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

Last updated date: 13th Jun 2024
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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.

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).
$\Rightarrow \sin {{40}^{\circ }}-\cos {{70}^{\circ }}=\sqrt{3}\cos {{80}^{\circ }}$.