Question

How do you evaluate sine, cosine and tangent of $-\dfrac{25\pi }{4}$ 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 start with bringing the term $-\dfrac{25\pi }{4}$under $\pi $ by taking out the unit circle for each $2\pi $ or $360{}^\circ $ . From which we can get to a result where we are left with $-\dfrac{\pi }{4}$after considering 3 whole circles. Then we will consider 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, $-\dfrac{25\pi }{4}$
And also, we can write $-\dfrac{25\pi }{4}$as, $-6\pi -\dfrac{\pi }{4}$,
So,$-\dfrac{25\pi }{4}=-6\pi -\dfrac{\pi }{4}$
And now, $6\pi $ is$(360{}^\circ \times 3)$,as $\pi =180{}^\circ $ ,
So we are left with only $-\dfrac{\pi }{4}$, as with every $360{}^\circ $we have every four quadrants fulfilled.
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, $\sin \left( -\dfrac{\pi }{4} \right)=-\sin \dfrac{\pi }{4},\cos \left( -\dfrac{\pi }{4} \right)=\cos \dfrac{\pi }{4},\tan \left( -\dfrac{\pi }{4} \right)=-\tan \dfrac{\pi }{4}$
As, we all know, $0<\dfrac{\pi }{4}<\dfrac{\pi }{2}$ ,we also get the conclusion that the angle is in the first quadrant, where the values of all trigonometric functions are all positive.
To start with, $\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$, from the trigonometric table.
Then, again from the trigonometric table the values are, $\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$and $\tan \dfrac{\pi }{4}=1$.
Thus, finally we reach to a conclusion that,
$\sin \left( -\dfrac{25\pi }{4} \right)=-\sin \left( \dfrac{\pi }{4} \right)=-\dfrac{1}{\sqrt{2}}$ ,
Then,
$\cos \left( -\dfrac{25\pi }{4} \right)=\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$
And at the end,
$\tan \left( -\dfrac{25\pi }{4} \right)=-\tan \left( \dfrac{\pi }{4} \right)=-1$

So, the correct answer is “Option a”.

Note: To understand how the values of trigonometric ratios change in different quadrants, first we have to understand the 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.