PART 3: DEDUCTION AND APPLICATION: Do not use a calculator when answering Part 3. 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: \[ \begin{aligned} \sin 2 x & =\sin (x+x) \\ & =\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\ & =\ldots \ldots \ldots \ldots \end{aligned} \] \( \qquad \) \( \qquad \) 3.2 Expand cach of the following, using your answer from 3.1: 3.2 .1 sin \( 100^{\circ}= \), \( \qquad \) (I) \( 3.2 .2 \sin 46^{\circ}= \) \( \qquad \) (1) 3.2.3 \( \sin 40= \). \( \qquad \) (1) 3.3. Write each of the following expressions as a single trigonometric ratio: 3.3.1 \( 2 \sin 19^{\circ} \cdot \cos 19^{\circ}= \). \( \qquad \) (1) 3.3.2 \( 2 \cos 40^{\circ} \cdot \sin 40^{\circ}= \) \( \qquad \) (1) 3.3.3 \( \sin 25^{\circ} \cos 155^{\circ}= \) \( \qquad \) 3

\( \textbf{3.1} \) We start by writing: \[ \sin 2x = \sin (x+x) \] Using the compound angle formula: \[ \sin (x+x)=\sin x \cos x+\cos x \sin x \] Since the two terms are equal, we combine them: \[ \sin 2x=2\sin x \cos x \] \( \textbf{3.2} \) Using the result \( \sin 2x=2\sin x \cos x \): 1. For \( \sin 100^\circ \): Write \( 100^\circ=2\cdot 50^\circ \). Then: \[ \sin 100^\circ =2\sin 50^\circ \cos 50^\circ \] 2. For \( \sin 46^\circ \): Write \( 46^\circ=2\cdot 23^\circ \). Then: \[ \sin 46^\circ =2\sin 23^\circ \cos 23^\circ \] 3. For \( \sin 40^\circ \): Write \( 40^\circ=2\cdot 20^\circ \). Then: \[ \sin 40^\circ =2\sin 20^\circ \cos 20^\circ \] \( \textbf{3.3} \) 1. For \( 2\sin 19^\circ \cdot\cos 19^\circ \): Using the double angle formula with \( x=19^\circ \): \[ 2\sin 19^\circ \cos 19^\circ =\sin 38^\circ \] 2. For \( 2\cos 40^\circ \cdot\sin 40^\circ \): With \( x=40^\circ \): \[ 2\sin 40^\circ \cos 40^\circ =\sin 80^\circ \] 3. For \( \sin 25^\circ \cos 155^\circ \): First, note that \[ \cos 155^\circ = -\cos(180^\circ-155^\circ) = -\cos 25^\circ \] So: \[ \sin 25^\circ \cos 155^\circ = -\sin 25^\circ \cos 25^\circ \] Recognize that: \[ 2\sin 25^\circ \cos 25^\circ = \sin 50^\circ \] Hence: \[ -\sin 25^\circ \cos 25^\circ = -\frac{1}{2}\sin 50^\circ \] Thus, the final answers are: \[ \begin{aligned} \textbf{3.1:} \quad & \sin 2x=2\sin x\cos x \\ \\ \textbf{3.2.1:} \quad & \sin 100^\circ = 2\sin 50^\circ \cos 50^\circ \\ \textbf{3.2.2:} \quad & \sin 46^\circ = 2\sin 23^\circ \cos 23^\circ \\ \textbf{3.2.3:} \quad & \sin 40^\circ = 2\sin 20^\circ \cos 20^\circ \\ \\ \textbf{3.3.1:} \quad & 2\sin 19^\circ \cos 19^\circ = \sin 38^\circ \\ \textbf{3.3.2:} \quad & 2\cos 40^\circ \sin 40^\circ = \sin 80^\circ \\ \textbf{3.3.3:} \quad & \sin 25^\circ \cos 155^\circ = -\frac{1}{2}\sin 50^\circ \\ \end{aligned} \]