QUESTION 5 5.1 Prove, without the use of a calculator, that, \[ \cos 75^{\circ}+\cos 15^{\circ}=\frac{\sqrt{6}}{2} \] 5.2 Determine the general solution of: \[ 1+4 \sin ^{2} x-5 \sin x+\cos 2 x=0 \] 5.3 Prove the identity \[ \frac{\sin 2 A}{1+\cos 2 A}=\tan A \] 5.4 Simplify \[ \frac{\sin \left(450^{\circ}-x\right) \tan \left(x-180^{\circ}\right) \sin 23^{\circ} \cos 23^{\circ}}{\cos 44^{\circ} \sin (-x)} \]

Verify the identity by following steps: - step0: Verify: \(\frac{\frac{\sqrt{6}}{2}}{1+\frac{\sqrt{6}}{2}}=\tan\left(A\right)\) - step1: Choose a side to work on: \(\frac{\sqrt{6}}{2+\sqrt{6}}=\tan\left(A\right)\) - step2: Verify the identity: \(\textrm{false}\) Solve the equation \( 1+4y^2-5y+\frac{1-\cos(2x)}{2}=0 \). Solve the equation by following steps: - step0: Solve for \(y\): \(1+4y^{2}-5y+\frac{1-\cos\left(2x\right)}{2}=0\) - step1: Multiply both sides of the equation by LCD: \(\left(1+4y^{2}-5y+\frac{1-\cos\left(2x\right)}{2}\right)\times 2=0\times 2\) - step2: Simplify the equation: \(3+8y^{2}-10y-\cos\left(2x\right)=0\) - step3: Simplify: \(3-\cos\left(2x\right)+8y^{2}-10y=0\) - step4: Rewrite in standard form: \(8y^{2}-10y+3-\cos\left(2x\right)=0\) - step5: Solve using the quadratic formula: \(y=\frac{10\pm \sqrt{\left(-10\right)^{2}-4\times 8\left(3-\cos\left(2x\right)\right)}}{2\times 8}\) - step6: Simplify the expression: \(y=\frac{10\pm \sqrt{\left(-10\right)^{2}-4\times 8\left(3-\cos\left(2x\right)\right)}}{16}\) - step7: Simplify the expression: \(y=\frac{10\pm \sqrt{4+32\cos\left(2x\right)}}{16}\) - step8: Simplify the expression: \(y=\frac{10\pm 2\sqrt{1+8\cos\left(2x\right)}}{16}\) - step9: Separate into possible cases: \(\begin{align}&y=\frac{10+2\sqrt{1+8\cos\left(2x\right)}}{16}\\&y=\frac{10-2\sqrt{1+8\cos\left(2x\right)}}{16}\end{align}\) - step10: Simplify the expression: \(\begin{align}&y=\frac{5+\sqrt{1+8\cos\left(2x\right)}}{8}\\&y=\frac{10-2\sqrt{1+8\cos\left(2x\right)}}{16}\end{align}\) - step11: Simplify the expression: \(\begin{align}&y=\frac{5+\sqrt{1+8\cos\left(2x\right)}}{8}\\&y=\frac{5-\sqrt{1+8\cos\left(2x\right)}}{8}\end{align}\) Determine whether the expression \( \frac{\sin(2A)}{1+\cos(2A)}=\tan A \) is always true. Verify the identity by following steps: - step0: Verify: \(\frac{\sin\left(2A\right)}{1+\cos\left(2A\right)}=\tan\left(A\right)\) - step1: Choose a side to work on: \(\tan\left(A\right)=\tan\left(A\right)\) - step2: Verify the identity: \(\textrm{true}\) Let's solve each part of the problem step by step. ### 5.1 Prove that \(\cos 75^{\circ}+\cos 15^{\circ}=\frac{\sqrt{6}}{2}\) To prove this, we can use the cosine addition formula: \[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \] Let \(A = 75^{\circ}\) and \(B = 15^{\circ}\): \[ \cos 75^{\circ} + \cos 15^{\circ} = 2 \cos\left(\frac{75^{\circ} + 15^{\circ}}{2}\right) \cos\left(\frac{75^{\circ} - 