The value of sin
Question:

The value of $\sin \left(\frac{\pi}{4}+\theta\right)-\cos \left(\frac{\pi}{4}-\theta\right)$ is

(a) 2 cosθ

(b) 2 sinθ

(c) 1

(d) 0

Solution:

$\sin \left(\frac{\pi}{4}+\theta\right)-\cos \left(\frac{\pi}{4}-\theta\right)$

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

Using identities :

$\sin (a+b)=\sin a \cos b+\cos a \sin b$

$\cos (a-b)=\cos a \cos b+\sin a \sin b$

$=\frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \sin \theta-\frac{1}{\sqrt{2}} \cos \theta-\frac{1}{\sqrt{2}} \sin \theta$

= 0

i. e $\sin \left(\frac{\pi}{4}+\theta\right)-\cos \left(\frac{\pi}{4}-\theta\right)=0$.

Hence, the correct answer is option D.

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