Question

(csc(4x)-cot(4x))/(csc(4x)+cot(4x))

A tan^2(4x)
B tan(2x)
C tan^2(2x)
D cot(4x)

Answer 1

\(\bf \cfrac{csc(4x)-cot(4x)}{csc(4x)+cot(4x)}\implies
\cfrac{
\frac{1}{sin(4x)}-\frac{cos(4x)}{sin(4x)}
}{
\frac{1}{sin(4x)}+\frac{cos(4x)}{sin(4x)}
}\implies
\cfrac{\frac{1-cos(4x)}{sin(4x)}}
{\frac{1+cos(4x)}{sin(4x)}}
\\ \quad \\
\cfrac{1-cos(4x)}{\cancel{ sin(4x) }}\cdot \cfrac{\cancel{ sin(4x) }}{1+cos(4x)}\implies ?\)

http://jwilson.coe.uga.edu/EMT668/EMAT6680.2003.fall/Bismarck/6690%20folder/trigproofs/powerreduc.gif <--- what do you think?

Answer 2

Okay, I see how you got to (1-cos(4x))/(1+cos(4x)).

Because of the formula (1-cos(2x))/(1+cos(2x))=tan^2x, does that mean my answer would be tan^4x?

Answer 3

I'm getting confused with the 4x within the cosine.

Answer 4

the cosine function on the right-side, uses 2 * the angle
the left-side uses half that

Answer 5

So as for my answer, I would end up with tan^2(2x).. Right?

Answer 6

yeap

Answer 7

Thank you :)

Answer 8

yw