Mathematics, 26.04.2020 01:56 Braxton12
If sin theta = (4)/(7), theta in quadrant II, find the exact value of (a) cos theta (b) sin (theta + (pi) / (6) ) (c) cos (theta - (pi) / (3) ) (d) tan (theta + (pi) / (4) )
Answers: 3
![\cos(\theta) = \frac{\sqrt[]{33}}{7}](/tpl/images/0627/5175/05c12.png)
![\sin(\theta + \frac{\pi}{6})\frac{-3\sqrt[]{11}+4}{14}](/tpl/images/0627/5175/028bc.png)
![\cos(\theta-\pi)=\frac{\sqrt[]{33}}{7}](/tpl/images/0627/5175/87538.png)
![\tan(\theta + \frac{\pi}{4}) = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}](/tpl/images/0627/5175/47f70.png)
. 
![x=-\sqrt[]{49-16} = -\sqrt[]{33}](/tpl/images/0627/5175/1afdb.png)
. Recall that cos(theta) is defined by adjacent side/hypotenuse. So, we know that the hypotenuse is 7, then ![\cos(\theta) = \frac{-\sqrt[]{33}}{7}](/tpl/images/0627/5175/f0248.png)
![\sin(\frac{\pi}{6}) =\frac{1}{2} , \cos(\frac{\pi}{6}) = \frac{\sqrt[]{3}}{2}](/tpl/images/0627/5175/5569b.png)
, then using the identity from above, we have that ![\sin(\theta + \frac{\pi}{6}) = \sin(\theta)\cos(\frac{\pi}{6})+\cos(\alpha)\sin(\frac{\pi}{6}) = \frac{4}{7}\frac{1}{2}-\frac{\sqrt[]{33}}{7}\frac{\sqrt[]{3}}{2} = \frac{-3\sqrt[]{11}+4}{14}](/tpl/images/0627/5175/2f119.png)
. Then, ![\cos(\theta-\pi)=\cos(\theta)\cos(\pi)+\sin(\theta)\sin(\pi) = \frac{-\sqrt[]{33}}{7}\cdot(-1) + 0 = \frac{\sqrt[]{33}}{7}](/tpl/images/0627/5175/93fd1.png)
and ![\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}=\frac{-4}{\sqrt[]{33}}](/tpl/images/0627/5175/70c8c.png)
. Then ![\tan(\theta+\frac{\pi}{4}) = \frac{\tan(\theta)+\tan(\frac{\pi}{4})}{1-\tan(\theta)\tan(\frac{\pi}{4})} = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}](/tpl/images/0627/5175/06a3c.png)
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If sin theta = (4)/(7), theta in quadrant II, find the exact value of (a) cos theta (b) sin (theta +...
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