Trig. (4 problems)

[Guest]

1.) Find all solutions of sin(2x) + cos(x) = 0 in the interval [0,2pie).
No idea where to start on this one.

2.) How could you show that sin(n*pie + alpha) = (-1)^n * sin(alpha) ?
When I attempted this I did:
sin(n*pie + alpha) = sin(n*pie)*cos(alpha) + cos(n*pie)*sin(alpha). Now what?

3.) What is the exact value of tan(2alpha) given that the terminal side of angle alpha goes through the point (-1, 2)?
I have no idea on this one either.

4.) Find csc(alpha) given that tan(alpha) = -4/7 and alpha is in the 3rd quadrant.
First time I put csc(alpha) = -1/4, which is incorrect. I tried it again and I got negative sqrt(65) / 4.

 

[soroban]

Hello, bloodelf_ella!

1.) Find all solutions of \(\displaystyle \sin(2x)\,+\,\cos(x)\:=\:0\) in the interval \(\displaystyle [0,2\pi).\)

Use the identity: .\(\displaystyle \sin2x\:=\:2\cdot\sin x\cdot\cos x\)

Then we have: .\(\displaystyle 2\cdot\sin x\cdot\cos x\,+\,\cos x\:=\:0\)

Factor: .\(\displaystyle \cos x\,(2\cdot\sin x\,+\,1)\:=\:0\)

And we have two equations to solve:
. . . \(\displaystyle \cos x\,=\,0\quad\Rightarrow\quad x\,=\,\frac{\pi}{2},\;\frac{3\pi}{2}\)
. . . \(\displaystyle 2\cdot\sin x\,+\,1\,=\,0\quad\Rightarrow\quad \sin x\,=\,-\frac{1}{2}\quad\Rightarrow\quad x\,=\,\frac{7\pi}{6},\,\frac{11\pi}{6}\)

2.) Show that: \(\displaystyle \sin(n\pi + \alpha)\:=\-1)^n\sin\alpha\) ?

When I attempted this, I did:
\(\displaystyle \sin(n\pi + \alpha)\:=\:\sin(n\pi)\cdot\cos\alpha\,+\,\cos(n\pi)\cdot\sin\alpha\) . Now what?

Thank you for showing your work . . . you're almost there!

Note that: \(\displaystyle \sin(n\pi)\,=\,0\)
. . The sine of any multiple of \(\displaystyle 180^o\) is 0.

Note that: \(\displaystyle \cos(n\pi)\,=\,\pm1\)
. . The cosine of any multiple of \(\displaystyle 180^o\) is either +1 or -1,
. . depending on where \(\displaystyle n\) is even or odd.

So that: .\(\displaystyle \sin(n\pi)\cdot\cos\alpha\,+\,\cos(n\pi)\cdot\sin\alpha\)
. . . . . \(\displaystyle =\;\;\;\;0\cdot\cos\alpha\:\:\:+\;\;(-1)^n\cdot\sin\alpha\)
. . . . . \(\displaystyle =\;\;\;\;\;\;\;\;(-1)^n\sin\alpha\)

3.) What is the exact value of \(\displaystyle tan(2\alpha)\),
given that the terminal side of angle \(\displaystyle \alpha\) goes through the point (-1, 2)?

Since \(\displaystyle \alpha\) goes through \(\displaystyle (-1,2)\), then: .\(\displaystyle \tan\alpha\,=\,\frac{2}{-1}\,=\,-2\)

Use the formula: .\(\displaystyle \tan2\theta\:=\:\frac{2\tan\theta}{1 - \tan^2\theta}\)

Then: .\(\displaystyle \tan2\alpha\:=\:\frac{2\tan\alpha}{1 - \tan^2\alpha}\:= \:\frac{2(-2)}{1 - (-2)^2}\:=\:\frac{-4}{-3}\:=\:\frac{4}{3}\)

4.) Find \(\displaystyle \csc\alpha\) given that \(\displaystyle \tan\alpha\,=\,-\frac{4}{7}\) and \(\displaystyle \alpha\) is in the 3rd quadrant. . impossible!
First time I put \(\displaystyle \csc\alpha\,=\,-\frac{1}{4}\), which is incorrect.
I tried it again and I got: -[\sqrt{65}]/4

This time the deck is stacked against you . . . there must be a typo.
. . Tangent is <u>positive</u> in the 3rd quadrant.

But your answer and your reasoning is correct.

 

[Guest]

I just wanted to say thank you very much for your help soroban! The problems make sense now.

I did actually type problem 4 incorrectly. The original problem had tan(alpha) = 4/7. On my paper I had added the negatives in myself because sin and cos are negative in QIII [ tan(alpha) = -4/-7 ] . Oops =)