Question

Evaluate each of the following in the simplest form: $\cos {{60}^{0}}\cos {{30}^{0}}-\sin {{60}^{0}}\sin {{30}^{0}}$.

Answer 1

Hint: Trigonometric expressions are first simplified then values are substituted. Take ${{60}^{0}}$ as A and ${{30}^{0}}$ as B and try to use the identity $\cos A\cos B-\sin A\sin B=\cos \left( A+B \right)$ and get the desired result.

Complete step-by-step solution -
 We have to find the value of expression, $\cos {{60}^{0}}\cos {{30}^{0}}-\sin {{60}^{0}}\sin {{30}^{0}}$
In the expression let’s take ${{60}^{0}}$ as $A$ and ${{30}^{0}}$ as $B$.
So we can write expression,$\cos {{60}^{0}}\cos {{30}^{0}}-\sin {{60}^{0}}\sin {{30}^{0}}$ as $\cos A\cos B-\sin A\sin B$
Now by observing the expression $\cos A\cos B-\sin A\sin B$ we can see that$\cos A\text{ }\cos B-\sin A\text{ }\sin B$ as in form of $\cos \left( A+B \right)$ whose identity is,
$\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ Which is applicable for all values of $A$ and$B$.
$\Rightarrow $By the formula of $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ we can represent $\cos {{60}^{0}}\text{ }\cos {{30}^{0}}-\sin {{60}^{0}}\text{ }\sin {{30}^{0}}$ as $\cos \left( {{60}^{0}}+{{30}^{0}} \right)$ which is $\cos {{90}^{0}}$.
By the use of a trigonometric table of standard angle we can say that $\cos {{90}^{0}}$ is $0$.
Hence the value of expression is $0$.

Note: We can do the same problem by using two more methods:
In first method we will use formula $\cos \left( {{90}^{0}}-\theta \right)=\sin \theta $ and we will transform expression $\cos {{60}^{0}}\cos {{30}^{0}}-\sin {{60}^{0}}\text{ }\sin {{30}^{0}}$ as $\sin {{30}^{0}}\sin {{60}^{0}}-\sin {{60}^{0}}\text{ }\sin {{30}^{0}}$ which is equal to ‘$0$’.
In the second method we will directly substitute values which is $\sin {{30}^{0}}=\dfrac{1}{2}, \sin {{60}^{0}}\text{=}\dfrac{\sqrt{3}}{2}\text{ }, \cos {{30}^{0}}=\dfrac{\sqrt{3}}{2}$ and $\cos {{60}^{0}}=\dfrac{1}{2}$. So by putting the values in the expression we can get the answer ‘$0$’.