Prove that: sin(π/3 − x) cos(π/6 + x) + cos(π/3 − x) sin(π/6 + x) = 1

Question

Prove that:

\[ \sin\left(\frac{\pi}{3}-x\right)\cos\left(\frac{\pi}{6}+x\right) + \cos\left(\frac{\pi}{3}-x\right)\sin\left(\frac{\pi}{6}+x\right) =1 \]

Proof

Consider the left-hand side:

\[ \sin\left(\frac{\pi}{3}-x\right)\cos\left(\frac{\pi}{6}+x\right) + \cos\left(\frac{\pi}{3}-x\right)\sin\left(\frac{\pi}{6}+x\right) \]

Using the identity:

\[ \sin A\cos B+\cos A\sin B=\sin(A+B) \]

Let

\[ A=\frac{\pi}{3}-x, \qquad B=\frac{\pi}{6}+x \]

Then,

\[ = \sin\left[\left(\frac{\pi}{3}-x\right)+\left(\frac{\pi}{6}+x\right)\right] \]

\[ = \sin\left(\frac{\pi}{3}+\frac{\pi}{6}\right) \]

\[ = \sin\left(\frac{2\pi+\pi}{6}\right) \]

\[ = \sin\left(\frac{3\pi}{6}\right) \]

\[ = \sin\left(\frac{\pi}{2}\right) \]

We know that:

\[ \sin\frac{\pi}{2}=1 \]

Therefore,

\[ \sin\left(\frac{\pi}{3}-x\right)\cos\left(\frac{\pi}{6}+x\right) + \cos\left(\frac{\pi}{3}-x\right)\sin\left(\frac{\pi}{6}+x\right) =1 \]

Hence proved.