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- Joined
- 28 Oct 2006
- Posts
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Show that Sin3θ = 3sinθ - 4sin^3θ
Sin3θ = Sin(2θ + θ)
= Sin2θcosθ + cos2θsinθ
=2sinθcosθcosθ + (1 - 2sin^2θ)sinθ
=2sinθcos^2θ + sinθ - 2sin^3θ
Am i going wrong? If I'm not, what next?
Sin3θ = Sin(2θ + θ)
= Sin2θcosθ + cos2θsinθ
=2sinθcosθcosθ + (1 - 2sin^2θ)sinθ
=2sinθcos^2θ + sinθ - 2sin^3θ
Am i going wrong? If I'm not, what next?
