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Question: If \(\cos \theta = - \dfrac{4}{9}\) with \(\theta \) in quadrant 2, find \(\sin \theta \)?...

If cos⁡θ=−49\cos \theta = - \dfrac{4}{9} with θ\theta in quadrant 2, find sin⁡θ\sin \theta ?

Explanation

Solution

We are given the value of cos⁡θ\cos \theta . So, to find the value of sin⁡θ\sin \theta we will use the most fundamental formula of trigonometry that states sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1. Replacing the value of cos⁡θ\cos \theta in the formula and then operating accordingly we will get the value of sin⁡θ\sin \theta . Also, given that the angle θ\theta lies in the second quadrant. So, we will have to simplify that more accordingly. So, let us see how to solve the problem.

Complete step-by-step answer:
Given, to us is cos⁡θ=−49\cos \theta = - \dfrac{4}{9}.
Now, we know the most basic relation between sine and cosine function is sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1.
Therefore, now substituting the given value in the formula, we get,
sin⁡2θ+(−49)2=1{\sin ^2}\theta + {\left( { - \dfrac{4}{9}} \right)^2} = 1
Now, on opening the brackets and squaring, we get,
⇒sin⁡2θ+1681=1\Rightarrow {\sin ^2}\theta + \dfrac{{16}}{{81}} = 1
Now, subtracting 1681\dfrac{{16}}{{81}} both sides, we get,
⇒sin⁡2θ=1−1681\Rightarrow {\sin ^2}\theta = 1 - \dfrac{{16}}{{81}}
Now, taking the LCM on right hand side, we get,
⇒sin⁡2θ=81−1681\Rightarrow {\sin ^2}\theta = \dfrac{{81 - 16}}{{81}}
⇒sin⁡2θ=6581\Rightarrow {\sin ^2}\theta = \dfrac{{65}}{{81}}
Now, taking square root on both the sides of the equation, we get,
⇒sin⁡θ=±6581\Rightarrow \sin \theta = \pm \sqrt {\dfrac{{65}}{{81}}}
⇒sin⁡θ=±659\Rightarrow \sin \theta = \pm \dfrac{{\sqrt {65} }}{9}
Now, it is given to us in the question that the angle θ\theta lies in the second quadrant.
We know, the value of sin⁡θ\sin \theta in the second quadrant is always positive.
Therefore, we will take only the positive value of sin⁡θ\sin \theta .
Therefore, we can conclude that,
sin⁡θ=659\sin \theta = \dfrac{{\sqrt {65} }}{9}

Note: The value of all the trigonometric functions are repetitive in the Cartesian plane because they are periodic in nature. Among the four quadrants, all the trigonometric functions are positive in the first quadrant. In the second quadrant only sine and cosecant functions are positive. In the third quadrant tangent and cotangent functions are positive. And in the fourth quadrant cosine and secant functions are positive.