# [SOLVED] Double angle identities

• Feb 25th 2009, 12:51 AM
Beard
[SOLVED] Double angle identities
Stuck on all of these and can't see to get my answers anywhere close to the answers

a) Express $\displaystyle \frac{\sin2x}{1 - 2\cos2x}$ in terms of $\displaystyle \tan(x)$

b) Prove the identity $\displaystyle \frac{1 + \sin2A - \cos2A}{1 + \sin2A + \cos2A} \equiv \tan(A)$

and finally

c) Given that $\displaystyle \cos2A = \tan^{2}x$ show that $\displaystyle \cos2x = \tan^{2}A$
• Feb 25th 2009, 01:36 AM
Quote:

Originally Posted by Beard
Stuck on all of these and can't see to get my answers anywhere close to the answers

a) Express $\displaystyle \frac{\sin2x}{1 - 2\cos2x}$ in terms of $\displaystyle \tan(x)$

$\displaystyle sin(2x) =2sin(x)cos(x)$

$\displaystyle cos(2x) = cos^2(x) -sin^2(x)$

$\displaystyle \frac{\sin2x}{1 - 2\cos2x}$

$\displaystyle =\frac{2sin(x)cos(x)}{1-2(cos^2(x) -sin^2(x))}$

Divide numerator and denominator by $\displaystyle cos^2(x)$

$\displaystyle \frac{\frac{2sin(x)}{cos(x)}} {\frac{1}{cos^2(x)}-\frac{2cos^2(x)}{cos^2(x)}+\frac{2sin^2(x)}{cos^2( x)}}$

I wont tell you that $\displaystyle tan(x)= \frac{sin(x)}{cos(x)}$

and that $\displaystyle tan^2(x)+1 = sec^2(x)$

Neither will I tell you that

$\displaystyle \frac{2tan(x)} {1+tan^2(x)-2+2tan^2(x)}$

$\displaystyle = ~\frac{2tan(x)}{-1+3tan^2(x)}$
• Feb 25th 2009, 01:50 AM
Quote:

Originally Posted by Beard
Stuck on all of these and can't see to get my answers anywhere close to the answers

b) Prove the identity $\displaystyle \frac{1 + \sin2A - \cos2A}{1 + \sin2A + \cos2A} \equiv \tan(A)$

Using the formula

http://www.mathhelpforum.com/math-he...ca515035-1.gif

http://www.mathhelpforum.com/math-he...c1b4fa9f-1.gif

We will get
numerator = $\displaystyle 1 +2sin(A)cos(A) - cos^2(A) + sin^2(A)$

Now we use
$\displaystyle 1-cos^2(x) = sin^2(x)$

So numerator $\displaystyle = 2sin^2(A) +2sin(A)cos(A) = 2sin(A)[sin(A) +cos(A)]$

Similarly Denominator will become

$\displaystyle 1 +2sin(A)cos(A) +cos^2(A)-sin^2(A)$

$\displaystyle = 2cos^2(A) +2sin(A)cos(A)$

$\displaystyle = 2cos(A) (sin(A) +cos(A))$
So our term is now

$\displaystyle =\frac{2sin(A)(sin(A) +cos(A))}{2cos(A)(sin(A) +cos(A))}$

Now cancel the terms common in numerator and denominator
And that's the end of the Second Show(Party)(Dance)
• Feb 25th 2009, 02:05 AM
Quote:

Originally Posted by Beard
c) Given that $\displaystyle \cos2A = \tan^{2}x$ show that
$\displaystyle \cos2x = \tan^{2}A$

$\displaystyle cos(2A) = cos^2(A) - sin^2(A) = 2cos^2(x) - 1$

Hence equation now is
$\displaystyle 2cos^2(A) - 1 = tan^2(x)$

$\displaystyle 2cos^2(A)= sec^2(x)$

$\displaystyle cos^2(A) = sec^2(x)/2$

Hence $\displaystyle sec^2(A) = \frac{2}{sec^2(x) } = 2 cos^2(x)$

Thus
$\displaystyle sec^2(A)=2 cos^2(x)$

Substract one from both sides

And (Dance)