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.