Question

eliminate theta from the x= 4 cos +5sin : y=4 sin theta +5cos theta​

Answer 1

Answer:

$x^{2}+ y^{2} =41$

Step-by-step explanation:

Given the trigonometric relations,

$x= 4 cos\theta +5sin\theta$      ...(1)

$y=4 sin \theta -5cos \theta$      ...(2)

(To eliminate θ, one must be a negative one)

Squaring (1) and (2), we get

$x^{2} = (4 cos\theta +5sin\theta)^{2}$      ...(3)

$y^{2} =(4 sin \theta -5cos \theta)^{2}$      ...(4)

Adding the equation (3) and (4),

$x^{2}+ y^{2} = (4 cos\theta +5sin\theta)^{2}+(4 sin \theta -5cos \theta)^{2}$

Expanding the above equation using the algebraic identity,  $(a+b)^{2}=a^{2} +2ab+b^{2}$, we get

$x^{2}+ y^{2} = (4 cos\theta)^2 +2\times 4 cos\theta5sin\theta+(5sin\theta)^{2}+(4 sin \theta)^2 -2\times 4 sin\theta5cos\theta+(5cos \theta)^{2}$

$=16 cos^2\theta +40 cos\theta sin\theta+25sin^{2}\theta+16 sin^2 \theta -40 sin\theta cos\theta+25cos^{2} \theta$

$=16 (cos^2\theta+sin^2 \theta) +25(sin^{2}\theta+ cos^{2} \theta)$

Using the trigonometric identity, $sin^{2}\theta+ cos^{2} \theta=1$, we get

$x^{2}+ y^{2} =16 +25=41$

Therefore, $x^{2}+ y^{2} =41$