11 If $$ \cos 58^{\circ}=m $$, determine the value of the following in terms of $$ m $$, without the use o $$ 11.1 \sin 572^{\circ} $$ $$ 11.2 an 58^{\circ} $$ $$ 11.3 \cos 64^{\circ} $$ $$ 11.4 \sin 29^{\circ} \cos 29^{\circ} $$ $$ 11.5 \sin 2^{\circ} $$

Question

11 If $$ \cos 58^{\circ}=m $$, determine the value of the following in terms of $$ m $$, without the use o $$ 11.1 \sin 572^{\circ} $$ $$ 11.2 	an 58^{\circ} $$ $$ 11.3 \cos 64^{\circ} $$ $$ 11.4 \sin 29^{\circ} \cos 29^{\circ} $$ $$ 11.5 \sin 2^{\circ} $$

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**11.1**  
We first reduce the angle:  
$$
572^\circ-360^\circ=212^\circ.
$$  
Since  
$$
212^\circ=180^\circ+32^\circ,
$$  
we have  
$$
\sin 212^\circ=-\sin 32^\circ.
$$  
Recall that $$ \sin (90^\circ - 58^\circ)=\cos 58^\circ $$, and since  
$$
90^\circ-58^\circ=32^\circ,
$$  
it follows that  
$$
\sin 32^\circ=\cos 58^\circ = m.
$$  
Thus,  
$$
\sin 572^\circ = -m.
$$

**11.2**  
By definition,  
$$
\tan 58^\circ=\frac{\sin 58^\circ}{\cos 58^\circ}.
$$  
Since we are given $$ \cos 58^\circ=m $$ and using the Pythagorean identity we have  
$$
\sin 58^\circ=\sqrt{1-\cos^2 58^\circ}=\sqrt{1-m^2},
$$  
(with the positive square root because $$58^\circ$$ is in the first quadrant), it follows that  
$$
\tan 58^\circ=\frac{\sqrt{1-m^2}}{m}.
$$

**11.3**  
Notice that  
$$
\cos64^\circ=\cos(2\cdot32^\circ).
$$  
Using the double‐angle formula,  
$$
\cos(2\theta)=2\cos^2\theta-1,
$$  
with $$\theta=32^\circ$$, we need to express $$\cos32^\circ$$ in terms of $$m$$. Using the co-function identity:  
$$
\cos32^\circ=\sin(90^\circ-32^\circ)=\sin58^\circ,
$$  
and as before,  
$$
\sin58^\circ=\sqrt{1-m^2}.
$$  
Thus,  
$$
\cos64^\circ=2\left(\sqrt{1-m^2}\right)^2-1=2(1-m^2)-1=1-2m^2.
$$

**11.4**  
Write the product as a double-angle expression:  
$$
\sin29^\circ\cos29^\circ=\frac{1}{2}\sin(2\cdot29^\circ)=\frac{1}{2}\sin58^\circ.
$$  
Since  
$$
\sin 58^\circ=\sqrt{1-m^2},
$$  
we have  
$$
\sin29^\circ\cos29^\circ=\frac{1}{2}\sqrt{1-m^2}.
$$

**11.5**  
Notice that  
$$
2^\circ=60^\circ-58^\circ.
$$  
Using the sine difference formula:  
$$
\sin (A-B)=\sin A\cos B-\cos A\sin B,
$$  
with $$A=60^\circ$$ and $$B=58^\circ$$, we get  
$$
\sin2^\circ = \sin60^\circ\cos58^\circ-\cos60^\circ\sin58^\circ.
$$  
Recall that  
$$
\sin60^\circ=\frac{\sqrt{3}}{2},\quad \cos60^\circ=\frac{1}{2},\quad \cos58^\circ=m,\quad \sin58^\circ=\sqrt{1-m^2}.
$$  
Thus,  
$$
\sin2^\circ = \frac{\sqrt{3}}{2}\,m-\frac{1}{2}\sqrt{1-m^2}=\frac{\sqrt{3}\,m-\sqrt{1-m^2}}{2}.
$$
11.1 $$ \sin 572^\circ = -m $$ 11.2 $$ \tan 58^\circ = \frac{\sqrt{1 - m^2}}{m} $$ 11.3 $$ \cos 64^\circ = 1 - 2m^2 $$ 11.4 $$ \sin 29^\circ \cos 29^\circ = \frac{\sqrt{1 - m^2}}{2} $$ 11.5 $$ \sin 2^\circ = \frac{\sqrt{3}m - \sqrt{1 - m^2}}{2} $$

11 If , determine the value of the following in terms of , without the use o

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