DoE/Feb. - March 2010 Mathematics/P2 $$ \begin{array}{c} 10 \ \mathrm{NSC} \end{array} $$ QUESTION 10 10.1 If $$ \sin 36^{\circ} \cos 12^{\circ}=p $$ and $$ \cos 36^{\circ} \sin 12^{\circ}=q $$, determine in terms of $$ p $$ and $$ q $$ the value of: 10.1.1 $$ \sin 48^{\circ} $$ $$ 10.1 .2 \sin 24^{\circ} $$ $$ 10.1 .3 \quad \cos 24^{\circ} $$ (3) (3) (3) 10.2 Show that $$ \sin ^{2} 20^{\circ}+\sin ^{2} 40^{\circ}+\sin ^{2} 80^{\circ}=\frac{3}{2} $$ (HINT: $$ 40^{\circ}=60^{\circ}-20^{\circ} $$ and $$ 80^{\circ}=60^{\circ}+20^{\circ} $$ ) (7) 10.3 10.3.1 Prove: $$ \frac{\sin ^{4} x+\sin ^{2} x \cos ^{2} x}{1+\cos x}=1-\cos x $$ 10.3.2 For which values of $$ x $$ is $$ \frac{\sin ^{4} x+\sin ^{2} x \cos ^{2} x}{1+\cos x}=1-\cos x $$ NOT true? (2) [22]

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$$
\textbf{10.1}
$$

We are given:
$$
p=\sin 36^\circ \cos 12^\circ,\qquad q=\cos 36^\circ \sin 12^\circ.
$$

1. Using the sine addition formula we have:
$$
\sin 36^\circ \cos 12^\circ+\cos 36^\circ \sin 12^\circ=\sin(36^\circ+12^\circ)=\sin 48^\circ.
$$
That is:
$$
\sin 48^\circ=p+q.
$$

2. Similarly, using the sine subtraction formula,
$$
\sin 36^\circ \cos 12^\circ-\cos 36^\circ \sin 12^\circ=\sin(36^\circ-12^\circ)=\sin 24^\circ,
$$
so that:
$$
\sin 24^\circ=p-q.
$$

3. Notice that the double-angle identity for sine gives:
$$
\sin48^\circ=2\sin24^\circ\cos24^\circ.
$$
Since we already found $$\sin48^\circ=p+q$$ and $$\sin24^\circ=p-q$$ (provided $$p\neq q$$), we have:
$$
p+q=2(p-q)\cos24^\circ.
$$
Therefore:
$$
\cos24^\circ=\frac{p+q}{2(p-q)}.
$$

$$
\textbf{10.2}
$$

We wish to show:
$$
\sin^2 20^\circ+\sin^2 40^\circ+\sin^2 80^\circ=\frac{3}{2}.
$$

Express each term using the identity:
$$
\sin^2\theta=\frac{1-\cos2\theta}{2}.
$$
Thus,
$$
\begin{aligned}
\sin^2 20^\circ+\sin^2 40^\circ+\sin^2 80^\circ &=\frac{1-\cos40^\circ}{2}+\frac{1-\cos80^\circ}{2}+\frac{1-\cos160^\circ}{2} \
&=\frac{3-(\cos40^\circ+\cos80^\circ+\cos160^\circ)}{2}.
\end{aligned}
$$

Now note that
$$
\cos160^\circ=\cos(180^\circ-20^\circ)=-\cos20^\circ.
$$
Thus,
$$
\cos40^\circ+\cos80^\circ+\cos160^\circ=\cos40^\circ+\cos80^\circ-\cos20^\circ.
$$
But, using the cosine sum-to-product formula,
$$
\cos40^\circ+\cos80^\circ=2\cos\frac{40^\circ+80^\circ}{2}\cos\frac{40^\circ-80^\circ}{2}=2\cos60^\circ\cos(-20^\circ)=2\left(\frac{1}{2}\right)\cos20^\circ=\cos20^\circ.
$$
Hence,
$$
\cos40^\circ+\cos80^\circ-\cos20^\circ= \cos20^\circ-\cos20^\circ=0.
$$
It follows that:
$$
\sin^2 20^\circ+\sin^2 40^\circ+\sin^2 80^\circ=\frac{3-0}{2}=\frac{3}{2}.
$$

$$
\textbf{10.3.1}
$$

We need to prove:
$$
\frac{\sin^4x+\sin^2x\cos^2x}{1+\cos x}=1-\cos x.
$$

Factor $$\sin^2x$$ from the numerator:
$$
\frac{\sin^4x+\sin^2x\cos^2x}{1+\cos x}=\frac{\sin^2x\left(\sin^2x+\cos^2x\right)}{1+\cos x}.
$$
Using the Pythagorean identity $$\sin^2x+\cos^2x=1$$, we obtain:
$$
\frac{\sin^2x}{1+\cos x}.
$$

Now, recall that:
$$
\sin^2 x=1-\cos^2x=(1-\cos x)(1+\cos x).
$$
Substitute this into the previous expression:
$$
\frac{(1-\cos x)(1+\cos x)}{1+\cos x}.
$$
For $$1+\cos x\ne 0$$ (so that the cancellation is allowed) we have:
$$
\frac{(1-\cos x)(\cancel{1+\cos x})}{\cancel{1+\cos x}}=1-\cos x.
$$

$$
\textbf{10.3.2}
$$

The cancellation is only valid when the denominator is not zero. Hence, we require:
$$
1+\cos x\neq 0\quad \Longrightarrow \quad \cos x\neq -1.
$$
But $$\cos x=-1$$ when:
$$
x=(2k+1)\pi,\quad k\in\mathbb{Z}.
$$
So the equality
$$
\frac{\sin^4x+\sin^2x\cos^2x}{1+\cos x}=1-\cos x
$$
is not true when $$x=(2k+1)\pi$$, $$k\in\mathbb{Z}$$.
**10.1** - $$ \sin 48^\circ = p + q $$ - $$ \sin 24^\circ = p - q $$ - $$ \cos 24^\circ = \frac{p + q}{2(p - q)} $$ **10.2** $$ \sin^2 20^\circ + \sin^2 40^\circ + \sin^2 80^\circ = \frac{3}{2} $$ **10.3.1** $$ \frac{\sin^4 x + \sin^2 x \cos^2 x}{1 + \cos x} = 1 - \cos x $$ **10.3.2** The equation is not true when $$ x = (2k+1)\pi $$ for any integer $$ k $$.

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