Question

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) )

Answer 1

Answer:

a) $\cos(\theta) = \frac{\sqrt[]{33}}{7}$

b) $\sin(\theta + \frac{\pi}{6})\frac{-3\sqrt[]{11}+4}{14}$

c) $\cos(\theta-\pi)=\frac{\sqrt[]{33}}{7}$

d)$\tan(\theta + \frac{\pi}{4}) = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}$

Step-by-step explanation:

We will use the following trigonometric identities

$\sin(\alpha+\beta) = \sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)$

$\cos(\alpha+\beta) = \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$$\tan(\alpha+\beta) = \frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}$.

Recall that given a right triangle, the sin(theta) is defined by opposite side/hypotenuse. Since we know that the angle is in quadrant 2, we know that x should be a negative number. We will use pythagoras theorem to find out the value of x. We have that

$x^2+4^2 = 7 ^2$

which implies that $x=-\sqrt[]{49-16} = -\sqrt[]{33}$. Recall that cos(theta) is defined by adjacent side/hypotenuse. So, we know that the hypotenuse is 7, then

$\cos(\theta) = \frac{-\sqrt[]{33}}{7}$

b)Recall that $\sin(\frac{\pi}{6}) =\frac{1}{2} , \cos(\frac{\pi}{6}) = \frac{\sqrt[]{3}}{2}$, then using the identity from above, we have that

$\sin(\theta + \frac{\pi}{6}) = \sin(\theta)\cos(\frac{\pi}{6})+\cos(\alpha)\sin(\frac{\pi}{6}) = \frac{4}{7}\frac{1}{2}-\frac{\sqrt[]{33}}{7}\frac{\sqrt[]{3}}{2} = \frac{-3\sqrt[]{11}+4}{14}$

c) Recall that $\sin(\pi)=0, \cos(\pi)=-1$. Then,

$\cos(\theta-\pi)=\cos(\theta)\cos(\pi)+\sin(\theta)\sin(\pi) = \frac{-\sqrt[]{33}}{7}\cdot(-1) + 0 = \frac{\sqrt[]{33}}{7}$

d) Recall that $\tan(\frac{\pi}{4}) = 1$ and $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}=\frac{-4}{\sqrt[]{33}}$. Then

$\tan(\theta+\frac{\pi}{4}) = \frac{\tan(\theta)+\tan(\frac{\pi}{4})}{1-\tan(\theta)\tan(\frac{\pi}{4})} = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}$