Question

Help would be very much appreciated!

Answer 1

\[\cos \theta=\sin(90-\theta)\] Would I use this cofunction Identity? I'm getting nowhere with it.

Answer 2

\(\bf sin\left(\frac{\pi}{2}-{\color{red}{ \theta}}\right)=cos({\color{red}{ \theta}})\qquad

cos\left(\frac{\pi}{2}-{\color{red}{ \theta}}\right)=sin({\color{red}{ \theta}})\)

what do you think?

Answer 3

yes... you;d use the cofunctions

Answer 4

Yeah, see I've been doing that and this is what I have so far: \[\frac{ Piy ^{2}+piy-2y^{2} }{ 2y^{2}+y }\]

Answer 5

But these are the answer choices

Answer 6

It doesn't seem like I'm on the right track :(

Answer 7

hmmm wait a second heheh... I thimk I may hav mixed something

Answer 8

well..the cheap way to do it I'd think is using instead \(\bf sin^2(\theta)+cos^2(\theta)=1\)

that is \(\bf {\color{brown}{ sin^2(\theta)+cos^2(\theta)=1\to sin^2(\theta)=1-cos^2(\theta)\to sin(\theta)=\pm \sqrt{1-cos^2(\theta)}}}
\\ \quad \\ \quad \\

sin(\theta)=\cfrac{y}{y+1}\implies \pm \sqrt{1-cos^2(\theta)}=\cfrac{y}{y+1}\)

Answer 9

hmmm a bit truncated... anyhow \(\bf {\color{brown}{ sin^2(\theta)+cos^2(\theta)=1\to sin^2(\theta)=1-cos^2(\theta)\\\to sin(\theta)=\pm \sqrt{1-cos^2(\theta)}}}
\\ \quad \\ \quad \\

sin(\theta)=\cfrac{y}{y+1}\implies \pm \sqrt{1-cos^2(\theta)}=\cfrac{y}{y+1}\)

Answer 10

That makes a lot more sense lol so I need to isolate cosx right? Should I square both sides?

Answer 11

yeap... square both sides
then move the 1 over
and multiply by -1 to end up with \(cos(\theta)\)

Answer 12

well... actually you'd end up with a \(cos^2(\theta)\) so you'd ned to take root also

Answer 13

\(\bf sin(\theta)=\cfrac{y}{y+1}\implies \pm \sqrt{1-cos^2(\theta)}=\cfrac{y}{y+1}
\\ \quad \\
1-cos^2(\theta)=\left(\cfrac{y}{y+1}\right)^2\implies -cos^2(\theta)=\left(\cfrac{y}{y+1}\right)^2+1
\\ \quad \\
cos^2(\theta)=-\left(\cfrac{y}{y+1}\right)^2-1\implies cos(\theta)=\sqrt{-\left(\cfrac{y}{y+1}\right)^2-1}\)

Answer 14

Is the answer \[\pm \frac{ \sqrt{2y-1} }{ y+1 }\]

Answer 15

hmmm actaully just noticed... the 1 should be negative

Answer 16

\(\bf sin(\theta)=\cfrac{y}{y+1}\implies \pm \sqrt{1-cos^2(\theta)}=\cfrac{y}{y+1}
\\ \quad \\
1-cos^2(\theta)=\left(\cfrac{y}{y+1}\right)^2\implies -cos^2(\theta)=\left(\cfrac{y}{y+1}\right)^2-1
\\ \quad \\
cos^2(\theta)=-\left(\cfrac{y}{y+1}\right)^2+1\implies cos(\theta)=\sqrt{-\left(\cfrac{y}{y+1}\right)^2+1}\)

Answer 17

if you want to coimbine the terms inside the root
you'd first need the expand the fraction...but the bottom gives a trinomial
lemme expand

Answer 18

\(\bf cos(\theta)=\sqrt{-\left(\cfrac{y}{y+1}\right)^2+1}
\\ \quad \\
cos(\theta)=\sqrt{1-\left(\cfrac{y}{y+1}\right)^2}\quad
\begin{cases}
1-\left(\cfrac{y}{y+1}\right)^2
\\ \quad \\
1-\cfrac{y^2}{(y+1)^2}
\\ \quad \\
\cfrac{(y+1)^2-y^2}{(y+1)^2}
\\ \quad \\
\cfrac{\cancel{ y^2 }+2y+1^2-\cancel{ y^2 }}{(y+1)^2}
\end{cases}\)

yeap... you're correct

Answer 19

Thank you so much.

Answer 20

yw

Answer 21

what about the 1

Answer 22

@jdoe0001 what about 1 in the last simplification