Using identities

**Furyan** · December 18th 2011, 11:43 AM

Hello,

The question I’m working on is

$f(x) = \4cos^2{x} - \3sin^2{x}$

Show that $f(x) = \dfrac{1}{2} +\dfrac{7}{2}\cos2{x}$

I have tried using $\sin^2{x} + \cos^2{x} = 1$

And the double angle identity for $\cos2{x}$

The best I’ve got is

$f(x) = \3cos2{x} + \cos^2{x}$ , which it might not be.

Maybe I should be using the half angle identity, but I'm not sure how.

I know I have to work at these, but if someone could point me in the right direction with this I’d be very grateful.

Thank you

**ILikeSerena** · December 18th 2011, 11:50 AM

Hi Furyan!

Almost there!
What is the double angle identity for $\cos 2x$ in terms of $\cos^2 x$ ?

**Furyan** · December 18th 2011, 12:10 PM

Hi ILikeSerena

Almost there, really? Cool

It's $\cos2{x} = \2cos^2{x} - 1$ ?

I'll try using it and see what I get.

Thank you.

**Siron** · December 18th 2011, 12:25 PM

We have given:
$f(x)=4\cos^2(x)-3\sin^2(x)$
which we can write as:
$f(x)=4\cos^2(x)-3[1-\cos^2(x)]$
$=4\cos^2(x)-3+3\cos^2(x)$
$=7\cos^2(x)-3$
$=\frac{7}{2}\left[2\cos^2(x)-\frac{6}{7}]$
$=\frac{7}{2}\left[2\cos^2(x)-1+\frac{1}{7}\right]$
$=\frac{7}{2}\left[\cos(2x)+\frac{1}{7}\right]$
$=\frac{7}{2}\cos(2x)+\frac{1}{2}$

**Furyan** · December 18th 2011, 01:34 PM

Hi LikeSerena

Thank you for letting me know I was on the right track, it really helped. I think I have it.

$\cos2{x} = \2cos^2{x} - 1$

$\cos2{x} + 1 = \2cos^2{x}$

$\dfrac{1}{2}\(cos2{x}+1) =\cos^2{x}$

Substituting I get

$\3cos2{x}+ \dfrac{1}{2}(\cos2{x}+1)$

$\3cos2{x}+ \dfrac{1}{2}\cos2{x}+ \dfrac{1}{2}$

$\dfrac{7}{2}\cos2{x}+ \dfrac{1}{2}$

That appears to be it. Yipee, thank you!

**ILikeSerena** · December 18th 2011, 01:48 PM

Good!

**Furyan** · December 18th 2011, 01:51 PM

Hi Siron

$=7\cos^2(x)-3$

This is another approach I had tried, but I had got completely stuck, my factoring is not so good. I am very grateful for your reply and I will enjoy working through your method. Thank you very much for taking the time to reply, I am very grateful. It is enormously helpful.

**Furyan** · December 18th 2011, 02:03 PM

**Archie Meade** · December 18th 2011, 02:07 PM

You could also try

$cos2x=cos^2x-sin^2x$

$4cos^2x-3sin^2x=cos^2x+3\left(cos^2x-sin^2x\right)=cos^2x+3cos2x$

$=\frac{2cos^2x}{2}-\frac{cos2x}{2}+\frac{7cos2x}{2}$

$=\frac{2cos^2x-cos^2x+sin^2x}{2}+\frac{7cos2x}{2}=\frac{cos^2x+si n^2x}{2}+\frac{7cos2x}{2}$

$=\frac{1}{2}+\frac{7}{2}cos2x$

**Furyan** · December 18th 2011, 02:18 PM

Thank you ILikeSerena You are great teacher!

**Furyan** · December 18th 2011, 02:29 PM

Thank you very much Archie Meade I will try that too.

Awesome!

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