====== 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. 


=====Question 7(i)=====
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*}
GOOD




=====Question 7(ii)=====
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*}

=====Question 7(iii)=====
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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