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Determine whether the equation is an identity. cos(3 theta) = 4(cos(theta))^3 - 3 cos(theta)
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\[\cos(3 \theta) =\cos(2 \theta+\theta)=\cos(2 \theta)\cos(\theta)-\sin(2 \theta)\sin(\theta)\]
\[you know that \cos(2 \theta)=\cos^2(\theta)-\sin^2(\theta)\]
\[\cos(2 \theta + \theta)=(\cos^2(\theta)-\sin^2(\theta))\cos(\theta)-2\sin(\theta)\cos(\theta)\sin(\theta)\]
\[i used other trig identity: \sin(2 \theta)=2\sin(\theta)\cos(\theta)\]
further:\[\cos^3(\theta)-(1-\cos^2(\theta))\cos(\theta)-2(1-\cos^2(\theta))\cos(\theta)\]
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I used other trig identity:\[\sin^2(\theta)=1-\cos^2(\theta)\]
so finally opening all brakets we have:\[\cos^3(\theta)-\cos(\theta)+\cos^3(\theta)-2\cos(\theta)+2\cos^3(\theta)=4\cos^3(\theta)-3\cos(\theta)\]
good luck
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