Trigonometry TricksDate: 11/14/2002 at 11:09:31 From: Tanya Riston Subject: Calculus 1 To find how (1/4)cos^4x-(1/6)cos^6x = -(1/4)sin^4x+(1/6)sin^6x, I've tried to change all the cos to sin by using sin^2x+cos^2x = 1 and then using double angle formulas. I am trying to prove that they are equal to each other. Date: 11/15/2002 at 10:47:17 From: Doctor Ian Subject: Re: Calculus 1 Hi Tanya, The first thing that occurs to me is that you can get rid of the fractions: (1/4)cos^4x - (1/6)cos^6x = -(1/4)sin^4x + (1/6)sin^6x cos^4x - (4/6)cos^6x = -sin^4x + (4/6)sin^6x 6cos^4x - 4cos^6x = -6sin^4x + 4sin^6x Now I'd look for symmetries that I can exploit. For example, can I get the same exponents together? That would let me use the distributive property. (At this point, I honestly don't know if this will work. I'm just trying to let you see how I'm thinking about it.) 6cos^4x + 6sin^4x = 4cos^6x + 4sin^6x 6(cos^4x + sin^4x) = 4(cos^6x + sin^6x) Now at this point, I don't see anything obvious, so I'll go look (in our FAQ) for some identities that I might be able to use. These look promising: sin^4(x) = (3 - 4 cos[2x] + cos[4x])/8, cos^4(x) = (3 + 4 cos[2x] + cos[4x])/8, sin^6(x) = (10 - 15 cos[2x] + 6 cos[4x] - cos[6x])/32, cos^6(x) = (10 + 15 cos[2x] + 6 cos[4x] + cos[6x])/32, Now, instead of substituting into the equation, which would give me a huge mess, I'll just try adding one pair together: sin^4(x) + cos^4(x) = (3 - 4 cos[2x] + cos[4x])/8 + (3 + 4 cos[2x] + cos[4x])/8 = (1/8)(3 - 4cos[2x] + cos[4x] + 3 + 4cos[2x] + cos[4x]) = (1/8)(6 + 2cos[4x]) So that doesn't look to bad! So the left side of my equation becomes 6(1/8)(6 + 2cos^4x) = 4(cos^6x + sin^6x) Just looking at the other pair, I can see that the [2x] terms and [6x] terms are going to kill each other, leaving me with only the constant and [4x] terms. So it looks like smooth sailing from here. Can you finish up? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/