Prove that $\sin(\frac{\pi}{3}+x)=\cos(\frac{\pi}{6}-x)$

Question

How to prove that $\sin(\frac{\pi}{3}+x)=\cos(\frac{\pi}{6}-x)$ without using calculus just trigonometry?

Answer 1

Hint: Write: $$\sin\left(\dfrac\pi3+x\right)=\sin\left(\dfrac\pi2-\dfrac\pi6+x\right)$$ and use the fact that: $$\sin\left(\dfrac\pi2+\theta\right)=\cos(\theta)=\cos(-\theta).$$

Answer 2

Hint: $\sin t =\cos(\frac\pi 2-t)$.

Answer 3

Trigonometric identities:

$$\sin\left(\frac\pi3+x\right)=\sin\frac\pi3\cos x+\sin x\cos\frac\pi3$$

$${}$$

$$\cos\left(\frac\pi6-x\right)=\cos\frac\pi6\cos x+\sin\frac\pi6\sin x$$

But

$$\sin\frac\pi3=\cos\frac\pi6=\frac{\sqrt3}2\;\;,\;\;\sin\frac\pi6=\cos\frac\pi3=\frac12\;,\;\;\text{so}\;\ldots$$

Answer 4

With the trigonometric identity $$\cos\left(\frac{\pi}{2}+\theta\right)= -\sin(\theta)=\sin(-\theta)$$ we have $$\cos\left(\frac{\pi}{6}-x\right)=\cos\left(\frac{\pi}{2} -\left(\frac{\pi}{3}+x\right)\right)=\sin\left(\frac{\pi}{3}+x\right)$$