Sin^4Ө =3/8-3/8cos(2Ө) Prove the following trigonometric identity

  • #1
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Homework Statement


Prove the following trigonometric identity. The question is sin^4Ө =3/8-3/8cos(2Ө)

Homework Equations


I think I'm supposed to use the power reducing formulas for trigonometric identities which are
sin^2(u)= (1- cos(2u))/2
cos^2(u)=(1+cos(2u))/2
*Let u represent any integer/value*

The Attempt at a Solution


To expand the equation I separated the equation into (1- cos^2Ө)(1- cos^2Ө) = 1- 2cos^2Ө + (cos^2Ө)^2. I reduced it because sin^2Ө = 1 – cos^2Ө and after this I'm confused on what the next steps are.
 

Answers and Replies

  • #2
BvU
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So, a much better post than the first try. Good !

Your other entry is from the righthand side: what have you got for ##\cos(2\theta)## that might be useful here ?

[edit] ah: your second relevant equation !
 
  • #3
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So you applied the sin^2(u) to the lefthand side of the equation and got (1/4) * (1 - cos(2u) )^2 right?

Next expand the

(1/4)*(1 - cos(2u) )^2 = (1/4) * ( 1 - 2*cos(2u) - cos(2u)^2 )

next apply the cos^2(u) rule to the last term and see what you get.
 
  • #4
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As well as the power reducing formulas, you'll want to glance at the double angle formulas.
 
  • #5
BvU
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Homework Statement


Prove the following trigonometric identity. The question is sin^4Ө =3/8-3/8cos(2Ө)

Homework Equations


I think I'm supposed to use the power reducing formulas for trigonometric identities which are
sin^2(u)= (1- cos(2u))/2
cos^2(u)=(1+cos(2u))/2
*Let u represent any integer/value*

The Attempt at a Solution


To expand the equation I separated the equation into (1- cos^2Ө)(1- cos^2Ө) = 1- 2cos^2Ө + (cos^2Ө)^2. I reduced it because sin^2Ө = 1 – cos^2Ө and after this I'm confused on what the next steps are.
But now I see a problem coming up:
let ##\theta = \pi/2\ \ ## then ##(sin\theta)^4 = 1\ \ ## and ##\ \ 3/8-3/8\cos(2\theta)=3/4## !?
So no identity at all !
Or did I read the original thingy in the wrong way ?
 
  • #6
haruspex
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But now I see a problem coming up:
let ##\theta = \pi/2\ \ ## then ##(sin\theta)^4 = 1\ \ ## and ##\ \ 3/8-3/8\cos(2\theta)=3/4## !?
So no identity at all !
Or did I read the original thingy in the wrong way ?
I believe the identity ought to read ##\sin^4(\theta)=\frac 38-\frac 12\cos(2\theta)+\frac 18\cos(4\theta)##.
Looks like someone turned that ##4\theta## into another ##2\theta##.
 
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  • #7
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Why would u have to be an integer?
 

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