Question

The value of $\sin \left(\frac{\mathrm{\pi }}{4}+\theta \right)-\cos \left(\frac{\mathrm{\pi }}{4}-\theta \right)$ is
(a) 2 cosθ
(b) 2 sinθ
(c) 1
(d) 0

Answer 1

$$\begin{gathered} \sin \left(\frac{\mathrm{\pi }}{4}+\theta \right)-\cos \left(\frac{\mathrm{\pi }}{4}-\theta \right) \\ =\sin \frac{\mathrm{\pi }}{4}\cos \theta +\cos \frac{\mathrm{\pi }}{4}\sin \theta -\left[\cos \frac{\mathrm{\pi }}{4}\cos \theta +\sin \frac{\mathrm{\pi }}{4}\sin \theta \right] \\ \mathrm{Using}\mathrm{identities}: \\ \sin (a+b)=\sin a\cos b+\cos a\sin b \\ \cos (a-b)=\cos a\cos b+\sin a\sin b \\ =\frac{1}{\sqrt{2}}\cos \theta +\frac{1}{\sqrt{2}}\sin \theta -\frac{1}{\sqrt{2}}\cos \theta -\frac{1}{\sqrt{2}}\sin \theta \\ =0 \\ \mathrm{i}.\mathrm{e}\sin \left(\frac{\mathrm{\pi }}{4}+\theta \right)-\cos \left(\frac{\mathrm{\pi }}{4}-\theta \right)=0. \\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{option}\mathrm{D}. \end{gathered}$$