Question 7 Exercise 8.2

Solutions of Question 7 of Exercise 8.2 of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Rewrite in term of an expression containing only cosines to the power 1: $$\sin ^{2} \alpha \cos ^{2} \alpha$$

Solution.

$$\begin{align*} \sin ^{2} \alpha \cos ^{2} \alpha &= \left(\frac{1-\cos 2\alpha}{2} \right)\left(\frac{1+\cos 2\alpha}{2} \right)\\ &= \frac{1}{4}(1-\cos^2 2\alpha) \\ &=\frac{1}{4}\left(1-\frac{1+\cos 4\alpha}{2} \right) \\ &=\frac{1}{4}\left(\frac{2-1-\cos 4\alpha}{2} \right) \\ &=\frac{1-\cos 4\alpha}{8} \end{align*}$$

Rewrite in terms of an expression containing only cosine to the power $1$: $$\sin ^{4} \alpha \cos ^{2} \alpha$$

Solution.

$$\begin{align*} \sin^4 \alpha \cos^2 \alpha &= \left(\frac{1-\cos 2\alpha}{2}\right)^2 \left(\frac{1+\cos 2\alpha}{2} \right) \\ &= \frac{1}{4} \left( \frac{1-\cos 2\alpha}{2} \right)^2 (1+\cos 2\alpha) \\ &= \frac{1}{16} (1-\cos 2\alpha)^2 (1+\cos 2\alpha) \\ &= \frac{1}{16} \left(1 - 2\cos 2\alpha + \cos^2 2\alpha \right)(1 + \cos 2\alpha) \\ &= \frac{1}{16} \left(1 - 2\cos 2\alpha + \cos^2 2\alpha + \cos 2\alpha - 2\cos^3 2\alpha + \cos^2 2\alpha\right) \\ &= \frac{1}{16} \left(1 - \cos 2\alpha + 2\cos^2 2\alpha - 2\cos^3 2\alpha \right)\\ &= \frac{1}{16} \left(1 - \cos 2\alpha + 2\left(\frac{1+\cos 4\alpha}{2}\right) - 2\cos 2\alpha \left( \frac{1+\cos 4\alpha}{2} \right)\right)\\ &= \frac{1}{16} \left(1 - \cos 2\alpha + 1+\cos 4\alpha - \cos 2\alpha \left(1+\cos 4\alpha \right)\right)\\ &= \frac{1}{16} \left(1 - \cos 2\alpha + 1+\cos 4\alpha - \cos 2\alpha -\cos2 \alpha \cos 4\alpha\right)\\ &= \frac{1}{16} \left(2 - 2\cos 2\alpha +\cos 4\alpha -\cos2 \alpha \cos 4\alpha\right) \end{align*}$$

Rewrite in terms of an expression containing only cosine to the power $1$: $\sin ^{4} \alpha \cos ^{4} \alpha$

Solution.

$$\begin{align*} \sin^4 \alpha \cos^4 \alpha &= \left(\frac{1-\cos 2\alpha}{2} \right)^2 \left(\frac{1+\cos 2\alpha}{2} \right)^2 \\ &= \frac{1}{16} (1-\cos^2 2\alpha)^2 \\ &= \frac{1}{16} \left(1 - \frac{1+\cos 4\alpha}{2} \right)^2 \\ &= \frac{1}{16} \left(\frac{2 - 1 - \cos 4\alpha}{2} \right) \\ &= \frac{1}{16} \left(1 - \cos 4\alpha \right)^2 \\ &= \frac{1}{64}(1+cos^24 \alpha -2cos 4 \alpha)\\ &= \frac{1}{64}(1+(\frac{1+cos8 \alpha}{2}) -2cos 4 \alpha)\\ &= \frac{1}{128}(3+cos8 \alpha -4cos 4 \alpha) \end{align*}$$

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