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**number0001****Member**- Registered: 2006-01-06
- Posts: 3

How do you do cos(3x)=4cos^3(X)-3cos(X)? I mean, I got to somehow transform the left side of the = sign so it looks exactly like the right side. Kind of like a proof. I think I need some of the trig formulas for this but I can't really get it... Thanks for any help!

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Trigonometric identies:

http://www.sosmath.com/trig/Trig5/trig5/trig5.html

You should print out a reference sheet.

Ok, I'm going to use the four following identies to do this:

1. sin^2 (x) + cos^2 (x) = 1

2. sin(2x) = 2 sin x cos x

3. cos (2x) = 2 cos^2 (x) - 1

4. cos (A + B) = cos A cos B - sin A sin B

I'll refere to these as identities 1, 2, 3 and 4.

Ok. We begin with cos (3x) and note this can be written as cos (2x + 1) if we use identity 4 and let A = 2x and B = x we have:

cos(2x) cos(x) - sin(2x) sin(x)

Now I'll substitue cos(2x) using identity 3:

(2cos^2 (x) -1) cos (x) - sin(2x) sin(x)

multiplied:

2 cos^3 (x) - cos(x) - sin(2x) sin (x)

Now I'll replace sin(2x) using identity 2.

2 cos^3 (x) - cos(x) - 2sin x cos x sin x

simplified:

2 cos^3 (x) - cos(x) - 2 cos (x) sin^2 (x)

in indentiy 1, we can solve for sin^2 (x) to find sin^2(x) = 1 - cos^2(x) we replace sin^2 (x) with (cos^2 (x) -1)

2 cos^3 (x) - cos(x) - 2 cos (x)(1 -cos^2(x))

multiplied:

2 cos^3 (x) - cos(x) + - 2 cos (x) + 2 cos^3(x)

Added like terms:

4 cos^3(x) -3 cos(x) Quod Erat Demonstratum! That which was to be demonstrated:-)

A logarithm is just a misspelled algorithm.

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**number0001****Member**- Registered: 2006-01-06
- Posts: 3

Thank you so much! That was very helpful--now I get it!

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