Solving 8sin³θ − 6sinθ + 1 = 0 with double angle identities

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Homework Statement


8 sin3 θ – 6 sin θ + 1 = 0

The answer includes changing this to
-2sin3θ+1=0

Homework Equations


The double angle identities
Sin2θ=sinθcosθ+cosθsinθ

The Attempt at a Solution



I do not know how to get started with this question
 
lioric said:

Homework Statement


8 sin3 θ – 6 sin θ + 1 = 0

The answer includes changing this to
-2sin3θ+1=0

Homework Equations


The double angle identities
Sin2θ=sinθcosθ+cosθsinθ

The Attempt at a Solution



I do not know how to get started with this question

You could try looking at ##\sin(3\theta) = \sin(2\theta + \theta)##.
 
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PeroK said:
You could try looking at ##\sin(3\theta) = \sin(2\theta + \theta)##.
I am familiar with that
Which is the double angle identity
But I fail to see the the relation
I believe that I am missing the transition identity which brings them together
Could you point it out for me please
 
lioric said:
I am familiar with that
Which is the double angle identity
But I fail to see the the relation
I believe that I am missing the transition identity which brings them together
Could you point it out for me please

Expand the expression in post #2. And then have another think.
 
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PeroK said:
Expand the expression in post #2. And then have another think.
This what I came to when I expanded
What do I do from there
Handwritten derivation in blue ink on a wooden desk of sin 3θ = sin 2θ cos θ + cos 2θ sin θ, expanded step by step to 3 sin θ − 4 sin³ θ, with some terms crossed out.
 
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  • Handwritten derivation in blue ink on a wooden desk of sin 3θ = sin 2θ cos θ + cos 2θ sin θ, expanded step by step to 3 sin θ − 4 sin³ θ, with some terms crossed out.
    IMG_0400.JPG
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Ray Vickson said:
So, what do you get for ##2 \sin 3\theta ?##
Oh
You mean twice that huh ok let me see
 
lioric said:
Oh
You mean twice that huh ok let me see
Thank you got it now
Thank you very much
Handwritten derivation on a wooden desk proving 2 sin 3θ = 6 sin θ − 8 sin³θ, expanding with double angle identities and substituting cos²θ = 1 − sin²θ, with the final line boxed.
 
Attachments
  • Handwritten derivation on a wooden desk proving 2 sin 3θ = 6 sin θ − 8 sin³θ, expanding with double angle identities and substituting cos²θ = 1 − sin²θ, with the final line boxed.
    IMG_0401.JPG
    55.4 KB · Views: 475
lioric said:

Homework Statement


8 sin3 θ – 6 sin θ + 1 = 0

The answer includes changing this to
-2sin3θ+1=0

Homework Equations


The double angle identities
Sin2θ=sinθcosθ+cosθsinθ

The Attempt at a Solution



I do not know how to get started with this question
It can't be an identity, because when I substitute 90 degrees for theta it comes out to - 8 - 6 + 1 = 0. What is it you are supposed to do? Solve for theta?
 
PeroK said:
You could try looking at ##\sin(3\theta) = \sin(2\theta + \theta)##.

lioric said:
I am familiar with that
Which is the double angle identity
No, PeroK is setting up to use an addition formula, not a double angle identity.

You seem to be confused about the concepts of proving an identity versus solving a conditional equation. This problem, which you mistakenly titled "I need help with this identity," actually entails solving an equation.

The difference between these two kinds of equations is that an identity is true for all values of the variable; for example, ##\sin^2(\theta) + \cos^2(\theta) = 1##. A conditional equation is one that is true for a limited number of values of the variable; for example ##x^2 - 2x + 1 = 0##.