The value of sin 78° − sin 66° − sin 42° + sin 60° is

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

Solution:

(d) None of these

$\sin 78^{\circ}-\sin 66^{\circ}-\sin 42^{\circ}+\sin 60^{\circ}$

$=\sin 78^{\circ}-\sin 42^{\circ}-\sin 66^{\circ}+\sin 60^{\circ}$

$=2 \sin \left(\frac{78^{\circ}-42^{\circ}}{2}\right) \cos \left(\frac{78^{\circ}+42}{2}\right)-\sin 66^{\circ}+\sin 60^{\circ}\left[\because \sin \mathrm{A}-\sin \mathrm{B}=2 \sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right) \cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\right]$

$=2 \sin 18^{\circ} \cos 60^{\circ}-\sin 66^{\circ}+\sin 60^{\circ}$

$=2 \times \frac{1}{2} \sin 18^{\circ}-\sin 66^{\circ}+\frac{\sqrt{3}}{2}$

$=\sin 18^{\circ}-\sin 66^{\circ}+\frac{\sqrt{3}}{2}$

$=\frac{\sqrt{5}-1}{4}-0.914+\frac{\sqrt{3}}{2}$

$=0.309-0.914+0.866$

$=0.261$

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