Question

cos squared theta minus sin square theta
​

Answer 1

$sin^{2} x+cos^{2} x=1\\cos^{2}x =1-sin^{2} x\\.: cos^{2}x-sin^{2} x=1-sin^{2} x-sin^{2} x=1-2sin^{2} x=1-2(1-cos^{2}x)=2cos^{2}x -1$

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Answer 2

Correct Question: The Value of the function $cos^2 \theta -sin^2\theta$ is

Answer:

The values of $cos^2 \theta -sin^2\theta=2cos^2\theta -1$

                                             $=1-2sin^2\theta$

                                             $=cos2\theta$

Step-by-step explanation:

Recall the trigonometric identity of sine and cosine functions,

$sin^2\theta +cos^2\theta =1$

Rewrite as follows:

⇒ $sin^2\theta =1-cos^2\theta$ or $cos^2\theta =1-sin^2\theta$

Consider the given trigonometric function as follows:

$cos^2 \theta -sin^2\theta$ ...... (1)

Three cases arises.

Case1. When $sin^2\theta =1-cos^2\theta$.

Substitute the value $1-cos^2\theta$ for $sin^2\theta$ in the function (1) as follows:

⇒ $cos^2 \theta -sin^2\theta=cos^2 \theta -(1-cos^2\theta )$

Simplify as follows:

⇒ $cos^2 \theta -sin^2\theta=cos^2 \theta -1+cos^2\theta$

                          $=2cos^2 \theta -1$

Case2. When $cos^2\theta =1-sin^2\theta$.

Substitute the value $1-sin^2\theta$ for $cos^2\theta$ in the function (1) as follows:

⇒ $cos^2 \theta -sin^2\theta=(1-sin^2 \theta )-sin^2\theta$

Simplify as follows:

⇒ $cos^2 \theta -sin^2\theta=1-sin^2 \theta -sin^2\theta$

                          $=1-2sin^2 \theta$

Case3. As we know the sum identity of cosine function,

$cos(x+y)=cosx\ cosy-sinx\ siny$

                $=cos^2x-sin^2y$ ...... (2)

So, function (1) can also be written as follows:

$cos^2 \theta -sin^2\theta=cos\theta \ cos\theta - sin\theta \ sin\theta$

Comparing with the identity (1), we get

$cos^2 \theta -sin^2\theta=cos(\theta + \theta)$

                     $=cos2\theta$.

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