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Mila [183]
3 years ago
12

Given cos = -2/5 a in quadrant III and cos b = 1/4, b in quadrant I find

Mathematics
1 answer:
BigorU [14]3 years ago
5 0

I guess you mean \cos a=-\dfrac25. Since a is in quadrant III, we expect \sin a. Then

\sin a=-\sqrt{1-\cos^2a}=-\dfrac{\sqrt{21}}5

Since b is in quadrant I, we expect \sin b>0, so that

\sin b=\sqrt{1-\cos^2b}=\dfrac{\sqrt{15}}4

Now,

\sin(a+b)=\sin a\cos b+\cos a\sin b=-\dfrac{2\sqrt{15}+\sqrt{21}}{20}

\cos(a+b)=\cos a\cos b-\sin a\sin b=\dfrac{3\sqrt{35}-2}{20}

and

\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=-\dfrac{2\sqrt{15}+\sqrt{21}}{3\sqrt{35}-2}

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