b^2} (a' \cos \theta + b' \sin \theta) \\ &=: \sqrt{a^2+b^2} (\cos \phi \cos \theta + \sin \phi \sin \theta) \\ &=: \sqrt{a^2+b^2} \cos (\theta - \phi) \end{align*} where we can find $\phi=\arctan(b'/a') + \pi n = \arctan(b/a) +\pi n$ (note the $\pi n$ because, like $\sin(x)$ and $\cos(x)$, $\tan(x)$ is also a periodic function, so we have to account for all possible inverse values).

End trigonometric trick

We apply the trick to get $$\cos \theta \pm \sin \theta = 1/2 - 2k \\ \sqrt{1^2+1^2} \cos (\theta - \arctan (\pm1/1)) = 1/2 - 2k \\ \sqrt 2 \cos (\theta \mp \pi/4) = 1/2 - 2k$$ (You can verify that the rewritten forms do indeed evaluate to the original.)

We continue: $$\cos (\theta \mp \pi/4) = 1/(2\sqrt{2}) + \sqrt{2} k = \sqrt{2} (1/4 - k)$$ Notice that $-1 \le \cos(x) \le 1$, so only $k = 0$ is valid. Hence we get $$\cos (\theta \mp \pi/4) = \sqrt{2}/4$$ Next, we notice \begin{align*} \sin 2\theta &= \cos (\pi/2 - 2 \theta) \\ &= \cos (2 \theta - \pi/2) \\ &= \cos(2 (\theta - \pi/4)) \\ &= 2 \cos^2 (\theta - \pi/4) - 1 \\ &= 2 (\sqrt{2}/4)^2 - 1 \\ &= 1/4 - 1 \\ &= -3/4 \end{align*} We notice the other solution \begin{align*} \sin 2\theta &= -\cos (\pi/2 - 2 \theta) \\ &= \cos (2 \theta - \pi/2) \\ &= -\cos (2 \theta - \pi/2 + \pi) \\ &= -\cos (2 \theta + \pi/2) \\ &= -\cos(2 (\theta + \pi/4)) \\ &= -(2 \cos^2 (\theta + \pi/4) - 1) \\ &= -(2 (\sqrt{2}/4)^2 - 1) \\ &= -(1/4 - 1) \\ &= 3/4 \end{align*}