Question

$\sin {20}^0+\sin {40}^0-\sin {80}^0=$

• A. $-1$
• B. $1$
• C. $2$
• D. $0$

Answer 1

The correct option is D $0$

given, $\sin {20}^0+\sin {40}^0-\sin {80}^0$

$={\sin 40}^0+{\sin 20}^0-{\sin 80}^0$

$=2\sin \left(\frac{{40}^0+{20}^0}{2}\right)\cos \left(\frac{{40}^0-{20}^0}{2}\right)-{\sin 80}^0$ $[\because sinA+sinB=2\sin \left(\frac{A+B}{2}\right).\cos \left(\frac{A-B}{2}\right)]$

$={2\sin 30}^0.{\cos 10}^0-{\sin 80}^0$

$=2{\sin 30}^0.{\cos 10}^0-\sin ({90}^0-{10}^0)$ $[\because {\sin 30}^0=1/2,\&\sin ({90}^0-\theta )=cos\theta ]$

$=2\times \frac{1}{2}\times {\cos 10}^0-{\cos 10}^0$

$={\cos 10}^0-{\cos 10}^0$

$=0$