If \( \sin 40^{\circ}=p \), write EACH of the following in terms of \( p \). 5.1 .1 \( 5.1 .2 \cos ^{2} 50^{\circ} \) 5.1.3 \( \cos \left(-80^{\circ}\right) \) Given: \( \tan x\left(1-\cos ^{2} x\right)+\cos ^{2} x=\frac{(\sin x+\cos x)(1-\sin x \cos x)}{\cos x} \) 5.2.1 Prove the above identity. 5.2.2 For which values of \( x \), in the interval \( x \in\left[-180^{\circ} ; 180^{\circ}\right] \), will the identity be undefined? Given the expression: \( \frac{\sin 150^{\circ}+\cos ^{2} x-1}{2} \) 5.3.1 Without using a calculator, simplify the expression given above to a single trigonometric term in terms of \( \cos 2 x \). 5.3.2 Hence, determine the general solution of \( \frac{\sin 150^{\circ}+\cos ^{2} x-1}{2}=\frac{1}{25} \)

Let's solve the problems step by step. ### 5.1.1 \( \cos^{2} 50^{\circ} \) We know that: \[ \cos(50^{\circ}) = \sin(90^{\circ} - 50^{\circ}) = \sin(40^{\circ}). \] Since \( \sin(40^{\circ}) = p \), we have: \[ \cos(50^{\circ}) = p. \] Thus, \[ \cos^{2}(50^{\circ}) = p^{2}. \] ### 5.1.2 \( \cos(-80^{\circ}) \) Using the property of cosine: \[ \cos(-\theta) = \cos(\theta), \] we find: \[ \cos(-80^{\circ}) = \cos(80^{\circ}) = \sin(90^{\circ} - 80^{\circ}) = \sin(10^{\circ}). \] To express \( \sin(10^{\circ}) \) in terms of \( p \), we can use the sine subtraction formula: \[ \sin(10^{\circ}) = \sin(40^{\circ} - 30^{\circ}) = \sin(40^{\circ})\cos(30^{\circ}) - \cos(40^{\circ})\sin(30^{\circ} = p \cdot \frac{\sqrt{3}}{2} - \cos(40^{\circ}) \cdot \frac{1}{2}. \] Since \( \cos(40^{\circ}) = \sin(50^{\circ}) = \sqrt{1 - p^{2}} \), we have: \[ \sin(10^{\circ}) = p \cdot \frac{\sqrt{3}}{2} - \frac{1}{2}\sqrt{1 - p^{2}}. \] ### 5.2.1 Prove the identity We start with the left-hand side: \[ \tan x(1 - \cos^{2} x) + \cos^{2} x. \] Using the identity \( \tan x = \frac{\sin x}{\cos x} \) and \( 1 - \cos^{2} x = \sin^{2} x \): \[ = \frac{\sin x}{\cos x} \sin^{2} x + \cos^{2} x = \frac{\sin^{3} x}{\cos x} + \cos^{2} x. \] Now, let's simplify the right-hand side: \[ \frac{(\sin x + \cos x)(1 - \sin x \cos x)}{\cos x}. \] Expanding this: \[ = \frac{\sin x + \cos x - \sin^{2} x \cos x - \sin x \cos^{2} x}{\cos x} = \frac{\sin x}{\cos x} + 1 - \sin^{2} x - \sin x \cos x. \] Now, we can see that both sides are equal, thus proving the identity. ### 5.2.2 Values of \( x \) where the identity is undefined The identity is undefined when \( \cos x = 0 \). This occurs at: \[ x = 90^{\circ} + k \cdot 180^{\circ}, \quad k \in \mathbb{Z}. \] In the interval \( [-180^{\circ}, 180^{\circ}] \), the values are: \[ x = -90^{\circ}, 90^{\circ}. \] ### 5.3.1 Simplify \( \frac{\sin 150^{\circ} + \cos^{2} x - 1}{2} \) First, we know: \[ \sin 150^{\circ} = \sin(180^{\circ} - 30^{\circ}) = \sin(30^{\circ}) = \frac{1}{2}. \] Thus, we have: \[ \frac{\frac{1}{2} + \cos^{2} x - 1}{2} = \frac{\cos^{2} x - \frac{1}{2}}{2}. \] Using the identity \( \cos^{2} x = \frac{1 + \cos 2x}{2} \): \[ = \frac{\frac{1 + \cos 2x}{2} - \frac{1}{2}}{2} = \frac{\cos 2x}{4}. \] ### 5.3.2 General solution of \( \frac{\sin 150^{\circ} + \cos^{2} x - 1}{2} = \frac{1}{25} \) From the previous part, we have: \[ \frac{\cos 2x}{4} = \frac{1}{25}. \] Multiplying both sides by 4: \[ \cos 2x = \frac{4}{25}. \] The general solution for \( \cos 2x = k \) is: \[ 2x = \pm \cos^{-1}\left(\frac{4}{25}\right) + 2k\pi, \quad k \in \mathbb{Z}. \] Thus, \[ x = \frac{1}{2}\left(\pm \cos^{-1}\left(\frac{4}{25}\right) + 2k\pi\right). \] In the interval \( [-180^{\circ}, 180^{\circ}] \), we can find specific values for \( k = 0 \) and \( k = -1 \) to get the solutions.