If sin x = frac{3}{5} and cos y = -frac{12}{1 | vedclass

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

Question

(A) We know that the formula for $\sin (x+y)$ is:$\sin (x+y) = \sin x \cos y + \cos x \sin y$ ... $(1)$Since $x$ is in the second quadrant,$\cos x$ is

Vedclass · Accepted Answer

$-\frac{56}{65}$

Vedclass · Answer

(A) We know that the formula for $\sin (x+y)$ is:$\sin (x+y) = \sin x \cos y + \cos x \sin y$ ... $(1)$Since $x$ is in the second quadrant,$\cos x$ is negative.$\cos^2 x = 1 - \sin^2 x = 1 - (\frac{3}{5})^2 = 1 - \frac{9}{25} = \frac{16}{25}$$\cos x = -\frac{4}{5}$Since $y$ is in the second quadrant,$\sin y$ is positive.$\sin^2 y = 1 - \cos^2 y = 1 - (-\frac{12}{13})^2 = 1 - \frac{144}{169} = \frac{25}{169}$$\sin y = \frac{5}{13}$Substituting these values into equation $(1)$:$\sin (x+y) = (\frac{3}{5})(-\frac{12}{13}) + (-\frac{4}{5})(\frac{5}{13})$$\sin (x+y) = -\frac{36}{65} - \frac{20}{65} = -\frac{56}{65}$

If $\sin x = \frac{3}{5}$ and $\cos y = -\frac{12}{13}$,where $x$ and $y$ both lie in the second quadrant,find the value of $\sin (x+y)$.

Similar Questions

$\tan 15^\circ = $

Prove that $\cot 4x(\sin 5x + \sin 3x) = \cot x(\sin 5x - \sin 3x)$

Prove that $\frac{\sin (x+y)}{\sin (x-y)} = \frac{\tan x + \tan y}{\tan x - \tan y}$.

If $(1+\tan \alpha)(1+\tan 4 \alpha)=2$ and $\alpha \in \left(0, \frac{\pi}{16}\right)$,then $\alpha$ is equal to

Prove that $\frac{\cos 9x - \cos 5x}{\sin 17x - \sin 3x} = -\frac{\sin 2x}{\cos 10x}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module