3.1 Apply the compound angle expansion $$ \sin (x+y)=\sin x \cos y+\cos x \sin y $$ to $$ \sin 2 x $$ and simplify your answer: $$ \begin{array}{l} \sin 2 x=\sin (x+x) \ = \ = \end{array} $$ 3.2 Expand each of the following, using your answer from 3.1: 3.2.1 $$ \sin 100= $$ $$ \qquad $$ 3.2.2 $$ \sin 46= $$. $$ \qquad $$ $$ 3.2 .3 \sin 40= $$ $$ \qquad $$ (1) 3.3. Write each of the following expressions as a single trigonometric ratio: 3.3.1 $$ 2 \sin 19 \cdot \cos 19^{\circ}= $$. $$ \qquad $$ 3.3.2 $$ 2 \cos 40^{\circ} \cdot \sin 40^{\circ}= $$ $$ \qquad $$ $$ 3.3 .3 \sin 25^{\circ} \cos 155^{\circ}= $$ $$ \qquad $$

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3.1  
We start with the compound angle formula:  
$$
\sin (x+y)=\sin x \cos y+\cos x \sin y.
$$  
To write $$\sin 2x$$ in this form, set $$y=x$$:
$$
\sin 2x=\sin (x+x)=\sin x \cos x+\cos x \sin x.
$$  
Since $$\sin x \cos x$$ appears twice, we combine like terms:
$$
\sin 2x=2\sin x\cos x.
$$

3.2  
We now use the result from 3.1 ($$\sin 2x=2 \sin x \cos x$$) to expand the given expressions.

3.2.1 For $$\sin 100^\circ$$:  
Write $$100^\circ$$ as $$2\times50^\circ$$, so
$$
\sin 100^\circ = \sin (2\cdot50^\circ)=2\sin 50^\circ\cos 50^\circ.
$$

3.2.2 For $$\sin 46^\circ$$:  
Write $$46^\circ$$ as $$2\times23^\circ$$, so
$$
\sin 46^\circ = \sin (2\cdot23^\circ)=2\sin 23^\circ\cos 23^\circ.
$$

3.2.3 For $$\sin 40^\circ$$:  
Write $$40^\circ$$ as $$2\times20^\circ$$, so
$$
\sin 40^\circ = \sin (2\cdot20^\circ)=2\sin 20^\circ\cos 20^\circ.
$$

3.3  
We now express each given product as a single trigonometric ratio.

3.3.1 For $$2 \sin 19^\circ \cdot \cos 19^\circ$$:  
Using the double-angle identity from 3.1:
$$
2\sin 19^\circ\cos 19^\circ=\sin (2\cdot19^\circ)=\sin 38^\circ.
$$

3.3.2 For $$2 \cos 40^\circ \cdot \sin 40^\circ$$:  
The expression is similar to 3.3.1 so:
$$
2\cos 40^\circ\sin 40^\circ=\sin (2\cdot40^\circ)=\sin 80^\circ.
$$

3.3.3 For $$\sin 25^\circ \cos 155^\circ$$:  
Notice that the cosine of an angle can be rewritten using the supplementary angle:
$$
\cos 155^\circ = \cos (180^\circ-25^\circ) = -\cos 25^\circ.
$$  
Thus:
$$
\sin 25^\circ \cos 155^\circ = \sin 25^\circ(-\cos 25^\circ)=-\sin 25^\circ\cos 25^\circ.
$$  
Now use the double-angle formula (but with a factor of $$ \frac{1}{2} $$) by writing:
$$
\sin 25^\circ\cos 25^\circ=\frac{1}{2}\sin (2\cdot25^\circ)=\frac{1}{2}\sin 50^\circ.
$$
Therefore:
$$
\sin 25^\circ \cos 155^\circ = -\frac{1}{2}\sin 50^\circ.
$$
3.1 $$ \sin 2x = 2\sin x \cos x $$ 3.2 3.2.1 $$ \sin 100^\circ = 2\sin 50^\circ \cos 50^\circ $$ 3.2.2 $$ \sin 46^\circ = 2\sin 23^\circ \cos 23^\circ $$ 3.2.3 $$ \sin 40^\circ = 2\sin 20^\circ \cos 20^\circ $$ 3.3 3.3.1 $$ 2\sin 19^\circ \cos 19^\circ = \sin 38^\circ $$ 3.3.2 $$ 2\cos 40^\circ \sin 40^\circ = \sin 80^\circ $$ 3.3.3 $$ \sin 25^\circ \cos 155^\circ = -\frac{1}{2}\sin 50^\circ $$

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