is in quadrant I, so
.
is in quadrant II, so
.
Recall that for any angle
,
![\sin^2\alpha+\cos^2\alpha=1](https://tex.z-dn.net/?f=%5Csin%5E2%5Calpha%2B%5Ccos%5E2%5Calpha%3D1)
Then with the conditions determined above, we get
![\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35](https://tex.z-dn.net/?f=%5Ccos%5Ctheta%3D%5Csqrt%7B1-%5Cleft%28%5Cdfrac45%5Cright%29%5E2%7D%3D%5Cdfrac35)
and
![\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}](https://tex.z-dn.net/?f=%5Csin%20x%3D%5Csqrt%7B1-%5Cleft%28-%5Cdfrac5%7B13%7D%5Cright%29%5E2%7D%3D%5Cdfrac%7B12%7D%7B13%7D)
Now recall the compound angle formulas:
![\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta](https://tex.z-dn.net/?f=%5Csin%28%5Calpha%5Cpm%5Cbeta%29%3D%5Csin%5Calpha%5Ccos%5Cbeta%5Cpm%5Ccos%5Calpha%5Csin%5Cbeta)
![\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta](https://tex.z-dn.net/?f=%5Ccos%28%5Calpha%5Cpm%5Cbeta%29%3D%5Ccos%5Calpha%5Ccos%5Cbeta%5Cmp%5Csin%5Calpha%5Csin%5Cbeta)
![\sin2\alpha=2\sin\alpha\cos\alpha](https://tex.z-dn.net/?f=%5Csin2%5Calpha%3D2%5Csin%5Calpha%5Ccos%5Calpha)
![\cos2\alpha=\cos^2\alpha-\sin^2\alpha](https://tex.z-dn.net/?f=%5Ccos2%5Calpha%3D%5Ccos%5E2%5Calpha-%5Csin%5E2%5Calpha)
as well as the definition of tangent:
![\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}](https://tex.z-dn.net/?f=%5Ctan%5Calpha%3D%5Cdfrac%7B%5Csin%5Calpha%7D%7B%5Ccos%5Calpha%7D)
Then
1. ![\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}](https://tex.z-dn.net/?f=%5Csin%28%5Ctheta%2Bx%29%3D%5Csin%5Ctheta%5Ccos%20x%2B%5Ccos%5Ctheta%5Csin%20x%3D%5Cdfrac%7B16%7D%7B65%7D)
2. ![\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}](https://tex.z-dn.net/?f=%5Ccos%28%5Ctheta-x%29%3D%5Ccos%5Ctheta%5Ccos%20x%2B%5Csin%5Ctheta%5Csin%20x%3D%5Cdfrac%7B33%7D%7B65%7D)
3. ![\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}](https://tex.z-dn.net/?f=%5Ctan%28%5Ctheta%2Bx%29%3D%5Cdfrac%7B%5Csin%28%5Ctheta%2Bx%29%7D%7B%5Ccos%28%5Ctheta%2Bx%29%7D%3D-%5Cdfrac%7B16%7D%7B63%7D)
4. ![\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}](https://tex.z-dn.net/?f=%5Csin2%5Ctheta%3D2%5Csin%5Ctheta%5Ccos%5Ctheta%3D%5Cdfrac%7B24%7D%7B25%7D)
5. ![\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}](https://tex.z-dn.net/?f=%5Ccos2x%3D%5Ccos%5E2x-%5Csin%5E2x%3D-%5Cdfrac%7B119%7D%7B169%7D)
6. ![\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7](https://tex.z-dn.net/?f=%5Ctan2%5Ctheta%3D%5Cdfrac%7B%5Csin2%5Ctheta%7D%7B%5Ccos2%5Ctheta%7D%3D-%5Cdfrac%7B24%7D7)
7. A bit more work required here. Recall the half-angle identities:
![\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2](https://tex.z-dn.net/?f=%5Ccos%5E2%5Cdfrac%5Calpha2%3D%5Cdfrac%7B1%2B%5Ccos%5Calpha%7D2)
![\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2](https://tex.z-dn.net/?f=%5Csin%5E2%5Cdfrac%5Calpha2%3D%5Cdfrac%7B1-%5Ccos%5Calpha%7D2)
![\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}](https://tex.z-dn.net/?f=%5Cimplies%5Ctan%5E2%5Cdfrac%5Calpha2%3D%5Cdfrac%7B1-%5Ccos%5Calpha%7D%7B1%2B%5Ccos%5Calpha%7D)
Because
is in quadrant II, we know that
is in quadrant I. Specifically, we know
, so
. In this quadrant, we have
, so
![\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32](https://tex.z-dn.net/?f=%5Ctan%5Cdfrac%20x2%3D%5Csqrt%7B%5Cdfrac%7B1-%5Ccos%20x%7D%7B1%2B%5Ccos%20x%7D%7D%3D%5Cdfrac32)
8. ![\sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}](https://tex.z-dn.net/?f=%5Csin3%5Ctheta%3D%5Csin%28%5Ctheta%2B2%5Ctheta%29%3D%5Cdfrac%7B44%7D%7B125%7D)
Answer:
3t-9 >= 0
Step-by-step explanation:
Original function f(t)= square root 3t-9
There cannot be a negative in the radicand therefore 3t-9 cannot equal a negative number.
Answer:
there are 52 cards in a deck so the chance is a red card (or not a red card) is 26 out of 52.
Step-by-step explanation:
Answer:
4
Step-by-step explanation:
I took the quiz, here's proof
Answer: Solve for X
4x−5y<1
Step-by-step explanation:
step 1 add 5y to each side
4x−5y+5y<1+5y
4x<5y+1
Step 2: Divide both sides by 4.
4x4<5y+14
x<54y+14
Answer:
x<54y+14
Let's solve for x.
12y−x<3
Step 1: Add (-1)/2y to both sides.
−x+12y+−12y<3+−12y
−x<−12y+3
Step 2: Divide both sides by -1.
−x−1<−12y+3−1
x>12y−3
Answer:
x>12y−3
now just graph 4x-5y<1 shows the given inequalities in slope -intercep form.