Question

Determine which expression is equivalent to cos3 θ. (Multiple Choice)

Answer 1

A. 3sin^3 θ-4cos θ

B.3cos^4 θ-4cos θ

C.4cos^3 θ-3cos θ

D.4sin^3 θ-3cos θ

Answer 2

There's two ways you can do this one.
1) by the expansion cos (x+y) = cosx cosy - sinx siny
2) by equating De Moivre's Theorem with binomial theorem terms

The way most people do is method one:

\[\cos(3 \theta) = \cos(2\theta + \theta) = \cos(2 \theta)\cos(\theta) - \sin(2\theta) \sin(\theta)\]
Then substitute the cos(2 theta) and sin (2theta) double angle identities in and see what you get

Answer 3

So is it like this? cos(2θ)=cos(1θ+1θ)=cos(1θ)cos(1θ)-sin(1θ)sin(1θ)

Answer 4

keep going. Simplify further

Answer 5

I don't know what to do after that.

Answer 6

you should get \[\cos(2 \theta) = \cos^2(\theta) - \sin^2(\theta) = 2\cos^2 (\theta) - 1 = 1 - 2 \sin^2(\theta)\]

(the three are interchangeable)

And \[\sin(2 \theta) = 2\sin(\theta)\cos(\theta)\]

Answer 7

Ok. I think it's C.

Answer 8

then sub these two back into the orginal

\[cos(3 theta) = cos(2 theta + theta) = cos(2 theta) cos(theta) - sin(2theta) sin(theta)\]
\[= (2\cos^2\theta -1)\cos \theta - 2\sin \theta \cos \theta(\sin \theta) = 2\cos^3(\theta) - \cos \theta -2\sin^2(\theta)\cos(\theta)\]

When you get sin^2(theta), replace it with 1 - cos^2(theta) (Pythagorean identity)

\[2\cos^3\theta - \cos \theta - 2(1-\cos^2(\theta))\cos(\theta) = 2\cos^3 \theta - \cos \theta -2\cos \theta + 2\cos^3 \theta\]
\[= 4\cos^3 \theta - 3 \cos \theta\]

Answer 9

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