Question

Prove that $\sin {{50}^{\circ }}-\sin {{70}^{\circ }}+\sin {{10}^{\circ }}=0$.

Answer 1

Hint: Use the formula for $\sin A+\sin B$. Take $\sin 50+\sin 10$, simplify it using the formula and substitute it back in the equation. Use the cosine function of trigonometry to solve the rest.

Complete step-by-step answer:
We need to prove that, $\sin {{50}^{\circ }}-\sin {{70}^{\circ }}+\sin {{10}^{\circ }}=0-\left( 1 \right)$
We know the formula of $\sin A+\sin B$.
$\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$
Let us take $\sin 50+\sin 10$, where A=50 and B=10.
$\sin 50+\sin 10=2\sin \left( \dfrac{50+10}{2} \right)\cos \left( \dfrac{50-10}{2} \right)$

& \sin 50+\sin 10=2\sin \left( \dfrac{60}{2} \right)\cos \left( \dfrac{40}{2} \right) \\\ & \sin 50+\sin 10=2\sin 30\cos 20 \\\ \end{aligned}$$ We know the value of, $$\sin 30=\dfrac{1}{2}$$ $$\begin{aligned} &\therefore 2\sin 30\cos 20=2\times \dfrac{1}{2}\times \cos 20=\cos 20 \\\ &\therefore \sin 50+\sin 10=\cos 20-(2) \\\ \end{aligned}$$ Put, $$\sin 50+\sin 10=\cos 20$$in equation (1). $$\therefore \cos 20-\sin 70-(4)$$ By using the trigonometric cosine function, $$\begin{aligned} & \cos \left( 90-\theta \right)=\sin \theta \\\ & \cos 20=\cos \left( 90-70 \right)=\sin 70 \\\ \end{aligned}$$ $$\therefore $$We got the value of $$\cos 20=\sin 70$$. Substitute $$\cos 20=\sin 70$$in equation (4), we get $$\sin 70-\sin 70=0$$ $$\therefore $$We proved that $$\sin {{50}^{\circ }}-\sin {{70}^{\circ }}+\sin {{10}^{\circ }}=0$$ Note: We can also solve by using the formulae, $$\begin{aligned} & \sin \left( A+B \right)=\sin A\cos B+\cos A\sin B \\\ & \sin \left( A-B \right)=\sin A\cos B-\cos A\sin B \\\ \end{aligned}$$ $$\sin {{50}^{\circ }}-\sin {{70}^{\circ }}+\sin {{10}^{\circ }}$$can be written as, $$\begin{aligned} & \sin \left( 60-10 \right)=\sin 60\cos 10-\cos 60\sin 10 \\\ & \sin \left( 60+10 \right)=\sin 60\cos 10+\cos 60\sin 10 \\\ \end{aligned}$$ $$\therefore \sin \left( 60-10 \right)-\sin \left( 60+10 \right)+\sin 10=\sin 60\cos 10-\cos 60\sin 10-\sin 60\cos 10-\cos 60\sin 10+\sin 10$$ [Cancel out like terms] $$\begin{aligned} & =-2\cos 60\sin 10+\sin 10 \\\ & \because \cos 60=\dfrac{1}{2} \\\ & =-2\times \dfrac{1}{2}\sin 10+\sin 10=-\sin 10+\sin 10=0 \\\ & \therefore \sin 50-\sin 70+\sin 10=0 \\\ \end{aligned}$$