Question

Simplify. Work is expected as this is the written part of the course. Then, submit through your student upload folders.

df
cos4theta - 2cos2thetha + 1

sint+cottcost

Solve for theta.
12=9-6costheta

Answer 1

1. \(\cos^{4}\theta-2\cos^{2}\theta+1\)
Let \(\text{x}=cos^2\theta \)
\(x^{2}-2x+1\)
You will find this is a perfect square trinomial. Simplify
\((x-1)^2\)
Plug x back in
\(\left(\cos^{2}\theta-1\right)^{2}\)
Use the identity that \(sin^2 u + cos^2 u = 1\)
This is rearranged to \(cos^2u-1 = -sin^2u\)
\((-sin^2\theta)^2\)
Simplify again
\(sin^4\theta\)

2.
\(sin(t)+cot(t)cos(t)\)
\(sin(t)+\frac{cos(t)}{sin(t)}cos(t)\)
\(\sin\left(t\right)+\frac{\cos^{2}\left(t\right)}{\sin\left(t\right)}\)
Find their common denominators
\(\frac{\sin^{2}\left(t\right)+\cos^{2}\left(t\right)}{\sin\left(t\right)}\)
Use the identity that \(sin^2 u + cos^2 u = 1\)
\(\frac{1}{sin(t)}\)
And as you know with the reciprocal identity, \(\frac{1}{sin(u)}=csc(u)\)
Then you can do that.

3.
You would do basic algebra at first
\(12=9-6\cos\theta\)
\(3=-6\cos\theta\)
\(-\frac{1}{2}=\cos\left(\theta\right)\)
Take cos inverse
\(\cos^{-1}\left(-\frac{1}{2}\right)=\theta\)
If you know your unit circle well, cos is -1/2 at 2pi/3. But since your question didn't tell any bounds, we can put the answer as \(\frac{2\pi}{3}+2n\pi~\) where n is an integer

Answer 2

By the way I believe your username is a bit wrong lol. I think you meant it to be Secure *Contain Protect* if you want it to mean the SCP foundation xD

Answer 3

oh I didnt realize I typed it wrong lmao

Answer 4

Oh well, I suppose it's SPC foundation now ¯\_(ツ)_/¯

Answer 5

lol hahaha