Statement (a) Prove the triple-angle formula cos3θ≡4cos3θ−3cosθ\cos 3\theta\equiv 4\cos^3\theta-3\cos\thetacos3θ≡4cos3θ−3cosθ. (b) Use it to find the exact value of cos3θ\cos 3\thetacos3θ given that cosθ=23\cos\theta=\dfrac{2}{3}cosθ=32. Solution (a) cos3θ=cos(2θ+θ)=cos2θcosθ−sin2θsinθ\cos 3\theta=\cos(2\theta+\theta)=\cos2\theta\cos\theta-\sin2\theta\sin\thetacos3θ=cos(2θ+θ)=cos2θcosθ−sin2θsinθ. With cos2θ=2cos2θ−1\cos2\theta=2\cos^2\theta-1cos2θ=2cos2θ−1 and sin2θ=2sinθcosθ\sin2\theta=2\sin\theta\cos\thetasin2θ=2sinθcosθ: cos3θ=(2cos2θ−1)cosθ−2sin2θcosθ=(2cos2θ−1)cosθ−2(1−cos2θ)cosθ=4cos3θ−3cosθ.\cos3\theta=(2\cos^2\theta-1)\cos\theta-2\sin^2\theta\cos\theta=(2\cos^2\theta-1)\cos\theta-2(1-\cos^2\theta)\cos\theta=4\cos^3\theta-3\cos\theta.cos3θ=(2cos2θ−1)cosθ−2sin2θcosθ=(2cos2θ−1)cosθ−2(1−cos2θ)cosθ=4cos3θ−3cosθ. (b) cos3θ=4(23)3−3⋅23=3227−2=−2227≈−0.8148\cos 3\theta=4\big(\tfrac23\big)^3-3\cdot\tfrac23=\dfrac{32}{27}-2=-\dfrac{22}{27}\approx -0{.}8148cos3θ=4(32)3−3⋅32=2732−2=−2722≈−0.8148. cos3θ=−2227≈−0.8148\boxed{\cos 3\theta=-\tfrac{22}{27}\approx -0{.}8148}cos3θ=−2722≈−0.8148