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Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Double angle formulae trig equation

jessyjellytot14

Solve sin4θ = cos2θ

What do you do with the sin4θ? I tried turning it into 2sin2θcos2θ using the double angle formulae but I don't know where to go from there...

Because say if I then did:

2(2sinθcosθ)(1-2sin²θ) = 1-2sin²θ

or...

2(2sinθcosθ)(2cos²θ-1) = wtf shall I use

I don't even know what to do- am I supposed to make everything so it's all in terms of sinθ or cosθ?? Nothing is seeming to work...

(edited 7 years ago)

Absolute first thing you do: set

\alpha = 2\theta

, so your equation becomes

\sin 2\alpha = \cos \alpha

. Solve for alpha, and then divide by 2 to find theta.

jessyjellytot14

OP

Original post by DFranklin

Absolute first thing you do: set

\alpha = 2\theta

, so your equation becomes

\sin 2\alpha = \cos \alpha

. Solve for alpha, and then divide by 2 to find theta.

Where would you get α from?

Study Forum Helper

Original post by jessyjellytot14

Where would you get α from?

It's just a substitution for the sake of simplicity, you can choose

\lambda = 2\theta

if you want, doesn't matter.

However, I do not see why you are expressing

\cos(2\theta)

in a different form. Simply get everything on one side at

2\sin(2\theta)\cos(2\theta)

stage, and just factor out a

\cos(2\theta)

. You can then solve for theta.

(edited 7 years ago)

jessyjellytot14

OP

Original post by RDKGames

Simply get everything on one side at

2\sin(2\theta)\cos(2\theta)

stage, and just factor out a

\cos(2\theta)

. You can then solve for theta.

Oooooh, that is the logical thing to do! How did I not think of that? :facepalm:

:facepalm:

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