Question

How do you evaluate $\cos {75^ \circ }$ ?

Answer 1

Hint: We will use the cosine sum and difference formula to find the exact functional value of $\cos {75^ \circ }$. So, here we will use $\cos \left( {A + B} \right)$ i.e., cosine sum identity or $\cos \left( {A - B} \right)$ i.e., cosine difference identity which are defined as \[\cos A\cos B - \sin A\sin B\] and $\cos A\cos B + \sin A\sin B$ using one of these we will get the required value.

Complete Step by Step Solution: -
We’ll solve this by using two methods one by using the cosine sum identity and another by using the cosine difference identity.
Method – 1: Using the cosine sum identity:
We have to find the value of $\cos {75^ \circ }$. So, we can also write $\cos {75^ \circ }$ as –
$\cos {75^ \circ } = \cos \left( {{{45}^ \circ } + {{30}^ \circ }} \right)$
We know that, cosine sum identity is –
$\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$
Using the above identity for $\cos \left( {{{45}^ \circ } + {{30}^ \circ }} \right)$ , we get –
Here, $A = {45^ \circ }$ and $B = {30^ \circ }$. Substituting these values in the identity, we get –
$ \Rightarrow \cos \left( {45 + 30} \right) = \cos 45\cos 30 - \sin 45\sin 30$
By using the specified cosine and cosine angle i.e., $\cos 45 = \dfrac{1}{{\sqrt 2 }},\cos 30 = \dfrac{{\sqrt 3 }}{2},\sin 45 = \dfrac{1}{{\sqrt 2 }}$ and $\sin 30 = \dfrac{1}{2}$ , we get –
$\therefore \cos \left( {75} \right) = \dfrac{1}{{\sqrt 2 }}.\dfrac{{\sqrt 3 }}{2} - \dfrac{1}{{\sqrt 2 }}.\dfrac{1}{2}$
On simplification, we get –
$ \Rightarrow \cos 75 = \dfrac{{\sqrt 3 }}{{2\sqrt 2 }} - \dfrac{1}{{2\sqrt 2 }}$
Taking $2\sqrt 2 $ common from the denominator, we get –
$ \Rightarrow \cos 75 = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$
Hence, the exact functional value of $\cos 75$ is $\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$.
Method – 2: Using the cosine difference identity:
We have to find the value of $\cos {75^ \circ }$. So, we can also write $\cos {75^ \circ }$ as –
$\cos {75^ \circ } = \cos \left( {{{135}^ \circ } - {{60}^ \circ }} \right)$
We know that, cosine sum identity is –
$\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$
Using the above identity for $\cos \left( {{{135}^ \circ } - {{60}^ \circ }} \right)$ , we get –
Here, $A = {135^ \circ }$ and $B = {60^ \circ }$. Substituting these values in the identity, we get –
\[ \Rightarrow \cos \left( {135 - 60} \right) = \cos 135\cos 60 - \sin 135\sin 60\]
By using the specified cosine and cosine angle i.e., $\cos 135 = - \dfrac{1}{{\sqrt 2 }},\cos 60 = \dfrac{1}{2},\sin 135 = \dfrac{1}{{\sqrt 2 }}$ and $\sin 60 = \dfrac{{\sqrt 3 }}{2}$ , we get –
$\therefore \cos \left( {75} \right) = \dfrac{{ - 1}}{{\sqrt 2 }}.\dfrac{1}{2} + \dfrac{1}{{\sqrt 2 }}.\dfrac{{\sqrt 3 }}{2}$
On simplification, we get –
$ \Rightarrow \cos 75 = - \dfrac{1}{{2\sqrt 2 }} + \dfrac{{\sqrt 3 }}{{2\sqrt 2 }}$
Taking $2\sqrt 2 $ common from the denominator, we get –
$ \Rightarrow \cos 75 = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$

Hence, the exact functional value of $\sin 75$ is $\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$.

Note:
The values of cosine and sine can be determined by using the other methods such as double angle formula, half angle formula. In this question, we found the value of $\cos 75$ by using the cosine sum and difference formula. Here, we used the value of trigonometry ratios of standard angles. That’s why we can determine the solution for the question.