Given that θ \theta\,θ is small, use the small angle approximation of cosθ \cos \theta\,cosθ to show that
5cos(θ)−cos2(2θ)≈4+1.5θ2−4θ4 5 \cos(\theta) - \cos^2(2\theta) \approx 4 + 1.5\theta^2 - 4\theta^4 5cos(θ)−cos2(2θ)≈4+1.5θ2−4θ4Hence find an approximation of 5cos(θ)−cos2(2θ)5 \cos(\theta) - \cos^2(2\theta)5cos(θ)−cos2(2θ) when θ=2∘\theta = 2^\circθ=2∘
Calculate the percentage error in your approximation
