Question

if X equal to a sin theta + b cos theta and Y equal a cos theta - b sin theta then find the value of x square + Y square

Answer 1

Heya mate, Here is ur answer

The answer is

x^2+ y^2 = a^2 +b^2

$\mathbb{PLS\: REFER\: TO\: ATTACHMENT \:FOR \:SOLUTION}$


Identities used --

⏩ Sin^2 theta + cos^2 theta =1.

⏩(a+b)^2=a^2+b^2+2ab

⏩(a-b)^2= a^2 +b^2-2ab

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@Laughterqueen

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Answer 2

$X^2+Y^2=a^2+b^2$

Explanation:

Given that,

$X=a\sin\theta+b\cos\theta$

$Y=a\cos\theta-b\sin\theta$

We need to find the value of $X^2+Y^2$

Put the value of X and Y in $X^2+Y^2$. We get :

$X^2+Y^2=(a\sin\theta+b\cos\theta)^2+(a\cos\theta-b\sin\theta)^2\\\\=a^2\sin^2\theta+b^2\cos^2\theta+2ab\cos\theta\sin\theta+(b^2\sin^2\theta+a^2\cos^2\theta-2ab\cos\theta\sin\theta)\\\\=a^2(\sin^2\theta+\cos^2\theta)+b^2(\sin^2\theta+\cos^2\theta)$

Since, $(\sin^2\theta+\cos^2\theta)=1$

$X^2+Y^2=a^2+b^2$

Hence, this is the required solution.

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Trigonometric

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