Question

Solve the expression $\sin 10^\circ + \sin 20^\circ + \sin 40^\circ + \sin 50^\circ = \sin 70^\circ + \sin 80^\circ $.

Answer 1

Hint: In the given question, we are given a problem in the form of addition of sine functions. We can solve the given problem by using formula of addition of two $\sin $ functions, $\sin A + \sin B$. Take $\sin 50^\circ + \sin 10^\circ $ and $\sin 40^\circ + \sin 20^\circ $, simplify these two using the formula. Substitute values of trigonometric ratios wherever required and at the end convert the cosine function of trigonometry into the sine function of trigonometry.
Formula used: $\sin A + \sin B$

Complete step-by-step solution:
We have, $\sin 10^\circ + \sin 20^\circ + \sin 40^\circ + \sin 50^\circ = \sin 70^\circ + \sin 80^\circ $
Here, LHS: $\sin 10 + \sin 20 + \sin 40 + \sin 50$
We know the formula of addition of two $\sin $ functions, $\sin A + \sin B$.
$\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
Let us take the given problem in the form of addition of two $\sin $ functions.
$\left( {\sin 50 + \sin 10} \right) + \left( {\sin 40 + \sin 20} \right)$
Apply the above written formula,
$ \Rightarrow \left( {2\sin \left( {\dfrac{{50 + 10}}{2}} \right)\cos \left( {\dfrac{{50 - 10}}{2}} \right)} \right) + \left( {2\sin \left( {\dfrac{{40 + 20}}{2}} \right)\cos \left( {\dfrac{{40 - 20}}{2}} \right)} \right)$
On Addition and subtraction of angles, we get
$ \Rightarrow \left( {2\sin \left( {\dfrac{{60}}{2}} \right)\cos \left( {\dfrac{{40}}{2}} \right)} \right) + \left( {2\sin \left( {\dfrac{{60}}{2}} \right)\cos \left( {\dfrac{{20}}{2}} \right)} \right)$
On division of angles inside the brackets, we get
$ \Rightarrow 2\sin 30\cos 20 + 2\sin 30\cos 10$
Take $2\sin 30^\circ $ as common
$ \Rightarrow 2\sin 30\left( {\cos 20 + \cos 10} \right)$
We know the value of, $\sin 30^\circ = \dfrac{1}{2}$
On substituting value of $\sin x$, we get
$ \Rightarrow 2 \times \dfrac{1}{2}\left( {\cos 20 + \cos 10} \right)$
$ \Rightarrow \cos 20 + \cos 10$
By using the trigonometric cosine function $\cos \left( {90 - \theta } \right) = \sin \theta $ we get
$ \Rightarrow \cos \left( {90 - 70} \right) + \cos \left( {90 - 80} \right)$
$ \Rightarrow \sin 70 + \sin 80$
The value we obtained after solving the LHS is equal to RHS value, hence proved LHS = RHS.

Note: To solve these type of questions we should know all the required values of standard angles say, $0^\circ $, $30^\circ $, $60^\circ $, $90^\circ $, $180^\circ $, $270^\circ $, $360^\circ $ respectively for each trigonometric term such as $\sin $, $\cos $, $\tan $, $\cot $, $\sec $, $\cos ec$, etc. We should take care of the calculations so as to be sure of our final answer. In the given question we added two $\sin $ functions. Similarly we can subtract two $\sin $ functions using formula of subtraction of two $\sin $ functions. Also, subtract or add two $\cos $ functions by using formulae. Such as,
$\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$
$\cos A - \cos B = - 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$
$\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
We use them according to the given problem.