Question

Please help me for the love of God if i fail I have to repeat the class

Given : sin ∅= 4/5 | cos x = -5/13 (∅: quadrant I, x: quadrant II)

what is:


1. sin(∅+x)


2. cos(∅-x)


3. tan(∅+x)


4. sin2 ∅


5. cos 2x


6. tan2 ∅


7. tan 1/2x


8. sin 3 ∅

Answer 1

`\theta` is in quadrant I, so
`\cos\theta>0`.

`x` is in quadrant II, so
`\sin x>0`.

Recall that for any angle
`\alpha`,

`\sin^2\alpha+\cos^2\alpha=1`

Then with the conditions determined above, we get

`\cos\theta=√(1-\left(\frac45\right)^2)=\frac35`

and

`\sin x=\sqrt{1-\left(-\frac5{13}\right)^2}=(12)/(13)`

Now recall the compound angle formulas:

`\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta`

`\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta`

`\sin2\alpha=2\sin\alpha\cos\alpha`

`\cos2\alpha=\cos^2\alpha-\sin^2\alpha`

as well as the definition of tangent:

`\tan\alpha=(\sin\alpha)/(\cos\alpha)`

Then

1.
`\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=(16)/(65)`

2.
`\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=(33)/(65)`

3.
`\tan(\theta+x)=(\sin(\theta+x))/(\cos(\theta+x))=-(16)/(63)`

4.
`\sin2\theta=2\sin\theta\cos\theta=(24)/(25)`

5.
`\cos2x=\cos^2x-\sin^2x=-(119)/(169)`

6.
`\tan2\theta=(\sin2\theta)/(\cos2\theta)=-\frac{24}7`

7. A bit more work required here. Recall the half-angle identities:

`\cos^2\frac\alpha2=\frac{1+\cos\alpha}2`

`\sin^2\frac\alpha2=\frac{1-\cos\alpha}2`

`\implies\tan^2\frac\alpha2=(1-\cos\alpha)/(1+\cos\alpha)`

Because
`x` is in quadrant II, we know that
`\frac x2` is in quadrant I. Specifically, we know
`\frac\pi2<x<\pi`, so
`\frac\pi4<\frac x2<\frac\pi2`. In this quadrant, we have
`\tan\frac x2>0`, so

`\tan\frac x2=\sqrt{(1-\cos x)/(1+\cos x)}=\frac32`

8.
`\sin3\theta=\sin(\theta+2\theta)=(44)/(125)`