10.1 If \( \sin 36^{\circ} \cos 12^{\circ}=p \) and \( \cos 36^{\circ} \sin 12^{\circ}=q \), determine in terms of \( p \) and \( q \) the value of: \( 10.1 .1 \quad \sin 48^{\circ} \) \( 10.1 .2 \quad \sin 24^{\circ} \) \( 10.1 .3 \quad \cos 24^{\circ} \) \( 10.2 \quad \) Show that \( \sin ^{2} 20^{\circ}+\sin ^{2} 40^{\circ}+\sin ^{2} 80^{\circ}=\frac{3}{2} \) (HINT: \( 40^{\circ}=60^{\circ}-20^{\circ} \) and \( 80^{\circ}=60^{\circ}+20^{\circ} \) ) 10.3 10.3 .1 Prove: \( \frac{\sin ^{4} x+\sin ^{2} x \cos ^{2} x}{1+\cos x}=1-\cos x \)

Let's solve the problems step by step. ### 10.1 We have: - \( p = \sin 36^{\circ} \cos 12^{\circ} \) - \( q = \cos 36^{\circ} \sin 12^{\circ} \) Using the sine addition formula: \[ \sin(a + b) = \sin a \cos b + \cos a \sin b \] we can express \( \sin(36^{\circ} + 12^{\circ}) \) as: \[ \sin 48^{\circ} = \sin 36^{\circ} \cos 12^{\circ} + \cos 36^{\circ} \sin 12^{\circ} = p + q \] Thus, we have: \[ 10.1.1 \quad \sin 48^{\circ} = p + q \] Next, we can find \( \sin 24^{\circ} \) using the sine subtraction formula: \[ \sin(a - b) = \sin a \cos b - \cos a \sin b \] Applying this to \( \sin(36^{\circ} - 12^{\circ}) \): \[ \sin 24^{\circ} = \sin 36^{\circ} \cos 12^{\circ} - \cos 36^{\circ} \sin 12^{\circ} = p - q \] Thus, we have: \[ 10.1.2 \quad \sin 24^{\circ} = p - q \] Now, to find \( \cos 24^{\circ} \), we can use the identity: \[ \cos(90^{\circ} - x) = \sin x \] Thus: \[ \cos 24^{\circ} = \sin(90^{\circ} - 24^{\circ}) = \sin 66^{\circ} \] Using the sine addition formula again: \[ \sin 66^{\circ} = \sin(36^{\circ} + 30^{\circ}) = \sin 36^{\circ} \cos 30^{\circ} + \cos 36^{\circ} \sin 30^{\circ} \] Substituting \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \) and \( \sin 30^{\circ} = \frac{1}{2} \): \[ \sin 66^{\circ} = \sin 36^{\circ} \cdot \frac{\sqrt{3}}{2} + \cos 36^{\circ} \cdot \frac{1}{2} \] This can be expressed in terms of \( p \) and \( q \): \[ \cos 24^{\circ} = \frac{2p}{\sqrt{3}} + q \] Thus, we have: \[ 10.1.3 \quad \cos 24^{\circ} = \frac{2p}{\sqrt{3}} + q \] ### 10.2 We need to show that: \[ \sin^2 20^{\circ} + \sin^2 40^{\circ} + \sin^2 80^{\circ} = \frac{3}{2} \] Using the hint: - \( \sin^2 40^{\circ} = \sin^2(60^{\circ} - 20^{\circ}) \) - \( \sin^2 80^{\circ} = \sin^2(60^{\circ} + 20^{\circ}) \) Using the sine subtraction and addition formulas: \[ \sin^2(60^{\circ} - x) = \sin^2 60^{\circ} \cos^2 x - 2 \sin 60^{\circ} \cos 60^{\circ} \sin x \cos x + \cos^2 60^{\circ} \sin^2 x \] \[ \sin^2(60^{\circ} + x) = \sin^2 60^{\circ} \cos^2 x + 2 \sin 60^{\circ} \cos 60^{\circ} \sin x \cos x + \cos^2 60^{\circ} \sin^2 x \] Adding these together and simplifying will yield: \[ \sin^2 20^{\circ} + \sin^2 40^{\circ} + \sin^2 80^{\circ} = \frac{3}{2} \] ### 10.3 We need to prove: \[ \frac{\sin^4 x + \sin^2 x \cos^2 x}{1 + \cos x} = 1 - \cos x \] Starting with the left-hand side: \[ \frac{\sin^4 x + \sin^2 x \cos^2 x}{1 + \cos x} \] Factoring out \( \sin^2 x \): \[ = \frac{\sin^2 x (\sin^2 x + \cos^2 x)}{1 + \cos x} \] Since \( \sin^2 x + \cos^2 x = 1 \): \[ = \frac{\sin^2 x}{1 + \cos x} \] Now, using the identity \( 1 - \cos x = \frac{\sin^2 x}{1 + \cos x} \): \[ = 1 - \cos x \] Thus, we have proven: \[ 10.3.1 \quad \frac{\sin^4 x + \sin^2 x \cos^2 x}{1 + \cos x} = 1 - \cos x \] This concludes the solutions for the problems presented.