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Show that sin3θ = 3sinθ - 4sin^3θ , and cos3θ = 4cos^3θ - 3cosθ.

 Jul 22, 2021

Best Answer 

 #1
avatar+436 
+2

Note: make sure to know sin/cos angle addition/subtraction identities, sin/cos double angle identities, the Pythagorean identity, and the proofs for each one of them.

Using the sine addition rule,

\(\sin(3\theta)=\sin(\theta+2\theta)=\sin(\theta)\cos(2\theta)+\sin(2\theta)\cos(\theta)\)

And then apply the double angle identities:

\(\sin(\theta)(1-2\sin^2\theta)+2\sin(\theta )\cos(\theta)\cos(\theta)\\ =\sin(\theta)(1-2\sin^2\theta+2\cos^2(\theta))\)

Note that by the Pythagorean identity,

\(\sin^2(\theta)+\cos^2(\theta)=1\\2\sin^2(\theta)+2\cos^2(\theta)=2\\-2\sin^2(\theta)+2\cos^2(\theta)=2-4\sin^2(\theta)\)

Substituting the value above, we get:

\(\sin(\theta)(1+2-4\sin^2(\theta))\\=3\sin(\theta)-4\sin^3(\theta) \blacksquare\)

A very similar proof can be used to prove the latter.

 Jul 22, 2021
 #1
avatar+436 
+2
Best Answer

Note: make sure to know sin/cos angle addition/subtraction identities, sin/cos double angle identities, the Pythagorean identity, and the proofs for each one of them.

Using the sine addition rule,

\(\sin(3\theta)=\sin(\theta+2\theta)=\sin(\theta)\cos(2\theta)+\sin(2\theta)\cos(\theta)\)

And then apply the double angle identities:

\(\sin(\theta)(1-2\sin^2\theta)+2\sin(\theta )\cos(\theta)\cos(\theta)\\ =\sin(\theta)(1-2\sin^2\theta+2\cos^2(\theta))\)

Note that by the Pythagorean identity,

\(\sin^2(\theta)+\cos^2(\theta)=1\\2\sin^2(\theta)+2\cos^2(\theta)=2\\-2\sin^2(\theta)+2\cos^2(\theta)=2-4\sin^2(\theta)\)

Substituting the value above, we get:

\(\sin(\theta)(1+2-4\sin^2(\theta))\\=3\sin(\theta)-4\sin^3(\theta) \blacksquare\)

A very similar proof can be used to prove the latter.

textot Jul 22, 2021

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