If x=sin theta and y=sin(k theta),then (1-x^2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $x=\sin 	heta$ and $y=\sin(k 	heta)$.First,find $y_1 = \frac{dy}{dx}$:$\frac{dy}{d	heta} = k \cos(k 	heta)$ and $\frac{dx}{d	heta} = \c

Vedclass · Accepted Answer

$-k^2$

Vedclass · Answer

(C) Given $x=\sin 	heta$ and $y=\sin(k 	heta)$.First,find $y_1 = \frac{dy}{dx}$:$\frac{dy}{d	heta} = k \cos(k 	heta)$ and $\frac{dx}{d	heta} = \cos 	heta$.$y_1 = \frac{dy/d	heta}{dx/d	heta} = \frac{k \cos(k 	heta)}{\cos 	heta} \implies y_1 \cos 	heta = k \cos(k 	heta)$.Differentiating both sides with respect to $x$:$y_2 \cos 	heta - y_1 \sin 	heta \cdot \frac{d	heta}{dx} = -k^2 \sin(k 	heta) \cdot \frac{d	heta}{dx}$.Since $\frac{d	heta}{dx} = \frac{1}{\cos 	heta}$,we have:$y_2 \cos 	heta - y_1 \sin 	heta \cdot \frac{1}{\cos 	heta} = -k^2 \sin(k 	heta) \cdot \frac{1}{\cos 	heta}$.Multiplying by $\cos 	heta$:$y_2 \cos^2 	heta - y_1 \sin 	heta = -k^2 \sin(k 	heta)$.Since $1-x^2 = 1-\sin^2 	heta = \cos^2 	heta$ and $x = \sin 	heta$:$(1-x^2) y_2 - x y_1 = -k^2 y$.Comparing this with $(1-x^2) y_2 - x y_1 - \alpha y = 0$,we get $\alpha y = -k^2 y$,so $\alpha = -k^2$.

If $x=\sin \theta$ and $y=\sin(k \theta)$,then $(1-x^2) y_2-x y_1-\alpha y=0$,for $\alpha=$

Similar Questions

If $x=a$ is a root of multiplicity two of a polynomial equation $f(x)=0$,then

Find the second order derivative of the function $f(x) = x^{20}$.

$f(x) = e^x \sin x$,then $f^{(6)}(x)$ is equal to :

$A$ non-zero polynomial with real coefficients has the property that $f''(x) f'(x) = f(x)$. Then the value of $f'''(x)$ is

If $y=f(\cosh x)$ and $f^{\prime}(x)=\log \left(x+\sqrt{x^2-1}\right)$,then $\frac{d^2 y}{d x^2}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module