Question

Find the values of
i) $\sin 15^\circ $ ii) $\cos 75^\circ $ iii) $\tan 105^\circ $
iv) $\cot 225^\circ $

Answer 1

Hint: To solve this question we will use the identities of $\sin (A + B)$ and $\cos (A + B)$. Also, we will use the value of trigonometric functions at $30^\circ $, $45^\circ $, etc after converting the given angles in the known angles.

Complete step-by-step answer:
Now, we will first find the value of $\sin 15^\circ $. Now, $\sin 15^\circ $ can be written as $\sin 15^\circ = \sin (45^\circ - 30^\circ )$
Now, we will use the identity of $\sin (A + B)$. From trigonometry, we have
$\sin (A + B) = \sin A\cos B + \cos A\sin B$
Similarly, from the above identity, the value of $\sin (A - B)$ can be written as,
$\sin (A - B) = \sin A\cos B - \cos A\sin B$
So, $\sin 15^\circ = \sin (45^\circ - 30^\circ )$ = $\sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ $
Applying values of trigonometric functions at $30^\circ $ and $45^\circ $, we get
$\sin 15^\circ = \dfrac{1}{{\sqrt 2 }}.\dfrac{{\sqrt 3 }}{2} - \dfrac{1}{{\sqrt 2 }}.\dfrac{1}{2}$
So, $\sin 15^\circ = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$
Now, we will use the identity $\cos (A + B) = \cos A\cos B - \sin A\sin B$ to find the value of $\cos 75^\circ $.
Now, $\cos 75^\circ $ = $\cos (45^\circ + 30^\circ )$ = $\cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ $
$\cos 75^\circ = \dfrac{1}{{\sqrt 2 }}.\dfrac{{\sqrt 3 }}{2} - \dfrac{1}{{\sqrt 2 }}.\dfrac{1}{2}$
Therefore, $\cos 75^\circ = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$
Now, $\tan (A + B) = \dfrac{{\sin (A + B)}}{{\cos (A + B)}}$
So, $\tan 105^\circ = \tan (60^\circ + 45^\circ ) = \dfrac{{\sin (60^\circ + 45^\circ )}}{{\cos (60^\circ + 45^\circ )}}$
$\tan 105^\circ = \dfrac{{\dfrac{{\sqrt 3 }}{2}.\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{2}.\dfrac{1}{{\sqrt 2 }}}}{{\dfrac{1}{2}.\dfrac{1}{{\sqrt 2 }} - \dfrac{{\sqrt 3 }}{2}.\dfrac{1}{{\sqrt 2 }}}}$ = $\dfrac{{\sqrt 3 + 1}}{{1 - \sqrt 3 }}$
$\tan 105^\circ = - (\dfrac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}})$
Now, $\cot 225^\circ $ can be written as $\cot (180^\circ + 45^\circ )$, so it lies in the third quadrant . So, we get
$\cot 225^\circ = \cot 45^\circ $
Now, we will use the value of cot x at $45^\circ $. As, $\cot 45^\circ = 1$
Therefore, $\cot 225^\circ = 1$
So, $\sin 15^\circ = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$
$\cos 75^\circ = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$
$\tan 105^\circ = - (\dfrac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}})$
$\cot 225^\circ = 1$

Note: Whenever we come up with such types of questions, we will use the properties of trigonometry. Such questions can be easily solved when we remember the term Add Coffee To Sugar which represents all, sin, tan and cos. This term shows in which quadrant which function remains positive. For example, in the first quadrant, all functions are positive, in the second quadrant, only sin and cosec are positive and so on. This is used in almost every problem and helps in solving difficult problems easily. Also, we have used the identities of $\sin (A + B)$ and $\cos (A + B)$in this question. These identities are basic identities and all other identities can be derived from them.