Prove that: sin(3π/8 − θ) cos(π/8 + θ) + cos(3π/8 − θ) sin(π/8 + θ) = 1

Question

Prove that:

\[ \sin\left(\frac{3\pi}{8}-\theta\right)\cos\left(\frac{\pi}{8}+\theta\right) + \cos\left(\frac{3\pi}{8}-\theta\right)\sin\left(\frac{\pi}{8}+\theta\right) =1 \]

Proof

Consider the left-hand side:

\[ \sin\left(\frac{3\pi}{8}-\theta\right)\cos\left(\frac{\pi}{8}+\theta\right) + \cos\left(\frac{3\pi}{8}-\theta\right)\sin\left(\frac{\pi}{8}+\theta\right) \]

Using the identity:

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

Let

\[ A=\frac{3\pi}{8}-\theta, \qquad B=\frac{\pi}{8}+\theta \]

Then,

\[ = \sin\left[\left(\frac{3\pi}{8}-\theta\right)+\left(\frac{\pi}{8}+\theta\right)\right] \]

\[ = \sin\left(\frac{3\pi}{8}+\frac{\pi}{8}\right) \]

\[ = \sin\left(\frac{4\pi}{8}\right) \]

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

We know that:

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

Therefore,

\[ \sin\left(\frac{3\pi}{8}-\theta\right)\cos\left(\frac{\pi}{8}+\theta\right) + \cos\left(\frac{3\pi}{8}-\theta\right)\sin\left(\frac{\pi}{8}+\theta\right) =1 \]

Hence proved.