15^{\circ}}{2}\right) \] Calculating the angles: \[ \frac{75^{\circ} + 15^{\circ}}{2} = \frac{90^{\circ}}{2} = 45^{\circ} \] \[ \frac{75^{\circ} - 15^{\circ}}{2} = \frac{60^{\circ}}{2} = 30^{\circ} \] Thus, we have: \[ \cos 75^{\circ} + \cos 15^{\circ} = 2 \cos(45^{\circ}) \cos(30^{\circ}) \] Now substituting the known values: \[ \cos(45^{\circ}) = \frac{\sqrt{2}}{2}, \quad \cos(30^{\circ}) = \frac{\sqrt{3}}{2} \] So, \[ \cos 75^{\circ} + \cos 15^{\circ} = 2 \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{6}}{2} \] Thus, the statement is proven. ### 5.2 Determine the general solution of \(1+4 \sin^2 x - 5 \sin x + \cos 2x = 0\) Using the identity \(\cos 2x = 1 - 2\sin^2 x\), we can rewrite the equation: \[ 1 + 4 \sin^2 x - 5 \sin x + (1 - 2 \sin^2 x) = 0 \] This simplifies to: \[ 2 + 2 \sin^2 x - 5 \sin x = 0 \] Dividing the entire equation by 2: \[ 1 + \sin^2 x - \frac{5}{2} \sin x = 0 \] Let \(y = \sin x\): \[ y^2 - \frac{5}{2}y + 1 = 0 \] Now, we can solve this quadratic equation using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{\frac{5}{2} \pm \sqrt{\left(\frac{5}{2}\right)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \] Calculating the discriminant: \[ \left(\frac{5}{2}\right)^2 - 4 = \frac{25}{4} - 4 = \frac{25}{4} - \frac{16}{4} = \frac{9}{4} \] Thus, \[ y = \frac{\frac{5}{2} \pm \frac{3}{2}}{2} = \frac{8/2}{2} \text{ or } \frac{2/2}{2} = 2 \text{ or } \frac{1}{2} \] Since \(\sin x\) must be between -1 and 1, we discard \(y = 2\) and keep \(y = \frac{1}{2}\). The general solution for \(\sin x = \frac{1}{2}\) is: \[ x = n\pi + (-1)^n \frac{\pi}{6}, \quad n \in \mathbb{Z} \] ### 5.3 Prove the identity \(\frac{\sin 2A}{1+\cos 2A}=\tan A\) Using the double angle formulas: \[ \sin 2A = 2 \sin A \cos A, \quad \cos 2A = 1 - 2 \sin^2 A \] Substituting these into the left side: \[ \frac{2 \sin A \cos A}{1 + (1 - 2 \sin^2 A)} = \frac{2 \sin A \cos A}{2 - 2 \sin^2 A} = \frac{2 \sin A \cos A}{2(1 - \sin^2 A)} = \frac{\sin A \cos A}{1 - \sin^2 A} = \frac{\sin A \cos A}{\cos^2 A} = \tan A \] Thus, the identity is proven. ### 5.4 Simplify \(\frac{\sin(450^{\circ}-x) \tan(x-180^{\circ}) \sin 23^{\circ} \cos 23^{\circ}}{\cos 44^{\circ} \sin(-x)}\) First, simplify each component: - \(\sin(450^{\circ} - x) = \sin(90^{\circ} - x) = \cos x\) - \(\tan(x - 180^{\circ}) = \tan x\) - \(\sin(-x) = -\sin x\) Now substituting these into the expression: \[ \frac{\cos x \tan x \sin 23^{\circ} \cos 23^{\circ}}{\cos 44^{\circ} (-\sin x)} = -\frac{\cos x \cdot \frac{\sin x}{\cos x} \cdot \sin 23^{\circ} \cos 23^{\circ}}{\cos 44^{\circ}} = -\frac{\sin x \cdot \sin 23^{\circ} \cos 23^{\circ}}{\cos 44^{\circ}} \] Thus, the simplified expression is: \[ -\frac{\sin x \cdot \sin 23^{\circ} \cos 23^{\circ}}{\cos 44^{\circ}} \] This concludes the solutions for all parts of the problem.