Question

Can someone help me?

Answer 1

ill help

Answer 2

thats math?

Answer 3

http://www.purplemath.com/modules/idents.htm

Answer 4

hold the mayo

Answer 5

Hold the mayo?

Answer 6

yes.... just extra mustard =)

Answer 7

Haha okay. Are you able to help me again?

Answer 8

\(\bf \cfrac{1+cos(2\alpha)}{sin(2\alpha)}
\\ \quad \\
\textit{Double Angle Identities}
\\ \quad \\
sin(2\theta)=2sin(\theta)cos(\theta)
\\ \quad \\
cos(2\theta)=
\begin{cases}
cos^2(\theta)-sin^2(\theta)\\
1-2sin^2(\theta)\\
2cos^2(\theta)-1
\end{cases}\qquad thus
\\ \quad \\
\cfrac{\cancel{1}+[{\color{brown}{ 2cos^2(\alpha)\cancel{-1} }}]}{{\color{brown}{ 2sin(\alpha)cos(\alpha)}}}\implies
\cfrac{2cos^2(\alpha)}{2sin(\alpha)cos(\alpha)}
\\ \quad \\
\cfrac{\cancel{2} cos(\alpha)\cancel{cos(\alpha)}}{\cancel{2} sin(\alpha)\cancel{cos(\alpha)}}\implies ?\)

Answer 9

cos(α)/sin(α)

Answer 10

yeap

Answer 11

So it's not an identity?

Answer 12

eh? well check your basic identities

what's \(\bf \cfrac{cos(\theta)}{sin(\theta)} =?\)

Answer 13

ohhh it does equal cot

Answer 14

yeap

Answer 15

My bad. Thanks!!!

Answer 16

yw