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Parents Pricing Home A-Level Edexcel Maths Pure Differential Equations Given the function:
$$f(\theta) = 4\cos^2 \theta - 3\sin^2 \theta$$
(a) Show that $f(\theta) = \frac{1}{2} - \frac{7}{2}\cos 2\theta.$
(b) Hence, using calculus, find the exact value of \( \int_{0}^{\frac{\pi}{2}} \theta f(\theta) d\theta \$.
Given the function:
$$f(\theta) = 4\cos^2 \theta - 3\sin^2 \theta$$
(a) Show that $f(\theta) = \frac{1}{2} - \frac{7}{2}\cos 2\theta.$
(b) Hence, using calculus, find the exact value of \( \int_{0}^{\frac{\pi}{2}} \theta f(\theta) d\theta \$. - Edexcel - A-Level Maths Pure - Question 8 - 2010 - Paper 6 Question 8
View full question Given the function:
$$f(\theta) = 4\cos^2 \theta - 3\sin^2 \theta$$
(a) Show that $f(\theta) = \frac{1}{2} - \frac{7}{2}\cos 2\theta.$
(b) Hence, using calculus, ... show full transcript
View marking scheme Worked Solution & Example Answer:Given the function:
$$f(\theta) = 4\cos^2 \theta - 3\sin^2 \theta$$
(a) Show that $f(\theta) = \frac{1}{2} - \frac{7}{2}\cos 2\theta.$
(b) Hence, using calculus, find the exact value of \( \int_{0}^{\frac{\pi}{2}} \theta f(\theta) d\theta \$. - Edexcel - A-Level Maths Pure - Question 8 - 2010 - Paper 6
Show that $f(\theta) = \frac{1}{2} - \frac{7}{2}\cos 2\theta$. Only available for registered users.
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To show that the equation holds, we can start by expressing cos 2 θ \cos^2 \theta cos 2 θ sin 2 θ \sin^2 \theta sin 2 θ cos 2 θ \cos 2\theta cos 2 θ
Use the identites:
cos 2 θ = 1 + cos 2 θ 2 \cos^2 \theta = \frac{1 + \cos 2\theta}{2} cos 2 θ = 2 1 + c o s 2 θ sin 2 θ = 1 − cos 2 θ 2 \sin^2 \theta = \frac{1 - \cos 2\theta}{2} sin 2 θ = 2 1 − c o s 2 θ
Substitute these identities into the function:
f ( θ ) = 4 ( 1 + cos 2 θ 2 ) − 3 ( 1 − cos 2 θ 2 ) f(\theta) = 4\left(\frac{1 + \cos 2\theta}{2}\right) - 3\left(\frac{1 - \cos 2\theta}{2}\right) f ( θ ) = 4 ( 2 1 + cos 2 θ ) − 3 ( 2 1 − cos 2 θ )
Simplifying this expression:
f ( θ ) = 2 ( 1 + cos 2 θ ) − 3 2 ( 1 − cos 2 θ ) = 2 + 2 cos 2 θ − 3 2 + 3 2 cos 2 θ = 4 2 − 3 2 + ( 2 + 3 2 ) cos 2 θ = 1 2 + 7 2 cos 2 θ f(\theta) = 2(1 + \cos 2\theta) - \frac{3}{2}(1 - \cos 2\theta) \
= 2 + 2\cos 2\theta - \frac{3}{2} + \frac{3}{2}\cos 2\theta \
= \frac{4}{2} - \frac{3}{2} + \left(2 + \frac{3}{2}\right)\cos 2\theta \
= \frac{1}{2} + \frac{7}{2}\cos 2\theta f ( θ ) = 2 ( 1 + cos 2 θ ) − 2 3 ( 1 − cos 2 θ ) = 2 + 2 cos 2 θ − 2 3 + 2 3 cos 2 θ = 2 4 − 2 3 + ( 2 + 2 3 ) cos 2 θ = 2 1 + 2 7 cos 2 θ
Thus, we can rewrite it as:
f ( θ ) = 1 2 − 7 2 cos 2 θ . f(\theta) = \frac{1}{2} - \frac{7}{2}\cos 2\theta. f ( θ ) = 2 1 − 2 7 cos 2 θ .
