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QUESTION 4 Given that \( \cos 42^{\circ}=k \), determine the following in terms of \( k \). 4.1.1 \( \sin 42^{\circ} \) (3) 4.1.2 \( \cos 21^{\circ} \) (3) Simplify the following expression to a single trigonometric ratio. \[ \frac{\cos (-\theta) \cdot \sin \left(90^{\circ}+\theta\right)}{\cos \left(360^{\circ}-\theta\right) \cdot \sin ^{2}\left(540^{\circ}+\theta\right)} \] March Control Test Grade 12 \[ \begin{array}{l} \text { Grade } 12 \\ \text { Prove the identity : } \frac{\sin 2 x}{1+\cos 2 x}=\tan x \end{array} \] (3) [15] TOTAL: 50

Ask by Chen Campbell. in South Africa
Mar 17,2025

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**4.1.1** \[ \sin 42^\circ = \sqrt{1 - k^2} \] **4.1.2** \[ \cos 21^\circ = \sqrt{\frac{k + 1}{2}} \] **Simplified Expression** \[ \frac{\cos \theta}{\sin^2 \theta} = \cot \theta \csc \theta \] **Proven Identity** \[ \frac{\sin 2x}{1 + \cos 2x} = \tan x \]

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To find \( \sin 42^{\circ} \) in terms of \( k \), we can use the Pythagorean identity which states that \( \sin^2 \theta + \cos^2 \theta = 1 \). Thus, \( \sin 42^{\circ} = \sqrt{1 - k^2} \). It's important to remember that since \( 42^{\circ} \) is in the first quadrant, the sine value is positive! For \( \cos 21^{\circ} \), we can use the half-angle formula: \( \cos 21^{\circ} = \sqrt{\frac{1 + \cos 42^{\circ}}{2}} = \sqrt{\frac{1 + k}{2}} \). Now, you’ve got both \( \sin 42^{\circ} \) and \( \cos 21^{\circ} \) in terms of \( k \). In the given expression, we can simplify step-by-step. The cosine of a negative angle is the cosine of the angle itself, so \( \cos(-\theta) = \cos(\theta) \). For \( \sin(90^{\circ} + \theta) \), it equals \( \cos(\theta) \). The identities for \( \cos(360^{\circ} - \theta) \) and \( \sin^2(540^{\circ} + \theta) \) simplify to \( \cos(\theta) \) and \( \sin^2(\theta) \), respectively. So plugging these back gives you the final expression: \[ \frac{\cos(\theta) \cdot \cos(\theta)}{\cos(\theta) \cdot \sin^2(\theta)} = \frac{\cos^2(\theta)}{\cos(\theta) \cdot \sin^2(\theta)} = \frac{\cos(\theta)}{\sin^2(\theta)} = \cot(\theta) \cdot \csc(\theta). \] Lastly, for the identity \( \frac{\sin 2x}{1+\cos 2x} = \tan x \), recall that \( \sin 2x = 2 \sin x \cos x \) and \( 1 + \cos 2x = 1 + 2\cos^2 x - 1 = 2\cos^2 x \). By rewriting the left side, we have: \[ \frac{2 \sin x \cos x}{2 \cos^2 x} = \frac{\sin x}{\cos x} = \tan x. \] And voilà! You’ve proved the identity!

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