If cos theta = - frac{1}{sqrt{2}} and tan the | vedclass

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(B) Given $\cos 	heta = -\frac{1}{\sqrt{2}}$ and $	an 	heta = 1$.Since $\cos 	heta$ is negative and $	an 	heta$ is positive,$	heta$ must lie in

Vedclass · Accepted Answer

$(2n + 1)\pi + \frac{\pi}{4}$

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(B) Given $\cos 	heta = -\frac{1}{\sqrt{2}}$ and $	an 	heta = 1$.Since $\cos 	heta$ is negative and $	an 	heta$ is positive,$	heta$ must lie in the third quadrant.The principal value of $	heta$ in the third quadrant is $	heta = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$.The general solution for $	heta$ when $\cos 	heta = \cos \alpha$ and $\sin 	heta = \sin \alpha$ is $	heta = 2n\pi + \alpha$.Thus,the general value is $	heta = 2n\pi + \frac{5\pi}{4}$.Since $\frac{5\pi}{4} = \pi + \frac{\pi}{4}$,we can write this as $	heta = 2n\pi + \pi + \frac{\pi}{4} = (2n + 1)\pi + \frac{\pi}{4}$.

If $\cos \theta = - \frac{1}{\sqrt{2}}$ and $\tan \theta = 1$,then the general value of $\theta$ is

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