Question

Prove the given result: $ \sin {{40}^{\circ }}-\cos {{70}^{\circ }}=\sqrt{3}.\cos {{80}^{\circ }} $ .

Answer 1

Hint:
We start solving the problem by considering the L.H.S (Left Hand side) of the given equation. We then make use of the result $ \sin \left( {{90}^{\circ }}-\alpha \right)=\cos \alpha $ to proceed through the problem. We then make use of the result $ \cos A-\cos B=-2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right) $ to proceed through further through the problem. We then make the necessary calculations and make use of the results $ \sin \left( -x \right)=-\sin x $ , $ \sin \left( {{90}^{\circ }}-\alpha \right)=\cos \alpha $ to complete the proof of given result.

Complete step by step answer:
According to the problem, we are asked to prove the given result: $ \sin {{40}^{\circ }}-\cos {{70}^{\circ }}=\sqrt{3}.\cos {{80}^{\circ }} $ .
Let us consider L.H.S (Left Hand side) of the given result i.e., $ \sin {{40}^{\circ }}-\cos {{70}^{\circ }} $ .
\[\Rightarrow \sin {{40}^{\circ }}-\cos {{70}^{\circ }}=\sin \left( {{90}^{\circ }}-{{50}^{\circ }} \right)-\cos {{70}^{\circ }}\] ---(1).
We know that $ \sin \left( {{90}^{\circ }}-\alpha \right)=\cos \alpha $ . Let us use this result in equation (1).
\[\Rightarrow \sin {{40}^{\circ }}-\cos {{70}^{\circ }}=\cos {{50}^{\circ }}-\cos {{70}^{\circ }}\] ---(2).
We know that $ \cos A-\cos B=-2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right) $ . Let us use this result in equation (2).
\[\Rightarrow \sin {{40}^{\circ }}-\cos {{70}^{\circ }}=-2\sin \left( \dfrac{{{50}^{\circ }}+{{70}^{\circ }}}{2} \right)\sin \left( \dfrac{{{50}^{\circ }}-{{70}^{\circ }}}{2} \right)\].
\[\Rightarrow \sin {{40}^{\circ }}-\cos {{70}^{\circ }}=-2\sin \left( \dfrac{{{120}^{\circ }}}{2} \right)\sin \left( \dfrac{-{{20}^{\circ }}}{2} \right)\].
\[\Rightarrow \sin {{40}^{\circ }}-\cos {{70}^{\circ }}=-2\sin \left( {{60}^{\circ }} \right)\sin \left( -{{10}^{\circ }} \right)\] ---(3).
We know that $ \sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2} $ and $ \sin \left( -x \right)=-\sin x $ . Let us use these results in equation (3).
\[\Rightarrow \sin {{40}^{\circ }}-\cos {{70}^{\circ }}=-2\left( \dfrac{\sqrt{3}}{2} \right)\left( -\sin {{10}^{\circ }} \right)\].
\[\Rightarrow \sin {{40}^{\circ }}-\cos {{70}^{\circ }}=\sqrt{3}\sin {{10}^{\circ }}\].
\[\Rightarrow \sin {{40}^{\circ }}-\cos {{70}^{\circ }}=\sqrt{3}\sin \left( {{90}^{\circ }}-{{80}^{\circ }} \right)\] ---(4).
We know that $ \sin \left( {{90}^{\circ }}-\alpha \right)=\cos \alpha $ . Let us use this result in equation (4).
\[\Rightarrow \sin {{40}^{\circ }}-\cos {{70}^{\circ }}=\sqrt{3}\cos {{80}^{\circ }}\] ---(5).
From equation (5), we can see that the L.H.S (Left Hand Side) is equal to the given R.H.S (Right-hand side).
 $ \therefore $ We have proved the given result \[\sin {{40}^{\circ }}-\cos {{70}^{\circ }}=\sqrt{3}\cos {{80}^{\circ }}\].

Note:
 We can also solve the given problem as shown below:
We have \[\sin {{40}^{\circ }}-\cos {{70}^{\circ }}\].
\[\Rightarrow \sin {{40}^{\circ }}-\cos {{70}^{\circ }}=\sin {{40}^{\circ }}-\cos \left( {{90}^{\circ }}-{{20}^{\circ }} \right)\] ---(6).
We know that $ \cos \left( {{90}^{\circ }}-\alpha \right)=\sin \alpha $ . Let us use this result in equation (6).
\[\Rightarrow \sin {{40}^{\circ }}-\cos {{70}^{\circ }}=\sin {{40}^{\circ }}-\sin {{20}^{\circ }}\] ---(7).
We know that $ \sin A-\sin B=2\cos \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right) $ . Let us use this result in equation (7).
\[\Rightarrow \sin {{40}^{\circ }}-\cos {{70}^{\circ }}=2\cos \left( \dfrac{{{40}^{\circ }}+{{20}^{\circ }}}{2} \right)\sin \left( \dfrac{{{40}^{\circ }}-{{20}^{\circ }}}{2} \right)\].
\[\Rightarrow \sin {{40}^{\circ }}-\cos {{70}^{\circ }}=2\cos \left( \dfrac{{{60}^{\circ }}}{2} \right)\sin \left( \dfrac{{{20}^{\circ }}}{2} \right)\].
\[\Rightarrow \sin {{40}^{\circ }}-\cos {{70}^{\circ }}=2\cos \left( {{30}^{\circ }} \right)\sin \left( {{10}^{\circ }} \right)\] ---(8).
We know that $ \cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2} $ . Let us use this result in equation (8).
\[\Rightarrow \sin {{40}^{\circ }}-\cos {{70}^{\circ }}=2\left( \dfrac{\sqrt{3}}{2} \right)\left( \sin {{10}^{\circ }} \right)\].
\[\Rightarrow \sin {{40}^{\circ }}-\cos {{70}^{\circ }}=\sqrt{3}\sin {{10}^{\circ }}\].
\[\Rightarrow \sin {{40}^{\circ }}-\cos {{70}^{\circ }}=\sqrt{3}\sin \left( {{90}^{\circ }}-{{80}^{\circ }} \right)\] ---(9).
We know that $ \sin \left( {{90}^{\circ }}-\alpha \right)=\cos \alpha $ . Let us use this result in equation (9).
\[\Rightarrow \sin {{40}^{\circ }}-\cos {{70}^{\circ }}=\sqrt{3}\cos {{80}^{\circ }}\].