Proving trigonometric equations

[peterspencers]

Homework Statement

Hi there, I am attempting to prove a trigonometric equation using the half angle and double angle formulae

Homework Equations

See image one...

The Attempt at a Solution

See image two...

I get stuck after the second line and can't see how to continue, please help.[/B]

 

[_N3WTON_]

Your denominator is a difference of squares, and can be factored...

 

[Brian_H]

1 / cosA + sin2A / cos2A = (1+sin2A) / cos2A . And you've got to change 1 = (sinx)^2 + (cosx)^2.

Then, 1+sin2A = (sinA) ^2 + (cosA)^2 + sin2A = (sinA)^2 + 2sinAcosA + (cosA)^2 = (sinA+ cosA)^2

and cos2A = (cosA)^2 - (sinA)^2 =(cosA-sinA)(cosA+sinA)

so (1+sin2A) / cos2A = (sinA + cosA) ^2 / (cosA-sinA)(cosA+sinA) = (sinA + cosA ) / (cosA - sinA)

 

[_N3WTON_]

1 / cosA + sin2A / cos2A = (1+sin2A) / cos2A . And you've got to change 1 = (sinx)^2 + (cosx)^2.

Then, 1+sin2A = (sinA) ^2 + (cosA)^2 + sin2A = (sinA)^2 + 2sinAcosA + (cosA)^2 = (sinA+ cosA)^2

and cos2A = (cosA)^2 - (sinA)^2 =(cosA-sinA)(cosA+sinA)

so (1+sin2A) / cos2A = (sinA + cosA) ^2 / (cosA-sinA)(cosA+sinA) = (sinA + cosA ) / (cosA - sinA)

This is hard for me to understand, but I'm pretty sure this is incorrect, I believe you made an error copying the problem:
The actual problem was:
\frac{1}{\cos(2x)} + \frac{\sin(2x)}{\cos(2x)}
While you wrote:
\frac{1}{\cos(x)} + \frac{\sin(2x)}{\cos(2x)}
As with most problems, I'm sure there are multiple ways to do it, but this is not very close to what I got when I did this problem, but it may just be because I am having a hard time reading your solution