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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 \( 2 \cos 40^{\circ} \cdot \sin 40^{\circ}= \) \( \qquad \) \( 3.3 .3 \sin 25^{\circ} \cos 155^{\circ}= \) \( \qquad \)

Ask by Wood Ayala. in South Africa
Mar 12,2025

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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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\[ \begin{array}{l} \sin 2x = \sin(x+x) \\ = \sin x \cos x + \cos x \sin x \\ = 2 \sin x \cos x \end{array} \] 3.2.1 Using the identity for angles, \( \sin 100^\circ = \sin(90^\circ + 10^\circ) = \cos 10^\circ \). 3.2.2 Following the same method, \( \sin 46^\circ = \sin(45^\circ + 1^\circ) = \sin 45^\circ \cos 1^\circ + \cos 45^\circ \sin 1^\circ = \frac{\sqrt{2}}{2} (\cos 1^\circ + \sin 1^\circ) \). 3.2.3 Applying the formula, \( \sin 40^\circ = \sin(30^\circ + 10^\circ) = \sin 30^\circ \cos 10^\circ + \cos 30^\circ \sin 10^\circ = \frac{1}{2} \cos 10^\circ + \frac{\sqrt{3}}{2} \sin 10^\circ \). 3.3.1 Using the double angle identity, \( 2 \sin 19^\circ \cos 19^\circ = \sin 38^\circ \). 3.3.2 Likewise, \( 2 \cos 40^\circ \sin 40^\circ = \sin 80^\circ \). 3.3.3 Since \( \cos 155^\circ = -\sin 25^\circ \), we get \( \sin 25^\circ \cos 155^\circ = \sin 25^\circ (-\sin 25^\circ) = -\sin^2 25^\circ \).

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