Question

cos^2theta + cos^2(alpha+theta)-2cosalpha×cos theta ×cos(alpha+theta) independent of theta

Answer 1

Answer:

$cos^2\theta+cos^2(\alpha+\theta)-2\:cos\alpha\:cos\theta\:cos(\alpha+\theta)$ is independent of $\theta$

Step-by-step explanation:

The given expression is simplified by using the following formulae

Formula used;

$1.\:cos(A+B)+cos(A-B)=2\;cosA\:cosB$

$2.\:cos(A+B).cos(A-B)=cos^2A-sin^2B$

Given:

$cos^2\theta+cos^2(\alpha+\theta)-2\:cos\alpha\:cos\theta\:cos(\alpha+\theta)$

$=cos^2\theta+cos^2(\alpha+\theta)-[2\:cos\theta\:cos\alpha]\:cos(\theta+\alpha)$

$=cos^2\theta+cos^2(\alpha+\theta)-[cos(\theta+\alpha)+cos(\theta-\alpha)]\:cos(\theta+\alpha)$

(using formula (1))

$=cos^2\theta+cos^2(\alpha+\theta)-[cos^2(\theta+\alpha)+cos(\theta+\alpha)\:cos(\theta-\alpha)]$

$=cos^2\theta+cos^2(\alpha+\theta)--cos^2(\theta+\alpha)-cos(\theta+\alpha)\:cos(\theta-\alpha)$

$=cos^2\theta-cos(\theta+\alpha)\:cos(\theta-\alpha)$

$=cos^2\theta-(cos^2\theta-sin^2\alpha)$      (using formula (2))

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

$=sin^2\alpha$

It is independent of $\theta$

Answer 2

Given :

${ \cos }^{2} \theta + { \cos}^{2} ( \alpha + \theta) - 2 \cos( \alpha )\cos( \theta) \cos( \alpha + \theta)$

Independent of $\theta$

Solution :

${ \cos }^{2} \theta + { \cos}^{2} ( \alpha + \theta) - 2 \cos \alpha \cos \theta \cos( \alpha + \theta)$

${ \cos}^{2} \theta + { \cos }^{2} ( \alpha + \theta) - [<strong>2 \cos( \alpha ) \cos( \theta)</strong> ]\cos( \alpha + \theta)$

${ \cos }^{2} \theta + { \cos}^{2} ( \alpha + \theta) - { \cos}^{2} ( \theta + \alpha ) + \cos (\theta - \alpha )\cos( \theta + \alpha )$

${ \cos}^{2} \theta - [\cos( \theta + \alpha ) \cos( \theta + \alpha ) ]$

${ \cos }^{2} \theta - { \cos}^{2} \theta + { \sin}^{2} \alpha$

${ \sin }^{2} \alpha$

Which is Independent of $\theta$

Identities Used :-

$\boxed{\cos( \alpha + \theta) + \cos( \theta - \alpha ) = 2 \cos( \alpha ) \cos( \theta)}$

$\boxed{ \cos( \theta - \alpha ) . \cos( \theta + \alpha ) = { \cos }^{2} \theta + { \sin}^{2} \alpha}$