Question

Prove the given trigonometric expression $\sin {{10}^{0}}+\sin {{20}^{0}}+\sin {{40}^{0}}+\sin {{50}^{0}}=\sin {{70}^{0}}+\sin {{80}^{0}}$.

Answer 1

Hint: First group the expression as,$\left( \sin {{50}^{0}}+\sin {{10}^{0}} \right)+\left( \sin {{20}^{0}}+\sin {{40}^{0}} \right)$ then use the formula $\sin c+\sin d=2\sin \dfrac{\left( c+d \right)}{2}\cos \dfrac{\left( c-d \right)}{2}$ and then after using fact that $\sin {{30}^{0}}=\dfrac{1}{2}$ use the identity$\cos \left( {{90}^{0}}-\theta \right)=\sin \theta $.

Complete step-by-step answer:
In the question we have to prove that value of $\sin {{10}^{0}}+\sin {{20}^{0}}+\sin {{40}^{0}}+\sin {{50}^{0}}$ is equal to $\sin {{70}^{0}}+\sin {{80}^{0}}$ value.
In the equation left the left hand side of equation it is given,
$LHS=\sin {{10}^{0}}+\sin {{20}^{0}}+\sin {{40}^{0}}+\sin {{50}^{0}}$
We will group or rearrange the expression as,
$LHS=\left( \sin {{50}^{0}}+\sin {{10}^{0}} \right)+\left( \sin {{40}^{0}}+\sin {{20}^{0}} \right)$
Now we will apply formula,
$\sin \left( C \right)+\sin \left( D \right)=2\sin \dfrac{\left( C+D \right)}{2}\cos \dfrac{\left( C-D \right)}{2}$
We can write it as,
$LHS=2\sin \left( \dfrac{50+10}{2} \right)\cos \left( \dfrac{50-10}{2} \right)+2\sin \left( \dfrac{40+20}{2} \right)\cos \left( \dfrac{40-20}{2} \right)$
On solving this we get
$LHS=2\sin {{30}^{0}}\cos {{20}^{0}}+2\sin {{30}^{0}}\cos {{10}^{0}}$
 Now using $\sin {{30}^{0}}=\dfrac{1}{2}$, the above equation can be written as,
$LHS=2\times \dfrac{1}{2}\times \cos {{20}^{0}}+2\times \dfrac{1}{2}\times \cos {{10}^{0}}=\cos {{20}^{0}}+\cos {{10}^{0}}$
Now using formula $\cos \left( {{90}^{0}}-\theta \right)=\sin \theta $, the above equation can be written as
$LHS=\cos \left( {{90}^{0}}-{{70}^{0}} \right)+\cos \left( {{90}^{0}}-{{80}^{0}} \right)$
On simplifying, we get
$LHS=\sin {{70}^{0}}+\sin {{80}^{0}}$
This is equal to the right side of the expression which is $\sin {{70}^{0}}+\sin {{80}^{0}}$.
Hence proved.

Note: While solving the expression $\sin {{10}^{0}}+\sin {{20}^{0}}+\sin {{40}^{0}}+\sin {{50}^{0}}$, one can also pair up as $\left( \sin {{10}^{0}}+\sin {{20}^{0}} \right)+\left( \sin {{40}^{0}}+\sin {{50}^{0}} \right)$ and use the formula of $\sin C+\sin D$ and get the desired result, we will get same result.