# Trig+Algerbra question

1. Aug 10, 2008

### H0T_S0UP

1. The problem statement, all variables and given/known data
Express [(Cos^3)x][(Sin^4)x] in terms of cos(x) and its powers:

2. Relevant equations
All I used was Sin^2+Cos^2=1 but apparently it didn't work! I probably should have used more identities but I don't know.

3. The attempt at a solution
I broke the equation into two sections, the [(Cos^3)x] and then [(Sin^4)x] multiplied them together

For [(Cos^3)x]

cos(x)(1-sin^2 (x))
cos(x)-cos(x)(sin^2 (x))
cos(x)-cos(x)[1-cos^2 (x)]
=
cos^3 (x)

For [(Sin^4)x]

(1-cos^2 (x))(1-cos^2 (x))
1-(cos^2 (x)) - (cos^2 (x)) + (cos^4 (x))
1 - (2cos^2 (x)) + {[1-sin^2 (x)][1-sin^2 (x)]}
1 - (2cos^2 (x)) + 1-(2sin^2 (x)) + (sin^4 (x))
{(sin^4 (x)) - 2(cos^2 (x)) - 2(sin^2 (x)) +2}
And I stop here because the same pattern could go on forever and ever!
Where did I go wrong and can someone please show me the work the review sheet does not have it obviously.

2. Aug 10, 2008

### Chrisas

If I'm reading the problem right, you need to reduce the product of cos^3 * sin^4 into just powers of cos? right? Well you are making the problem too hard by trying to do anything with cos^3. It is already a power of cos, so you don't need to do anything with it. This is why you are coming around full circle back to cos^3 on that part. In your solution for sin^4, you have a line that is all cos. Stop there. Don't try to put sin back in after that. All that's left is to mutiply that through by cos^3.

3. Aug 10, 2008

### H0T_S0UP

Oh wow thanks I cant believe the mistake was that simple. If I did multiply 1-(cos^2 (x)) - (cos^2 (x)) + (cos^4 (x)) by cos^3 (x) I would have gotten:
1-(cos^2 (x)) - (cos^2 (x)) + (cos^4 (x))
cos^3 - 2cos^5 + cos^7 which was the correct answer.
Thanks.