If sin (x+y)+cos (x+y)=sin left[cos ^{-1}left | vedclass

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

Question

(B) Let $\cos ^{-1}\left(\frac{1}{3}ight)=	heta$,then $\cos 	heta=\frac{1}{3}$.Since $\sin^2 	heta + \cos^2 	heta = 1$,we have $\sin 	heta = \s

Vedclass · Accepted Answer

-$1$

Vedclass · Answer

(B) Let $\cos ^{-1}\left(\frac{1}{3}ight)=	heta$,then $\cos 	heta=\frac{1}{3}$.Since $\sin^2 	heta + \cos^2 	heta = 1$,we have $\sin 	heta = \sqrt{1-\left(\frac{1}{3}ight)^2} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$.Thus,the given equation is $\sin (x+y)+\cos (x+y)=\frac{2\sqrt{2}}{3}$.Since $\frac{2\sqrt{2}}{3} \approx 0.9428$,and the maximum value of $\sin(x+y)+\cos(x+y)$ is $\sqrt{2} \approx 1.414$,this equation represents a constant value for the expression $(x+y)$.Let $x+y = C$,where $C$ is a constant.Differentiating both sides with respect to $x$:$\frac{d}{dx}(x+y) = \frac{d}{dx}(C)$$1 + \frac{dy}{dx} = 0$$\frac{dy}{dx} = -1$.

If $\sin (x+y)+\cos (x+y)=\sin \left[\cos ^{-1}\left(\frac{1}{3}\right)\right]$,then $\frac{d y}{d x}=$

Similar Questions

The two curves $x^{3}-3xy^{2}+2=0$ and $3x^{2}y-y^{3}=2$:

Let $f$ be a differentiable function satisfying $f(x + 2y) = 2yf(x) + xf(y) - 3xy + 1$ for all $x, y \in R$ such that $f'(0) = 1$. Then $f(2)$ is equal to:

Let $f : R \to R$ be a twice differentiable function satisfying $f(0) = f(1) = 0$ and $f'(x) = f^2(x)$ for all $x \in R$. Then $\lim_{x \to 2} (f(x) + xf'(x) + x^2f''(x))$ is equal to:

If $a(4+x^2)=x$ and $y-x^3=a^2$,then $\frac{dy}{dx}$ at $x=1$ is ...

If $x^2+y^2=t-\frac{1}{t}$ and $x^4+y^4=t^2+\frac{1}{t^2}$, then $\frac{dy}{dx}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module