Question

The value of sin 78° − sin 66° − sin 42° + sin 60° is
(a) $\frac{1}{2}$

(b) $-\frac{1}{2}$

(c) −1

(d) None of these

Answer 1

(d) None of these

$$\begin{gathered} \sin 78°-\sin 66°-\sin 42°+\sin 60° \\ =\sin 78°-\sin 42°-\sin 66°+\sin 60° \\ =2\sin \left(\frac{78°-42°}{2}\right)\cos \left(\frac{78°+42}{2}\right)-\sin 66°+\sin 60°\left[\because \mathrm{sinA}-\mathrm{sinB}=2\sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\cos \left({\displaystyle \frac{\mathrm{A}+\mathrm{B}}{2}}\right)\right] \\ =2\sin 18°\cos 60°-\sin 66°+\sin 60° \\ =2\times \frac{1}{2}\sin 18°-\sin 66°+\frac{\sqrt{3}}{2} \\ =\sin 18°-\sin 66°+\frac{\sqrt{3}}{2} \\ =\frac{\sqrt{5}-1}{4}-0.914+\frac{\sqrt{3}}{2} \\ =0.309-0.914+0.866 \\ =0.261 \end{gathered}$$