theta

**ImKo** · January 3rd 2009, 09:13 PM

How do I express cos3 theta in terms of sin theta?

So far I GOT
=cos (2 theta + theta)
=cos 2 theta cos theta - sin 2 theta sin theta
=((2(1-sin^2 theta))-1)cos theta - sin 2 theta sin theta
=((2-2sin^2 theta)-1)cos theta - sin 2 theta sin theta

Now..... how do I eliminate the cos theta in order for it to become all sin theta

And

How do I solve this sin5 theta

So far I GOT
=sin(2 theta + 3 theta)
=sin 2 theta cos 3 theta + cos 2 theta sin 3 theta
=(2sin theta cos theta)(4 cos^3 theta - cos theta)+(1-2sin^2 theta)(3sin theta-4ain^3theta)

I want to know how to express cos 3 theta in terms of sin theta in order for me to solve, I think, sin 5 by substitution theta.Please help me also how to simplify the expression sin 5 theta.

thanks

My edit: Added "by substitution"

**TwistedOne151** · January 3rd 2009, 10:09 PM

Try reading about these: Chebyshev Polynomials.

--Kevin C.

**Soroban** · January 4th 2009, 06:47 AM

Hello, ImKo!

We need a bunch of identities and formulas:

. . $\sin^2\!A + \cos^2\!A \:=\:1\quad\Rightarrow\quad \cos A \:=\:\sqrt{1 - \sin^2\!A}$

. . $\sin2A \:=\:2\sin A\cos A$
. . $\cos2A \:=\:\cos^2\!A - \sin^2\!A \:=\:2\cos^2\!A - 1 \:=\:1 - 2\sin^2\!A$

. . $\sin(A \pm B) \:=\:\sin A\cos B \pm \sin B\cos A$
. . $\cos(A \pm B) \:=\:\cos A\cos B \mp \sin A\sin B$

How do I express $\cos3\theta$ in terms of $\sin\theta$ ?

It can be done . . . if you allow a square root in your result.

$\cos3\theta \;=\;\cos(2\theta + \theta)$

. . . . $= \qquad\underbrace{\cos2\theta}\cos\theta \quad\; - \quad\; \underbrace{\sin2\theta}\sin\theta$

. . . . $= \;\overbrace{\left(1 - 2\sin^2\!\theta\right)}\cos\theta - \overbrace{\left(2\sin\theta\cos\theta\right)}\sin \theta$

. . . . $= \;\cos\theta - 2\sin^2\!\theta\cos\theta - 2\sin^2\!\theta\cos\theta$

. . . . $= \;\cos\theta - 4\sin^2\!\theta\cos\theta$

. . . . $= \;\left(1-4\sin^2\!\theta\right)\cos\theta$

. . . . $= \;\left(1 - 4\sin^2\!\theta\right)\sqrt{1 - \sin^2\!\theta}$

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