# Compound angles help

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I can't figure this compound angles question out for the life of me.

Question:

Determine for what range of values of θ between 0 and 2π, Sinθ > Sin3θ

A walkthrough of how to do it would be much appreciated. Thanks in advance.

Question:

Determine for what range of values of θ between 0 and 2π, Sinθ > Sin3θ

A walkthrough of how to do it would be much appreciated. Thanks in advance.

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#3

(Original post by

I can't figure this compound angles question out for the life of me.

Question:

Determine for what range of values of θ between 0 and 2π, Sinθ > Sin3θ.

**TheGibbsy**)I can't figure this compound angles question out for the life of me.

Question:

Determine for what range of values of θ between 0 and 2π, Sinθ > Sin3θ.

You should now be able to see where these intervals are, from where the first curve is higher up than the second curve.

Now the tricky part! You need to find the points of intersection of the two curves, that is, you need to solve sin θ = sin 3θ. Any ideas?

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(Original post by

Start by sketching y = sin θ and y = sin 3θ on the same set of axes. The former should be very easy, and the latter only requires you to know the effect of transformations of thie kind y = f(x) to y = f(3x).

You should now be able to see where these intervals are, from where the first curve is higher up than the second curve.

Now the tricky part! You need to find the points of intersection of the two curves, that is, you need to solve sin θ = sin 3θ. Any ideas?

**Pangol**)Start by sketching y = sin θ and y = sin 3θ on the same set of axes. The former should be very easy, and the latter only requires you to know the effect of transformations of thie kind y = f(x) to y = f(3x).

You should now be able to see where these intervals are, from where the first curve is higher up than the second curve.

Now the tricky part! You need to find the points of intersection of the two curves, that is, you need to solve sin θ = sin 3θ. Any ideas?

Sin(2θ+θ) = Sinθ and from there using the rule,

Sin2θcosθ + Cos2θsinθ = sinθ

However, I've hit a wall again as to where to go from there.

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#5

(Original post by

Thanks for the pointer. Using Sin (A+B) i've changed the equation to:

Sin(2θ+θ) = Sinθ and from there using the rule,

Sin2θcosθ + Cos2θsinθ = sinθ

However, I've hit a wall again as to where to go from there.

**TheGibbsy**)Thanks for the pointer. Using Sin (A+B) i've changed the equation to:

Sin(2θ+θ) = Sinθ and from there using the rule,

Sin2θcosθ + Cos2θsinθ = sinθ

However, I've hit a wall again as to where to go from there.

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(Original post by

You need to express the LHS purely in terms of sin θ. Start by using the identites for sin 2θ and cos 2θ, and see if you can then get rid of any remaining cosines that you have.

**Pangol**)You need to express the LHS purely in terms of sin θ. Start by using the identites for sin 2θ and cos 2θ, and see if you can then get rid of any remaining cosines that you have.

Substituting the identities in:

(1-2sin^2θ)sinθ + 2sinθcosθcosθ = sinθ - Expand and clean it up to get

Sinθ(2sin^2θ-1) = 0 and solve for Sinθ and 2sin^2θ resulting in:

0, π/4, 3π/4, π, 5π/4, 7π/4 and 2π

Thank you! You're a life saver!

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#7

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Thank you! I never even considered the Identities.

Substituting the identities in:

(1-2sin^2θ)sinθ + 2sinθcosθcosθ = sinθ - Expand and clean it up to get

Sinθ(2sin^2θ-1) = 0 and solve for Sinθ and 2sin^2θ resulting in:

0, π/4, 3π/4, π, 5π/4, 7π/4 and 2π

Thank you! You're a life saver!

**TheGibbsy**)Thank you! I never even considered the Identities.

Substituting the identities in:

(1-2sin^2θ)sinθ + 2sinθcosθcosθ = sinθ - Expand and clean it up to get

Sinθ(2sin^2θ-1) = 0 and solve for Sinθ and 2sin^2θ resulting in:

0, π/4, 3π/4, π, 5π/4, 7π/4 and 2π

Thank you! You're a life saver!

So clearly one solution is and there are other solutions if you consider the graph / CAST etc.

Let us know if you want to have a go at this method and if you need help with it.

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#8

**TheGibbsy**)

Thank you! I never even considered the Identities.

Substituting the identities in:

(1-2sin^2θ)sinθ + 2sinθcosθcosθ = sinθ - Expand and clean it up to get

Sinθ(2sin^2θ-1) = 0 and solve for Sinθ and 2sin^2θ resulting in:

0, π/4, 3π/4, π, 5π/4, 7π/4 and 2π

Thank you! You're a life saver!

sin 3θ - sin θ = 0 then use the formula to turn this into a product

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#9

(Original post by

Following on from Notnek's suggestion you could look at:

sin 3θ - sin θ = 0 then use the formula to turn this into a product

**Muttley79**)Following on from Notnek's suggestion you could look at:

sin 3θ - sin θ = 0 then use the formula to turn this into a product

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