Trig simplification (this one is tough)

kaplac · #1 $Old$ November 5th, 2009, 12:19 AM

f(x)= (4cos (x) -cos (3x))^2 + (3sin (x) -sin (3x))^2
f(x) has to be simplified to the form: f(x)= a cos ^2 (2x) + b cos (2x) + c where a,b,and c are real numbers.

so far I expanded the two parentheticals to: f(x)= 16 cos ^2 (x) - 8 cos(x) cos(3x) + cos ^2 (3x) + 9 sin^2 (x) -6 sin (x) sin (3x) + sin ^2 (3x)

after this I'm stumped.

I tried merging the 16cos ^2 (x) + sin ^2 (3x) to a 7 cos ^2(x) + 9, and doing the same with the sin ^2 (3x) + cos ^2 (3x) = 1. But it isn't taking me anywhere.

If someone can solve this with work, I would highly appreciate it!

Thanks
-kaplac

Grandad · #2 $Old$ November 5th, 2009, 10:22 AM

Hello kaplac

Quote:

Originally Posted by kaplac $View Post$

f(x)= (4cos (x) -cos (3x))^2 + (3sin (x) -sin (3x))^2
f(x) has to be simplified to the form: f(x)= a cos ^2 (2x) + b cos (2x) + c where a,b,and c are real numbers.

so far I expanded the two parentheticals to: f(x)= 16 cos ^2 (x) - 8 cos(x) cos(3x) + cos ^2 (3x) + 9 sin^2 (x) -6 sin (x) sin (3x) + sin ^2 (3x)

after this I'm stumped.

I tried merging the 16cos ^2 (x) + sin ^2 (3x) to a 7 cos ^2(x) + 9, and doing the same with the sin ^2 (3x) + cos ^2 (3x) = 1. But it isn't taking me anywhere.

If someone can solve this with work, I would highly appreciate it!

Thanks
-kaplac

Using the identities

$\cos3x = 4\cos^3x-3\cos x$ ; and

$\sin 3x = 3\sin x - 4\sin^3x$ , we get:

$f(x) = (7\cos x - 4\cos^3x)^2+(4\sin^3x)^2$

$=16\cos^6x-56\cos^4x+49\cos^2x+16(1-\cos^2x)^3$

$=16\cos^6x-56\cos^4x+49\cos^2x+16(1-3\cos^2x+3\cos^4x-\cos^6x)$

$= -8\cos^4x+\cos^2x+16$

Now $a\cos^22x+b\cos2x+c = a(2\cos^2x-1)^2 +b(2\cos^2x-1) + c$

$=4a\cos^4x+(2b-4a)\cos^2x+a-b+c$

Equating coefficients gives

$a = -2, b = -\frac72, c=\frac{29}{2}$

But check my working!

Grandad

kaplac · #3 $Old$ November 5th, 2009, 02:54 PM

Thank you so much Grandad!