The value of sin(π/4 + θ) – cos(π/4 – θ) is (a) 2 cos θ (b) 2 sin θ (c) 1 (d) 0

Find the Value of sin(π/4 + θ) − cos(π/4 − θ)

Question:
The value of \[ \sin\left(\frac{\pi}{4}+\theta\right) – \cos\left(\frac{\pi}{4}-\theta\right) \] is

(a) \(2\cos\theta\)
(b) \(2\sin\theta\)
(c) \(1\)
(d) \(0\)

Solution

Using the identity:

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

Therefore,

\[ \sin\left(\frac{\pi}{4}+\theta\right) = \sin\frac{\pi}{4}\cos\theta + \cos\frac{\pi}{4}\sin\theta \]

\[ = \frac{1}{\sqrt2}\cos\theta + \frac{1}{\sqrt2}\sin\theta \]

Similarly, using

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

we get

\[ \cos\left(\frac{\pi}{4}-\theta\right) = \cos\frac{\pi}{4}\cos\theta + \sin\frac{\pi}{4}\sin\theta \]

\[ = \frac{1}{\sqrt2}\cos\theta + \frac{1}{\sqrt2}\sin\theta \]

Hence,

\[ \sin\left(\frac{\pi}{4}+\theta\right) – \cos\left(\frac{\pi}{4}-\theta\right) = 0 \]

\[ \boxed{ \sin\left(\frac{\pi}{4}+\theta\right) – \cos\left(\frac{\pi}{4}-\theta\right) =0 } \]

Correct Option: (d)

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