Answer:
C.
Step-by-step explanation:
Answer:
a) ![\cos(\theta) = \frac{\sqrt[]{33}}{7}](https://tex.z-dn.net/?f=%5Ccos%28%5Ctheta%29%20%3D%20%5Cfrac%7B%5Csqrt%5B%5D%7B33%7D%7D%7B7%7D)
b) ![\sin(\theta + \frac{\pi}{6})\frac{-3\sqrt[]{11}+4}{14}](https://tex.z-dn.net/?f=%5Csin%28%5Ctheta%20%2B%20%5Cfrac%7B%5Cpi%7D%7B6%7D%29%5Cfrac%7B-3%5Csqrt%5B%5D%7B11%7D%2B4%7D%7B14%7D)
c) ![\cos(\theta-\pi)=\frac{\sqrt[]{33}}{7}](https://tex.z-dn.net/?f=%5Ccos%28%5Ctheta-%5Cpi%29%3D%5Cfrac%7B%5Csqrt%5B%5D%7B33%7D%7D%7B7%7D)
d)![\tan(\theta + \frac{\pi}{4}) = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}](https://tex.z-dn.net/?f=%5Ctan%28%5Ctheta%20%2B%20%5Cfrac%7B%5Cpi%7D%7B4%7D%29%20%3D%20%5Cfrac%7B%5Cfrac%7B-4%7D%7B%5Csqrt%5B%5D%7B33%7D%7D%2B1%7D%7B1%2B%5Cfrac%7B4%7D%7B%5Csqrt%5B%5D%7B33%7D%7D%7D)
Step-by-step explanation:
We will use the following trigonometric identities
![\sin(\alpha+\beta) = \sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)](https://tex.z-dn.net/?f=%5Csin%28%5Calpha%2B%5Cbeta%29%20%3D%20%5Csin%28%5Calpha%29%5Ccos%28%5Cbeta%29%2B%5Ccos%28%5Calpha%29%5Csin%28%5Cbeta%29)
![\cos(\alpha+\beta) = \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)](https://tex.z-dn.net/?f=%5Ccos%28%5Calpha%2B%5Cbeta%29%20%3D%20%5Ccos%28%5Calpha%29%5Ccos%28%5Cbeta%29-%5Csin%28%5Calpha%29%5Csin%28%5Cbeta%29)
.
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](https://tex.z-dn.net/?f=x%5E2%2B4%5E2%20%3D%207%20%5E2)
which implies that
. 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}](https://tex.z-dn.net/?f=%5Ccos%28%5Ctheta%29%20%3D%20%5Cfrac%7B-%5Csqrt%5B%5D%7B33%7D%7D%7B7%7D)
b)Recall that
, 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}](https://tex.z-dn.net/?f=%5Csin%28%5Ctheta%20%2B%20%5Cfrac%7B%5Cpi%7D%7B6%7D%29%20%3D%20%5Csin%28%5Ctheta%29%5Ccos%28%5Cfrac%7B%5Cpi%7D%7B6%7D%29%2B%5Ccos%28%5Calpha%29%5Csin%28%5Cfrac%7B%5Cpi%7D%7B6%7D%29%20%3D%20%5Cfrac%7B4%7D%7B7%7D%5Cfrac%7B1%7D%7B2%7D-%5Cfrac%7B%5Csqrt%5B%5D%7B33%7D%7D%7B7%7D%5Cfrac%7B%5Csqrt%5B%5D%7B3%7D%7D%7B2%7D%20%3D%20%5Cfrac%7B-3%5Csqrt%5B%5D%7B11%7D%2B4%7D%7B14%7D)
c) Recall that
. Then,
![\cos(\theta-\pi)=\cos(\theta)\cos(\pi)+\sin(\theta)\sin(\pi) = \frac{-\sqrt[]{33}}{7}\cdot(-1) + 0 = \frac{\sqrt[]{33}}{7}](https://tex.z-dn.net/?f=%5Ccos%28%5Ctheta-%5Cpi%29%3D%5Ccos%28%5Ctheta%29%5Ccos%28%5Cpi%29%2B%5Csin%28%5Ctheta%29%5Csin%28%5Cpi%29%20%3D%20%5Cfrac%7B-%5Csqrt%5B%5D%7B33%7D%7D%7B7%7D%5Ccdot%28-1%29%20%2B%200%20%3D%20%5Cfrac%7B%5Csqrt%5B%5D%7B33%7D%7D%7B7%7D)
d) Recall that
and
. 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}}}](https://tex.z-dn.net/?f=%5Ctan%28%5Ctheta%2B%5Cfrac%7B%5Cpi%7D%7B4%7D%29%20%3D%20%5Cfrac%7B%5Ctan%28%5Ctheta%29%2B%5Ctan%28%5Cfrac%7B%5Cpi%7D%7B4%7D%29%7D%7B1-%5Ctan%28%5Ctheta%29%5Ctan%28%5Cfrac%7B%5Cpi%7D%7B4%7D%29%7D%20%3D%20%5Cfrac%7B%5Cfrac%7B-4%7D%7B%5Csqrt%5B%5D%7B33%7D%7D%2B1%7D%7B1%2B%5Cfrac%7B4%7D%7B%5Csqrt%5B%5D%7B33%7D%7D%7D)
Answer: 1/8
Step-by-step explanation:
-1/4 = -2/8
3/8 + (-2/8) = 3/8 - 2/8
3/8 - 2/8 = 1/8
Hope I helped!!
420 over 7 hours. no table shown
Answer:
22.5 or 22 1/20
Step-by-step explanation:
1/4 x 6 = 1.5 7 x 3 = 21 21 + 1.5 = 22.5