Question

How do you find the exact functional value of $\sin 405^\circ + \sin 120^\circ $ using the cosine sum or difference identity?

Answer 1

Hint: Here we have to find the exact functional value of $\sin 405^\circ + \sin 120^\circ $. To solve this trigonometric function first we will split each angle and then we will use the trigonometric identity $\sin (A + B) = \sin A\cos B + \sin B\cos A$. After applying this trigonometric identity we will put the values of angles of $\sin $ and $\cos $ such as $\sin (360^\circ ) = 0,\,\,\sin (45^\circ ) = \dfrac{1}{{\sqrt 2 }},\,\,\cos (360^\circ ) = 1,\,\,\cos (45^\circ ) = \dfrac{1}{{\sqrt 2 }}$.

Complete step by step answer:
To find the exact functional value of $\sin 405^\circ + \sin 120^\circ $. We will split the angles in the angles whose values are known to us.So, we can write,
$\sin 405^\circ = \sin (360^\circ + 45^\circ )$
We know that $\sin (A + B) = \sin A\cos B + \sin B\cos A$

Using the above formula to evaluate the value of $\sin 405^\circ $. We get,
$\sin (360^\circ + 45^\circ ) = \sin 360^\circ \cos 45^\circ + \sin 45^\circ \cos 360^\circ $
We know that,
$\sin (360^\circ ) = 0,\,\,\sin (45^\circ ) \\
\Rightarrow \sin (360^\circ ) = \dfrac{1}{{\sqrt 2 }},\,\,\cos (360^\circ ) \\
\Rightarrow \sin (360^\circ ) = 1,\,\,\cos (45^\circ ) \\
\Rightarrow \sin (360^\circ ) = \dfrac{1}{{\sqrt 2 }}$

Putting these values in the above equation. We get,
$ \Rightarrow \sin (360^\circ + 45^\circ ) = 0 \times \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }} \times 1$
$ \Rightarrow \sin (360^\circ + 45^\circ ) = \dfrac{1}{{\sqrt 2 }}$
Rationalizing the above value. We get,
$ \Rightarrow \sin (360^\circ + 45^\circ ) = \dfrac{{1 \times \sqrt 2 }}{{\sqrt 2 \times \sqrt 2 }} \\
\Rightarrow \sin (360^\circ + 45^\circ ) = \dfrac{{\sqrt 2 }}{2}$
Hence, the value of $\sin (405^\circ ) = \dfrac{{\sqrt 2 }}{2}$

Now, we will evaluate the value of value of $\sin (120^\circ )$
We can write $\sin (120^\circ ) = \sin (90^\circ + 30^\circ )$
Therefore, $\sin (90^\circ + 30^\circ ) = \sin 90^\circ \cos 30^\circ + \sin 30^\circ \cos 90^\circ $
We know that $\sin (90^\circ ) = 1,\,\,\sin (30^\circ ) = \dfrac{1}{2},\,\,\cos (90) = 0,\,\,\cos (30^\circ ) = \dfrac{{\sqrt 3 }}{2}$

Putting these values in the above equation. We get,
$ \Rightarrow $$\sin (90^\circ + 30^\circ ) = 1 \times \dfrac{{\sqrt 3 }}{2} + \dfrac{1}{2} \times 0$
$ \Rightarrow \sin (90^\circ + 30^\circ ) = \dfrac{{\sqrt 3 }}{2}$
Hence, the value of $\sin (120^\circ )$$ = \dfrac{{\sqrt 3 }}{2}$
Now put the values of $\sin 405^\circ $ and $\sin (120^\circ )$ in $\sin 405^\circ + \sin 120^\circ $ to calculate the exact value.
So, $\sin 405^\circ + \sin 120^\circ = \dfrac{{\sqrt 2 }}{2} + \dfrac{{\sqrt 3 }}{2}$

Hence, $\sin 405^\circ + \sin 120^\circ = \dfrac{{\sqrt 2 + \sqrt 3 }}{2}$.

Note: Some students are confused between trigonometric identities such as $\sin (A + B)$ and $\sin A + \operatorname{Sin} B$. These both are different identities in one there is only a sum of angles and in second there is a sum of angles of $\sin $. In this question we can also use the identity $\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$ to find the exact value of the trigonometric function.