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Question: If \( \cot \theta = - 2 \) and \( \cos \theta < 0 \) , how do you find \( \sin \theta \)...

If cot⁡θ=−2\cot \theta = - 2 and cos⁡θ<0\cos \theta < 0 , how do you find sin⁡θ\sin \theta

Explanation

Solution

Hint : From reading the question, we got to know that we have to find the sine value from the two relations given. Firstly, square the first relation and then use this identity 1+cot⁡2θ=cos⁡ec2θ1 + {\cot ^2}\theta = \cos e{c^2}\theta and then find the sine from the second relation, you will come to know about the quadrant.

Complete step-by-step answer :
In this question, we are given two relations and from that we need to find the value of sin⁡θ\sin \theta
Using the first given relation cot⁡θ=−2\cot \theta = - 2
Squaring both sides
cot⁡2θ=4{\cot ^2}\theta = 4
Now, use the identity that 1+cot⁡2θ=cos⁡ec2θ1 + {\cot ^2}\theta = \cos e{c^2}\theta
1+4=cos⁡ec2θ1 + 4 = \cos e{c^2}\theta
Or cos⁡ec2θ=5\cos e{c^2}\theta = 5
As we all know that sin⁡θ\sin \theta and cos⁡ecθ\cos ec\theta are both reverse quantities that is sin⁡θ=1cos⁡ecθ\sin \theta = \dfrac{1}{{\cos ec\theta }}
Hence, sin⁡2θ=15{\sin ^2}\theta = \dfrac{1}{5}
Taking root,
sin⁡θ=±15\sin \theta = \pm \dfrac{1}{{\sqrt 5 }}
Now, which value of sine is our required answer for knowing that using the second relation given.
cos⁡θ<0\cos \theta < 0 which tells that the cosine is a negative quantity whereas cot⁡θ=−2\cot \theta = - 2 is also a negative quantity and so the sine quantity must be positive and quadrant must be second.
Hence, the required answer is sin⁡θ=+15\sin \theta = + \dfrac{1}{{\sqrt 5 }}
So, the correct answer is “ sin⁡θ=+15\sin \theta = + \dfrac{1}{{\sqrt 5 }} ”.

Note : Learn all the trigonometric functions carefully. I mean on tips for that you need to solve as many trigonometric questions as you can. The quadrants should be chosen carefully. As mostly students make silly mistakes while choosing the quadrants.