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Question: How do you verify \({\cos ^2}x - {\sin ^2}x = 1 - 2{\sin ^2}x\)?...

How do you verify cos⁡2x−sin⁡2x=1−2sin⁡2x{\cos ^2}x - {\sin ^2}x = 1 - 2{\sin ^2}x?

Explanation

Solution

Here we need to use the trigonometric properties where we must know that in the right angles triangle sin⁡θ=perpendicular hypotenuse\sin \theta = \dfrac{{{\text{perpendicular }}}}{{{\text{hypotenuse}}}} and cos⁡θ=basehypotenuse\cos \theta = \dfrac{{{\text{base}}}}{{{\text{hypotenuse}}}} where θ\theta is the angle between the base and the hypotenuse. So we can derive from it the property that cos⁡2x+sin⁡2x=1{\cos ^2}x + {\sin ^2}x = 1 and then get the desired result.

Complete step-by-step answer:
Here we are given that we need to verify cos⁡2x−sin⁡2x=1−2sin⁡2x{\cos ^2}x - {\sin ^2}x = 1 - 2{\sin ^2}x
So let us take the right angled triangle ABCABC right angles at BB

We know that sin⁡θ=perpendicular hypotenuse\sin \theta = \dfrac{{{\text{perpendicular }}}}{{{\text{hypotenuse}}}} and cos⁡θ=basehypotenuse\cos \theta = \dfrac{{{\text{base}}}}{{{\text{hypotenuse}}}}
So we can say that according to the figure:
sin⁡θ=perpendicular hypotenuse=ph\sin \theta = \dfrac{{{\text{perpendicular }}}}{{{\text{hypotenuse}}}} = \dfrac{p}{h} −−−−(1) - - - - (1)
cos⁡θ=basehypotenuse=bh−−−−−(2)\cos \theta = \dfrac{{{\text{base}}}}{{{\text{hypotenuse}}}} = \dfrac{b}{h} - - - - - (2)
Squaring (1) and (2) and adding we get:
⇒\Rightarrow sin⁡2θ+cos⁡2θ=p2+b2h2{\sin ^2}\theta + {\cos ^2}\theta = \dfrac{{{p^2} + {b^2}}}{{{h^2}}}
Now we know that according to the Pythagoras theorem we have:
⇒\Rightarrow p2+b2=h2{p^2} + {b^2} = {h^2}
So we can say that:
sin⁡2θ+cos⁡2θ=h2h2=1 ⇒sin⁡2θ+cos⁡2θ=1  {\sin ^2}\theta + {\cos ^2}\theta = \dfrac{{{h^2}}}{{{h^2}}} = 1 \\\ \Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = 1 \\\
So we get that sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1
So we get that cos⁡2x=1−sin⁡2x−−−−−(3){\cos ^2}x = 1 - {\sin ^2}x - - - - - (3)
Now we need to find the value of cos⁡2x−sin⁡2x{\cos ^2}x - {\sin ^2}x
Put the value in it from the equation (3):
So we will get:
1−sin⁡2x−sin⁡2x=1−2sin⁡2x1 - {\sin ^2}x - {\sin ^2}x = 1 - 2{\sin ^2}x
Hence we get the required verification by just using the trigonometric identities and using them to convert into the type as we need in the proof. So we need to utilise our mind in order to know and judge which formula we need to apply in order to get the result that is required.
For example: If we are given that we need to prove cos⁡2x−sin⁡2x=2cos⁡2x−1{\cos ^2}x - {\sin ^2}x = 2{\cos ^2}x - 1
So we just need to think that we need to get the terms in cosine in our proof so we will try to eliminate the terms that contain sin and we will get the desired proof.
Here we will substitute sin⁡2x=1−cos⁡2x{\sin ^2}x = 1 - {\cos ^2}x instead of cos⁡2x=1−sin⁡2x{\cos ^2}x = 1 - {\sin ^2}x.

Note: In these types of problems where we need to prove or verify the left hand side and the right hand side of the problem, we need to just know the properties and formulae of the trigonometric identities and get the desired proof by using substitution in the correct way.