Hence, using calculus, find the exact value of $\int_{0}^{\frac{\pi}{2}} \theta f(\theta) d\theta$. Only available for registered users.
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We first express the integral:
∫ 0 π 2 θ f ( θ ) d θ = ∫ 0 π 2 θ ( 1 2 − 7 2 cos 2 θ ) d θ \int_{0}^{\frac{\pi}{2}} \theta f(\theta) d\theta = \int_{0}^{\frac{\pi}{2}} \theta \left(\frac{1}{2} - \frac{7}{2}\cos 2\theta\right) d\theta ∫ 0 2 π θ f ( θ ) d θ = ∫ 0 2 π θ ( 2 1 − 2 7 cos 2 θ ) d θ
Splitting the integral:
= 1 2 ∫ 0 π 2 θ d θ − 7 2 ∫ 0 π 2 θ cos 2 θ d θ = \frac{1}{2}\int_{0}^{\frac{\pi}{2}} \theta d\theta - \frac{7}{2}\int_{0}^{\frac{\pi}{2}} \theta \cos 2\theta d\theta = 2 1 ∫ 0 2 π θ d θ − 2 7 ∫ 0 2 π θ cos 2 θ d θ
Calculate the first integral:
∫ 0 π 2 θ d θ = [ θ 2 2 ] 0 π 2 = ( π 2 ) 2 2 = π 2 8 \int_{0}^{\frac{\pi}{2}} \theta d\theta = \left[\frac{\theta^2}{2}\right]_{0}^{\frac{\pi}{2}} = \frac{\left(\frac{\pi}{2}\right)^2}{2} = \frac{\pi^2}{8} ∫ 0 2 π θ d θ = [ 2 θ 2 ] 0 2 π = 2 ( 2 π ) 2 = 8 π 2
For the second integral, we use integration by parts:
Let:
u = θ u = \theta u = θ d v = cos 2 θ d θ dv = \cos 2\theta d\theta d v = cos 2 θ d θ d u = d θ du = d\theta d u = d θ v = 1 2 sin 2 θ v = \frac{1}{2}\sin 2\theta v = 2 1 sin 2 θ
Applying integration by parts:
∫ θ cos 2 θ d θ = u v − ∫ v d u = θ ⋅ 1 2 sin 2 θ − ∫ 1 2 sin 2 θ d θ \int \theta \cos 2\theta d\theta = u v - \int v du = \theta \cdot \frac{1}{2}\sin 2\theta - \int \frac{1}{2}\sin 2\theta d\theta ∫ θ cos 2 θ d θ = uv − ∫ v d u = θ ⋅ 2 1 sin 2 θ − ∫ 2 1 sin 2 θ d θ The second integral becomes:
Substituting back in:
Now evaluate the bound:
= [ π 2 ⋅ 1 2 ⋅ 0 − 0 ] + 1 4 = 0 + 1 4 = \left[\frac{\pi}{2} \cdot \frac{1}{2} \cdot 0 - 0\right] + \frac{1}{4} = 0 + \frac{1}{4} = [ 2 π ⋅ 2 1 ⋅ 0 − 0 ] + 4 1 = 0 + 4 1
Thus, putting everything together:
∫ 0 π 2 θ f ( θ ) d θ = 1 2 ⋅ π 2 8 − 7 2 ⋅ ( 1 4 + 0 ) = π 2 16 − 7 8 \int_{0}^{\frac{\pi}{2}} \theta f(\theta) d\theta = \frac{1}{2}\cdot \frac{\pi^2}{8} - \frac{7}{2} \cdot \left(\frac{1}{4} + 0\right)= \frac{\pi^2}{16} - \frac{7}{8} ∫ 0 2 π θ f ( θ ) d θ = 2 1 ⋅ 8 π 2 − 2 7 ⋅ ( 4 1 + 0 ) = 16 π 2 − 8 7
The final answer is:
π 2 16 − 7 8 \frac{\pi^2}{16} - \frac{7}{8} 16 π 2 − 8 7
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