1) Simplify the following expressions: a) \( \sin ^{2} 108^{\circ}+\sin ^{2} 18^{\circ} \) (b) \( \frac{\sin \left(360^{\circ}-\theta\right) \cos \left(180^{\circ}+\theta\right)}{\sin \left(270^{\circ}\right) \cos \left(360^{\circ}-\theta\right) \sin (90+\theta)} \) 2) Prove the following trigonometric identities. You \( m \) a) \( \sin x \tan x+\cos x=\frac{1}{\cos x} \) b) \( \frac{\cos x}{1-\sin x}-\frac{\cos x}{1+\sin x}=2 \tan x \) 3) Determine the value of \( \sin \theta \cos \theta \) if \( \tan \theta=\frac{3}{4} \) and Compound Angles

Solve the equation by following steps: - step0: Solve for \(\theta\): \(\tan\left(\theta \right)=\frac{3}{4}\) - step1: Find the domain: \(\tan\left(\theta \right)=\frac{3}{4},\theta \neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\) - step2: Use the inverse trigonometric function: \(\theta =\arctan\left(\frac{3}{4}\right)\) - step3: Add the period: \(\theta =\arctan\left(\frac{3}{4}\right)+k\pi ,k \in \mathbb{Z}\) - step4: Check if the solution is in the defined range: \(\theta =\arctan\left(\frac{3}{4}\right)+k\pi ,k \in \mathbb{Z},\theta \neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\) - step5: Find the intersection: \(\theta =\arctan\left(\frac{3}{4}\right)+k\pi ,k \in \mathbb{Z}\) Calculate or simplify the expression \( \sin(108)^2 + \sin(18)^2 \). Calculate the value by following steps: - step0: Calculate: \(\sin\left(108^{2}\right)+\sin\left(18^{2}\right)\) - step1: Transform the expression: \(2\sin\left(5994\right)\cos\left(5670\right)\) - step2: Transform the expression: \(\sin\left(11664\right)+\sin\left(324\right)\) Determine whether the expression \( \sin(x) * \tan(x) + \cos(x) = 1/\cos(x) \) is always true. Verify the identity by following steps: - step0: Verify: \(\sin\left(x\right)\tan\left(x\right)+\cos\left(x\right)=\frac{1}{\cos\left(x\right)}\) - step1: Choose a side to work on: \(\frac{1}{\cos\left(x\right)}=\frac{1}{\cos\left(x\right)}\) - step2: Verify the identity: \(\textrm{true}\) Solve the equation \( \tan(\theta) = 3/4 \). Solve the equation by following steps: - step0: Solve for \(\theta\): \(\tan\left(\theta \right)=\frac{3}{4}\) - step1: Find the domain: \(\tan\left(\theta \right)=\frac{3}{4},\theta \neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\) - step2: Use the inverse trigonometric function: \(\theta =\arctan\left(\frac{3}{4}\right)\) - step3: Add the period: \(\theta =\arctan\left(\frac{3}{4}\right)+k\pi ,k \in \mathbb{Z}\) - step4: Check if the solution is in the defined range: \(\theta =\arctan\left(\frac{3}{4}\right)+k\pi ,k \in \mathbb{Z},\theta \neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\) - step5: Find the intersection: \(\theta =\arctan\left(\frac{3}{4}\right)+k\pi ,k \in \mathbb{Z}\) Determine whether the expression \( (\cos(x)/(1-\sin(x))) - (\cos(x)/(1+\sin(x))) = 2*\tan(x) \) is always true. Verify the identity by following steps: - step0: Verify: \(\left(\frac{\cos\left(x\right)}{\left(1-\sin\left(x\right)\right)}\right)-\left(\frac{\cos\left(x\right)}{\left(1+\sin\left(x\right)\right)}\right)=2\tan\left(x\right)\) - step1: Choose a side to work on: \(2\tan\left(x\right)=2\tan\left(x\right)\) - step2: Verify the identity: \(\textrm{true}\) Calculate or simplify the expression \( (\sin(360 - \theta) * \cos(180 + \theta)) / (\sin(270) * \cos(360 - \theta) * \sin(90 + \theta)) \). Simplify the expression by following