Question

Find the value of $\sin (105^\circ ) + \cos (105^\circ )$.
A) $\dfrac{1}{2}$
B) $\dfrac{3}{2}$
C) $\sqrt 2 $
D) $\dfrac{1}{{\sqrt 2 }}$

Answer 1

Hint: We have trigonometric equations for finding $\sin (A + B)$ and $\cos (A + B)$. The given angle, $105^\circ $ can be written as the sum of $60^\circ $ and $45^\circ $. Thus we can apply the sum equations and substitute the known values. Then by simplifying the equation we get the answer.

Formula used: For any angles $A,B$we have,
$\sin (A + B) = \sin A\cos B + \cos A\sin B$
$\cos (A + B) = \cos A\cos B - \sin A\sin B$
$\sin 45^\circ = \dfrac{1}{{\sqrt 2 }}$
$\sin 60^\circ = \dfrac{{\sqrt 3 }}{2}$
$\cos 45^\circ = \dfrac{1}{{\sqrt 2 }}$
$\cos 60^\circ = \dfrac{1}{2}$

Complete step-by-step answer:
We are asked to find the value of $\sin (105^\circ ) + \cos (105^\circ )$.
We can write $105$as the sum of $60^\circ $ and $45^\circ $.
Thus we get,
$\sin (105^\circ ) = \sin (60^\circ + 45^\circ )$
$\cos (105^\circ ) = \cos (60^\circ + 45^\circ )$
For any angles $A,B$we have,
$\sin (A + B) = \sin A\cos B + \cos A\sin B$
$\cos (A + B) = \cos A\cos B - \sin A\sin B$
We can substitute $A = 60^\circ ,B = 45^\circ $ in the above two equations.
So we have,
$\sin (60^\circ + 45^\circ ) = \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ - - - (i)$
$\cos (60^\circ + 45^\circ ) = \cos 60^\circ \cos 45^\circ - \sin 60^\circ \sin 45^\circ - - - (ii)$
Also we know these trigonometric values.
$\sin 45^\circ = \dfrac{1}{{\sqrt 2 }}$
$\sin 60^\circ = \dfrac{{\sqrt 3 }}{2}$
$\cos 45^\circ = \dfrac{1}{{\sqrt 2 }}$
$\cos 60^\circ = \dfrac{1}{2}$
Substituting these values in the above equations we have,
$(i) \Rightarrow \sin (60^\circ + 45^\circ ) = \dfrac{{\sqrt 3 }}{2} \times \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{2} \times \dfrac{1}{{\sqrt 2 }}$
Simplifying the expression we get,
$ \Rightarrow \sin (60^\circ + 45^\circ ) = \dfrac{{\sqrt 3 }}{{2\sqrt 2 }} + \dfrac{1}{{2\sqrt 2 }}$
$ \Rightarrow \sin (60^\circ + 45^\circ ) = \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}$
$(ii) \Rightarrow \cos (60^\circ + 45^\circ ) = \dfrac{1}{2} \times \dfrac{1}{{\sqrt 2 }} - \dfrac{{\sqrt 3 }}{2} \times \dfrac{1}{{\sqrt 2 }}$
Simplifying the expression we get,
$ \Rightarrow \cos (60^\circ + 45^\circ ) = \dfrac{1}{{2\sqrt 2 }} - \dfrac{{\sqrt 3 }}{{2\sqrt 2 }}$
$ \Rightarrow \cos (60^\circ + 45^\circ ) = \dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }}$
We need to find the value of $\sin (105^\circ ) + \cos (105^\circ )$.
We have,
$\sin (105^\circ ) = \sin (60^\circ + 45^\circ )$
$\cos (105^\circ ) = \cos (60^\circ + 45^\circ )$
And
$\sin (60^\circ + 45^\circ ) = \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}$
$\cos (60^\circ + 45^\circ ) = \dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }}$
Combining these two results, we get
$\sin (105^\circ ) = \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}$
$\cos (105^\circ ) = \dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }}$
So we have,
$\sin (105^\circ ) + \cos (105^\circ ) = \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }} + \dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }}$
Simplifying the right hand side we get,
$\sin (105^\circ ) + \cos (105^\circ ) = \dfrac{{\sqrt 3 + 1 + 1 - \sqrt 3 }}{{2\sqrt 2 }}$
$ \Rightarrow \sin (105^\circ ) + \cos (105^\circ ) = \dfrac{2}{{2\sqrt 2 }}$
Dividing numerator and denominator by $2$ we get,
$ \Rightarrow \sin (105^\circ ) + \cos (105^\circ ) = \dfrac{1}{{\sqrt 2 }}$

$\therefore $ The answer is option D.

Note: Here we get the answer as $\dfrac{1}{{\sqrt 2 }}$ which belongs to the options. In some cases, instead of giving the answer directly, we may give it in another way. This value $\dfrac{1}{{\sqrt 2 }}$ is equal to $\sin 45^\circ $ and $\cos 45^\circ $. So, any of these answers, if there are options will be correct. Also the given angle $105^\circ $ can be split into a sum in different ways. But we particularly chose $60^\circ + 45^\circ $, since trigonometric values are known for these angles. So we could substitute them and simplify to get the answer.