Question

How do you evaluate sine, cosine and tangent of $-150{}^\circ $ without using a calculator?
(a) Use trigonometric table of special arcs and unit circle
(b) Use trigonometric identities
(c) Use linear solutions
(d) None of the above

Answer 1

Hint: We will start with considering the properties of each trigonometric function one by one and act accordingly. For sin and tan function the function value does not dissolve the minus sign whereas in case of cos function it dissolves the negative sign. Thus, by analyzing the problem in that way we reach our desired solution.

Complete step by step solution:
According to the question, we start with our given angle $-150{}^\circ $.
For sin function, we have, $\sin \left( -\theta \right)=-\sin \theta $ ,
For cosine function, we have, $\cos \left( -\theta \right)=\cos \theta $,
For tangent function, we have, $\tan \left( -\theta \right)=-\tan \theta $,
So, now, from the identities of trigonometric functions, $\sin \left( -150{}^\circ \right)=-\sin 150{}^\circ ,\cos \left( -150{}^\circ \right)=\cos 150{}^\circ ,\tan \left( -150{}^\circ \right)=-\tan 150{}^\circ $
As, $\dfrac{\pi }{2}<150{}^\circ <\pi $ , we also find that the angle is in the second quadrant, where the values of sin functions are only positive. Trigonometric Identities are useful whenever trigonometric functions are involved in an expression or an equation. Identity inequalities which are true for every value occurring on both sides of an equation. Geometrically, these identities involve certain functions of one or more angles. There are various distinct identities involving the side length as well as the angle of a triangle. The trigonometric identities hold true only for the right-angle triangle.
And now, the value of $\pi $ is $180{}^\circ $ , so we can write $150{}^\circ $as $\left( 180{}^\circ -30{}^\circ \right)$ .

To start with, $\sin 150{}^\circ =\sin \left( 180{}^\circ -30{}^\circ \right)=\sin 30{}^\circ $
This gives us the value of $\sin \left( -150{}^\circ \right)=-\sin 30{}^\circ =-\dfrac{1}{2}$
Then, $\cos 150{}^\circ =\cos \left( 180{}^\circ -30{}^\circ \right)=-\cos 30{}^\circ $
This gives us the value of $\cos \left( -150{}^\circ \right)=-\cos 30{}^\circ =-\dfrac{\sqrt{3}}{2}$
Again, $\tan 150{}^\circ =\tan \left( 180{}^\circ -30{}^\circ \right)=-\tan 30{}^\circ $
This gives us the value of $\tan \left( -150{}^\circ \right)=\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$

So, the correct answer is “Option a”.

Note: To understand how the values of trigonometric ratios like $150{}^\circ $ change in different quadrants, first we have to understand ASTC rule. The ASTC rule is nothing but the "all sin tan cos" rule in trigonometry. The angles which lie between 0° and 90° are said to lie in the first quadrant. The angles between 90° and 180° are in the second quadrant, angles between 180° and 270° are in the third quadrant and angles between 270° and 360° are in the fourth quadrant. In the first quadrant, the values for sin, cos and tan are positive. In the second quadrant, the values for sin are positive only. In the third quadrant, the values for tan are positive only. In the fourth quadrant, the values for cos are positive only.