Question

Verify the following:
$\sin {{60}^{0}}\cos {{30}^{0}}-\cos {{60}^{0}}\sin {{30}^{0}}=\sin {{30}^{0}}$

Answer 1

Hint: The trigonometric ratios table helps to find the values of trigonometric standard angles such as ${{0}^{0}},{{30}^{0}},{{45}^{0}},{{60}^{0}}$ and ${{90}^{0}}$. It consists of trigonometric ratios – sine, cosine, tangent, cosecant, secant and cotangent. These ratios can be written in short as sin, cos, tan, cosec, sec and cot.

Complete step-by-step answer:
The value of the trigonometric ratios by using the trigonometric table is given below.

$\sin {{60}^{0}}=\dfrac{\sqrt{3}}{2},\sin {{30}^{0}}=\dfrac{1}{2},\cos {{60}^{0}}=\dfrac{1}{2},\cos {{30}^{0}}=\dfrac{\sqrt{3}}{2}$

Let us consider the left side of the given expression

$\sin {{60}^{0}}\cos {{30}^{0}}-\cos {{60}^{0}}\sin {{30}^{0}}=\dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}-\dfrac{1}{2}\times \dfrac{1}{2}$

Multiplying the terms on the right side, we get

$\sin {{60}^{0}}\cos {{30}^{0}}-\cos {{60}^{0}}\sin {{30}^{0}}=\dfrac{3}{4}-\dfrac{1}{4}$

$\sin {{60}^{0}}\cos {{30}^{0}}-\cos {{60}^{0}}\sin {{30}^{0}}=\dfrac{3-1}{4}$

$\sin {{60}^{0}}\cos {{30}^{0}}-\cos {{60}^{0}}\sin {{30}^{0}}=\dfrac{2}{4}$

$\sin {{60}^{0}}\cos {{30}^{0}}-\cos {{60}^{0}}\sin {{30}^{0}}=\dfrac{1}{2}$

From the trigonometric table, $\dfrac{1}{2}=\sin {{30}^{0}}$

$\sin {{60}^{0}}\cos {{30}^{0}}-\cos {{60}^{0}}\sin {{30}^{0}}=\sin {{30}^{0}}$

Hence the given expression is verified


Note: Alternatively, the given question is verified by using a formula for the sine of the difference of two angles, $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$. Let $A={{60}^{0}}$ and $B={{30}^{0}}$, then $\sin {{60}^{0}}\cos {{30}^{0}}-\cos {{60}^{0}}\sin {{30}^{0}}=\sin \left( {{60}^{0}}-{{30}^{0}} \right)=\sin {{30}^{0}}$.