Question

Evaluate......................​

Answer 1

We use,

$\sin A\cos B-\cos A\sin B=\sin(A-B)\\\\\cos A\cos B-\sin A\sin B=\cos(A+B)$

So,

$\begin{aligned}(\mathrm{i})\quad\ &\sin\dfrac{7\pi}{12}\cos\dfrac{\pi}{4}-\cos\dfrac{7\pi}{12}\sin\dfrac{\pi}{4}\\\\\implies\ \ &\sin\left(\dfrac{7\pi}{12}-\dfrac{\pi}{4}\right)\\\\\implies\ \ &\sin\dfrac{\pi}{3}\\\\\implies\ \ &\dfrac{\sqrt3}{2}\end{aligned}$

And,

$\begin{aligned}(\mathrm{ii})\quad\ &\cos\dfrac{2\pi}{3}\cos\dfrac{\pi}{4}-\sin\dfrac{2\pi}{3}\sin\dfrac{\pi}{4}\\\\\implies\ \ &\cos\left(\dfrac{2\pi}{3}+\dfrac{\pi}{4}\right)\\\\\implies\ \ &\cos\dfrac{11\pi}{12}\\\\\implies\ \ &\cos\left(\pi-\dfrac{\pi}{12}\right)\\\\\implies\ \ &-\cos\dfrac{\pi}{12}\end{aligned}$

What about  $\cos\dfrac{\pi}{12}\ ?$

We know  $\cos\dfrac{\pi}{6}=\dfrac{\sqrt3}{2}.$

So we use another formula,

$\cos\theta=\sqrt{\dfrac{\cos(2\theta)+1}{2}}$

This is derived from  $\cos(2\theta)=2\cos^2\theta-1.$

[You can use this equation too if you feel trouble applying that equation.]

So,

$\cos\dfrac{\pi}{12}=\sqrt{\dfrac{\cos\dfrac{\pi}{6}+1}{2}}\\\\\\\cos\dfrac{\pi}{12}=\sqrt{\dfrac{\dfrac{\sqrt3}{2}+1}{2}}\\\\\\\\\cos\dfrac{\pi}{12}=\sqrt{\dfrac{2+\sqrt3}{4}}$

Can this be further simplified?!

Answer 2

Answer:

Hope it helps you.......