steps: - step0: Solution: \(\frac{\left(\sin\left(360-\theta \right)\cos\left(180+\theta \right)\right)}{\left(\sin\left(270\right)\cos\left(360-\theta \right)\sin\left(90+\theta \right)\right)}\) - step1: Remove the parentheses: \(\frac{\sin\left(360-\theta \right)\cos\left(180+\theta \right)}{\sin\left(270\right)\cos\left(360-\theta \right)\sin\left(90+\theta \right)}\) - step2: Rewrite the expression: \(\frac{\sin\left(360-\theta \right)\cos\left(\theta +180\right)}{\sin\left(270\right)\cos\left(360-\theta \right)\sin\left(90+\theta \right)}\) - step3: Rewrite the expression: \(\frac{\sin\left(360-\theta \right)\cos\left(\theta +180\right)}{\sin\left(270\right)\cos\left(360-\theta \right)\sin\left(\theta +90\right)}\) - step4: Transform the expression: \(\frac{\cos\left(\theta +180\right)\tan\left(360-\theta \right)}{\sin\left(270\right)\sin\left(\theta +90\right)}\) - step5: Transform the expression: \(\cos\left(\theta +180\right)\tan\left(360-\theta \right)\csc\left(270\right)\csc\left(\theta +90\right)\) - step6: Rewrite the expression: \(\csc\left(270\right)\cos\left(\theta +180\right)\tan\left(360-\theta \right)\csc\left(\theta +90\right)\) Let's go through each part of your problem step by step. ### 1) Simplify the following expressions: #### a) \( \sin ^{2} 108^{\circ}+\sin ^{2} 18^{\circ} \) Using the identity \( \sin(90^\circ + x) = \cos(x) \), we can rewrite \( \sin(108^\circ) \) as: \[ \sin(108^\circ) = \sin(90^\circ + 18^\circ) = \cos(18^\circ) \] Thus, we have: \[ \sin^2(108^\circ) = \cos^2(18^\circ) \] Now, using the Pythagorean identity \( \sin^2(x) + \cos^2(x) = 1 \): \[ \sin^2(108^\circ) + \sin^2(18^\circ) = \cos^2(18^\circ) + \sin^2(18^\circ) = 1 \] #### b) \( \frac{\sin \left(360^{\circ}-\theta\right) \cos \left(180^{\circ}+\theta\right)}{\sin \left(270^{\circ}\right) \cos \left(360^{\circ}-\theta\right) \sin (90+\theta)} \) Using trigonometric identities: - \( \sin(360^\circ - \theta) = -\sin(\theta) \) - \( \cos(180^\circ + \theta) = -\cos(\theta) \) - \( \sin(270^\circ) = -1 \) - \( \cos(360^\circ - \theta) = \cos(\theta) \) - \( \sin(90^\circ + \theta) = \cos(\theta) \) Substituting these into the expression gives: \[ \frac{-\sin(\theta)(-\cos(\theta))}{-1 \cdot \cos(\theta) \cdot \cos(\theta)} = \frac{\sin(\theta) \cos(\theta)}{\cos^2(\theta)} = \tan(\theta) \] ### 2) Prove the following trigonometric identities: #### a) \( \sin x \tan x+\cos x=\frac{1}{\cos x} \) Starting with the left side: \[ \sin x \tan x + \cos x = \sin x \cdot \frac{\sin x}{\cos x} + \cos x = \frac{\sin^2 x + \cos^2 x}{\cos x} = \frac{1}{\cos x} \] This proves the identity. #### b) \( \frac{\cos x}{1-\sin x}-\frac{\cos x}{1+\sin x}=2 \tan x \) Finding a common denominator: \[ \frac{\cos x(1+\sin x) - \cos x(1-\sin x)}{(1-\sin x)(1+\sin x)} = \frac{\cos x(1+\sin x - 1 + \sin x)}{1 - \sin^2 x} = \frac{2\cos x \sin x}{\cos^2 x} = 2 \tan x \] This proves the identity. ### 3) Determine the value of \( \sin \theta \cos \theta \) if \( \tan \theta=\frac{3}{4} \) From \( \tan \theta = \frac{3}{4} \), we can use the identity: \[ \sin \theta = \frac{3}{\sqrt{3^2 + 4^2}} = \frac{3}{5}, \quad \cos \theta = \frac{4}{\sqrt{3^2 + 4^2}} = \frac{4}{5} \] Thus, \[ \sin \theta \cos \theta = \frac{3}{5} \cdot \frac{4}{5} = \frac{12}{25} \] ### Summary of Results: 1. a) \( 1 \) b) \( \tan(\theta) \) 2. a) Proven b) Proven 3. \( \sin \theta \cos \theta = \frac{12}{25} \)