Question

If X=3sin theta + 4cos theta and y equals to 3cos theta -4 sin theta then prove that x2+y2=

Answer 1

Answer:

refer to the attachment

$x {}^{2} + y {}^{2} = 25$

hope it's helpful mark as brainlest

Answer 2

SOLUTION

GIVEN

x = 3 sin θ + 4 cos θ and y = 3 cos θ - 4 sin θ

TO DETERMINE

$\sf {x}^{2} + {y}^{2}$

EVALUATION

Here it is given that

x = 3 sin θ + 4 cos θ

y = 3 cos θ - 4 sin θ

Now we have

$\sf {x}^{2} + {y}^{2}$

$\sf = {(3 \sin \theta + 4 \cos \theta)}^{2} + {(3 \cos \theta - 4 \sin \theta)}^{2}$

$\sf = {(3 \sin \theta)}^{2} +2 \times 3 \sin \theta \times 4 \cos \theta + {(4 \cos \theta)}^{2} + {(3 \cos \theta ) }^{2} -2 \times 3 \cos \theta \times 4 \sin \theta + {(4 \sin \theta)}^{2}$

$\sf = 9{ \sin}^{2} \theta +24 \sin \theta \cos \theta +16 { \cos }^{2} \theta + 9{\cos}^{2} \theta -24\sin \theta \cos \theta +16 { \sin}^{2} \theta$

$\sf = 9{ \sin}^{2} \theta +16 { \cos }^{2} \theta + 9{\cos}^{2} \theta +16 { \sin}^{2} \theta$

$\sf = (9 + 16){ \sin}^{2} \theta +(16 + 9){ \cos }^{2} \theta$

$\sf = 25{ \sin}^{2} \theta +25{ \cos }^{2} \theta$

$\sf = 25({ \sin}^{2} \theta +{ \cos }^{2} \theta )$

$\sf = 25 \times 1$

$\sf = 25$

FINAL ANSWER

$\sf {x}^{2} + {y}^{2} = 25$

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