If sin theta = frac{12}{13} where 0

Question

(B) Given $\sin 	heta = \frac{12}{13}$ and $0 < 	heta < \frac{\pi}{2}$ (first quadrant),$\cos 	heta$ is positive.$\cos 	heta = \sqrt{1 - \sin^2

Vedclass · Accepted Answer

$\frac{-56}{65}$

Vedclass · Answer

(B) Given $\sin 	heta = \frac{12}{13}$ and $0 $\cos 	heta = \sqrt{1 - \sin^2 	heta} = \sqrt{1 - (\frac{12}{13})^2} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}$.Given $\cos \phi = -\frac{3}{5}$ and $\pi $\sin \phi = -\sqrt{1 - \cos^2 \phi} = -\sqrt{1 - (-\frac{3}{5})^2} = -\sqrt{1 - \frac{9}{25}} = -\sqrt{\frac{16}{25}} = -\frac{4}{5}$.Using the identity $\sin(	heta + \phi) = \sin 	heta \cos \phi + \cos 	heta \sin \phi$:$\sin(	heta + \phi) = (\frac{12}{13})(-\frac{3}{5}) + (\frac{5}{13})(-\frac{4}{5})$$= -\frac{36}{65} - \frac{20}{65} = -\frac{56}{65}$.

If $\sin \theta = \frac{12}{13}$ where $0 < \theta < \frac{\pi}{2}$ and $\cos \phi = -\frac{3}{5}$ where $\pi < \phi < \frac{3\pi}{2}$,then $\sin(\theta + \phi)$ is:

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