Question

How do you find the exact functional value of $\sin {75^ \circ }$ by using the cosine sum or difference identity?

Answer 1

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

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

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