MHB Proving a trigonometric identity II

[egillesp]

Hi,

I need help proving the following trig identity,
(2sinx)\overline{secxtan(2x)}=2cos^2x-csc^2x+cot^2x
I have tried starting from the left hand side, the right hand side, and doing both together, but nothing seems to work.

One of the ways I tried:
LHS: (2sinx)\overline{tan(2x)}\overline{cosx}
=2tanx\overline{tan(2x)}
=2tanx\overline{2tanx}\overline{1-tan^2x} using the tan(2x) identity
=1\overline{1-tan^2x}
=1\overline{1-(sin^x)/(cos^2x)}
=1\overline{(cos^2x-sin^2x)}\overline{cos^2x}
=1\overline{cos(2x)}\overline{cos^2x}
=[1/(cos(2x))][1/cos^2x]
=[1/(1-2sin^2x)][1/cos^2x]
=1/(cos^2x-2sin^2x*cos^2x)
=sec^2x/(1-2sin^2x)
=(tan^2x+1)/(1-2(1-cos^2x))
=(tan^2x+1)/(2cos^2x-1)
and I just don't get anywhere

Help would be greatly appreciated :)

 

[MarkFL]

Rather than us editing all of your posts, the $\LaTeX$ code for expressing a fraction is:

\frac{numerator}{denominator}

Also, you need to wrap your code with tags, such as the MATH tags, which are generated by clicking the $\sum$ button on our posting toolbar. :D

I would recommend editing your post so that it is easier to read...

 

[MarkFL]

This is what I would do...first state the identity:

$$\frac{2\sin(x)}{\sec(x)\tan(2x)}=2\cos^2(x)-\csc^2(x)+\cot^2(x)$$

Next, verify it is an identity before potentially wasting time:

>>click here<<

W|A says we're good to go. So, let's begin with the left side of the identity:

$$\frac{2\sin(x)}{\sec(x)\tan(2x)}$$

Let's move the secant function from the denominator to the numerator as a cosine function:

$$\frac{2\sin(x)\cos(x)}{\tan(2x)}$$

Next, in the numerator, let's apply the double-angle identity for sine:

$$\frac{\sin(2x)}{\tan(2x)}$$

Rewrite the tangent function in terms of sine and cosine:

$$\frac{\sin(2x)}{\frac{\sin(2x)}{\cos(2x)}}$$

Simplify algebraically:

$$\cos(2x)$$

Apply a double-angle identity for cosine:

$$2\cos^2(x)-1$$

Apply the Pythagorean identity $$1=\csc^2(x)-\cot^2(x)$$:

$$2\cos^2(x)-\csc^2(x)+\cot^2(x)$$

And we're done. :D