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

[bubbly16]

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

The first thing I did in this equation was break down sin^4(Ө) which turns into (sin^2Ө)^2. I then took the square root of sin making the new equation ((1-cos2x)/2)^2 next I multiplied it out to get (1-(2cos2x+(cos^2)2x)/4. I again had to take the square root of cos to get rid of the squares making it (1-2cos2x+(1+cos4x)/2) /4 and again had to multiply the top and bottom of the equation by 2 to get (2-4cos2x+1+cos4x)/8. Last but not least that will leave the final equation as (3-4cos2x+cos4x)/8 instead of sin^4Ө. After this point I am confused on what to do next???? Any help is greatly appreciated.

 

[stapel]

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

The first thing I did in this equation was break down sin^4(Ө) which turns into (sin^2Ө)^2.

What you've posted seems to give your first step as being the following:

. . . . .\(\displaystyle \sin^4(\theta)\, =\, \dfrac{3}{8}\, -\, \dfrac{3}{8}\cos(2\theta)\)

. . . . .\(\displaystyle \left(\sin^2(\theta)\right)^2\, =\, \dfrac{3}{8}\, -\, \dfrac{3}{8}\cos(2\theta)\)

But then you say:

I then took the square root of sin making the new equation ((1-cos2x)/2)^2

But this, having no "equals" in it, is not an "equation"; it is only an expression, and I am not even sure to which side of the original equation this expression is meant to correspond...?

Kindly please reply with clarification, showing both sides of each equation. Thank you!

 

[Mathematist]

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

Please check the question once again. Because for example, if you substitute Ө as 45 degree in both the sides of the equation, you will get 1/4 = 3/8. So this identity doesn't exist. I think it is sin^2 Ө instead of sin^4Ө. Anyway, check the question once again.