Question

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

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

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

Multiplying the terms on the right side, we get

$\cos {{60}^{0}}\cos {{30}^{0}}+\sin {{60}^{0}}\sin {{30}^{0}}=\dfrac{\sqrt{3}}{4}+\dfrac{\sqrt{3}}{4}$

$\cos {{60}^{0}}\cos {{30}^{0}}+\sin {{60}^{0}}\sin {{30}^{0}}=\dfrac{2\sqrt{3}}{4}$

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

From the trigonometric table, $\dfrac{\sqrt{3}}{2}=\cos {{30}^{0}}$

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


Hence the given expression is verified

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