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