Compound angles help

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#1
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.
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3 years ago
#2
Have u tried drawing the graphs of both of these functions on the same axis?
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3 years ago
#3
(Original post by 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.
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?
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#4
(Original post by 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?
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.
0
3 years ago
#5
(Original post by 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.
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.
1
#6
(Original post by 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.
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!
0
3 years ago
#7
(Original post by 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!
For solving , you can save a lot of time if you consider when two sine functions are equal instead of using identities to change the equation. This method confuses students in my experience but it's very useful.

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.
0
3 years ago
#8
(Original post by 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!
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
1
3 years ago
#9
(Original post by 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
Yes this is also a nice quick way to do it.
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