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Question: Prove that \[2{\sin ^2}\theta + 3{\cos ^2}\theta = 2 + {\cos ^2}\theta \]...

Prove that 2sin⁡2θ+3cos⁡2θ=2+cos⁡2θ2{\sin ^2}\theta + 3{\cos ^2}\theta = 2 + {\cos ^2}\theta

Explanation

Solution

In this question, we have to prove that the given relation is equal or not. The given problem is the relation to prove. By using the trigonometry relations in the given relation we will prove the required result. We have to apply the formula of sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1

Complete step-by-step answer:
It is stated in the question 2sin⁡2θ+3cos⁡2θ2{\sin ^2}\theta + 3{\cos ^2}\theta ….(i)(i)
Now we have to apply the formula of sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 from we here we can write sin⁡2θ=1−cos⁡2θ{\sin ^2}\theta = 1 - {\cos ^2}\theta which we have to put in the above equation (i)(i)
2sin⁡2θ+3cos⁡2θ2{\sin ^2}\theta + 3{\cos ^2}\theta
By using the relation sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 we get,
=2(1−cos⁡2θ)+3cos⁡2θ= 2(1 - {\cos ^2}\theta ) + 3{\cos ^2}\theta
Multiplying the terms we get,
=2−2cos⁡2θ+3cos⁡2θ= 2 - 2{\cos ^2}\theta + 3{\cos ^2}\theta
Simplifying we get,
=2+cos⁡2θ= 2 + {\cos ^2}\theta
∴\therefore We have proved 2sin⁡2θ+3cos⁡2θ=2+cos⁡2θ2{\sin ^2}\theta + 3{\cos ^2}\theta = 2 + {\cos ^2}\theta .

Note: For solving this questions of trigonometry you have to remember the formulas for all like if you have to find out the value of sin⁡θ\sin \theta or Cos⁡θ\operatorname{Cos} \theta you can apply the formula sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 if you have the value of either of them
Next if you need to find out the value of sec⁡θ\sec \theta or tan⁡θ\tan \theta then you can apply the formula sec⁡2θ−tan⁡2θ=1{\sec ^2}\theta - {\tan ^2}\theta = 1, if you have the value of either of them
Now if you want to find out the value of cot⁡θ\cot \theta or cos⁡ecθ\cos ec\theta , then you can apply the formula cos⁡ec2θ−cot⁡2θ=1\cos e{c^2}\theta - {\cot ^2}\theta = 1, but for this you must know the value of either of them.
Besides these there are some other formulas which are necessary for solving the questions of trigonometry like sin⁡θ=1cos⁡ecθ\sin \theta = \dfrac{1}{{\cos ec\theta }}, tan⁡θ=1cot⁡θ\tan \theta = \dfrac{1}{{\cot \theta }} and cos⁡θ=1sec⁡θ\cos \theta = \dfrac{1}{{\sec \theta }}
In some cases there are some questions where we have to find out the value of θ\theta by applying those formula like sin⁡θ=cos⁡(90∘−θ)\sin \theta = \cos ({90^ \circ } - \theta ), tan⁡θ=cot⁡(90∘−θ)\tan \theta = \cot ({90^ \circ } - \theta ) sand sec⁡θ=cos⁡ec(90∘−θ)\sec \theta = \cos ec({90^ \circ } - \theta )