Given question is : $$\sin(5\alpha + \theta) = \cos(\theta - 3\alpha)$$
We are to find the least positive value of $\alpha$ for which above equation holds.
The way I did is as, $$\sin5\alpha \cos\theta + \cos5\alpha \sin\theta = \cos\theta \cos3\alpha + \sin\theta \sin3\alpha$$
Now for this to be true $$\sin5\alpha = \cos3\alpha$$ and $$\cos5\alpha = \sin3\alpha$$
How do I find the value of $\alpha$ that satisfies the above criteria?
$\endgroup$ 153 Answers
$\begingroup$Divide the two relations, after noting that none among $\sin5\alpha$, $\cos5\alpha$, $\sin3\alpha$ and $\cos3\alpha$ can be zero, if the relations you have are to hold. Then $$ \tan5\alpha=\cot3\alpha=\tan\left(\frac{\pi}{2}-3\alpha\right) $$ Recall that $\tan x=\tan y$ if and only if $x=y+k\pi$, for some integer $k$.
$\endgroup$ 4 $\begingroup$My suggestion would be:
$$\sin(5\alpha + \theta) = \cos(\theta - 3\alpha)=\sin(\pi/2-\theta+3\alpha)$$
so, $$\sin(5\alpha + \theta) -\sin(\pi/2-\theta+3\alpha)=0$$
$$2\sin\left(\frac{(5\alpha + \theta)-(\pi/2-\theta+3\alpha)}{2}\right)\cdot \cos\left(\frac{(5\alpha + \theta)+(\pi/2-\theta+3\alpha)}{2}\right)=0$$
$$2\sin\left(\alpha + \theta-\frac{\pi}{4}\right)\cdot \cos\left(4\alpha+\frac{\pi}{4}\right)=0$$
So your full solution is:
$$\cos\left(4\alpha+\frac{\pi}{4}\right)=0\to 4\alpha+\frac{\pi}{4}=\frac{\pi}{2}+k\pi\to \alpha=\frac{\pi}{16}+\frac{k\pi}{4}$$ or
$$\sin\left(\alpha + \theta-\frac{\pi}{4}\right)=0\to \alpha + \theta-\frac{\pi}{4}=k\pi\to \alpha=\frac{\pi}{4}-\theta+k\pi$$
$\endgroup$ 10 $\begingroup$use then $$\sin(x)-\cos(y)=-2 \sin \left(-\frac{x}{2}-\frac{y}{2}+\frac{\pi }{4}\right) \sin \left(-\frac{x}{2}+\frac{y}{2}+\frac{\pi }{4}\right)$$
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