5.I Given that \( \cos (A-B)=\cos A \cdot \cos B+\sin A \cdot \sin B \), use \( \cos (A-B) \) to prove that \( \cos (A+B)=\cos A \cdot \cos B-\sin A \cdot \sin B \) 5.2 Simplify \( : \frac{\cos \left(\theta-900^{\circ}\right)}{\sin (90+0) \tan \left(720^{\circ}-\theta\right)} \) 5.3 If \( \cos 48^{\circ}=t_{1} \) determine in terms of \( t \) \( 5.3 .1 \cos 228^{\circ} \) \( 5.3 .2 \sin \left(-42^{\circ}\right) \) \( 5.3 .3 \cos 24^{\circ} \)

Simplify the expression by following steps: - step0: Solution: \(\frac{\cos\left(\theta -900^{\circ}\right)}{\sin\left(90+0\right)\tan\left(720^{\circ}-\theta \right)}\) - step1: Remove 0: \(\frac{\cos\left(\theta -900^{\circ}\right)}{\sin\left(90\right)\tan\left(720^{\circ}-\theta \right)}\) - step2: Rewrite the expression: \(\frac{\cos\left(\theta -900^{\circ}\right)}{\sin\left(90\right)\left(-\tan\left(\theta \right)\right)}\) - step3: Rewrite the expression: \(\frac{-\cos\left(\theta \right)}{\sin\left(90\right)\left(-\tan\left(\theta \right)\right)}\) - step4: Rewrite the expression: \(\frac{-\cos\left(\theta \right)}{-\sin\left(90\right)\tan\left(\theta \right)}\) - step5: Rewrite the fraction: \(\frac{\cos\left(\theta \right)}{\sin\left(90\right)\tan\left(\theta \right)}\) - step6: Rewrite the expression: \(\sin^{-1}\left(90\right)\tan^{-1}\left(\theta \right)\cos\left(\theta \right)\) - step7: Simplify: \(\frac{1}{\sin\left(90\right)}\times \tan^{-1}\left(\theta \right)\cos\left(\theta \right)\) - step8: Calculate: \(\frac{1}{\sin\left(90\right)}\times \frac{\cos\left(\theta \right)}{\sin\left(\theta \right)}\times \cos\left(\theta \right)\) - step9: Calculate: \(\frac{\cos^{2}\left(\theta \right)}{\sin\left(90\right)\sin\left(\theta \right)}\) - step10: Rewrite the expression: \(\cos^{2}\left(\theta \right)\sin^{-1}\left(90\right)\sin^{-1}\left(\theta \right)\) - step11: Rewrite the expression: \(\frac{1}{\sin\left(90\right)}\times \sin^{-1}\left(\theta \right)\cos^{2}\left(\theta \right)\) - step12: Rewrite the expression: \(\frac{1}{\sin\left(90\right)}\times \cos^{2}\left(\theta \right)\sin^{-1}\left(\theta \right)\) - step13: Simplify: \(\frac{1}{\sin\left(90\right)}\times \cos^{2}\left(\theta \right)\csc\left(\theta \right)\) - step14: Rewrite the expression: \(\frac{1}{\sin\left(90\right)}\times \csc\left(\theta \right)\cos^{2}\left(\theta \right)\) - step15: Calculate: \(\frac{\csc\left(\theta \right)\cos^{2}\left(\theta \right)}{\sin\left(90\right)}\) Determine whether the expression \( \frac{\cos(A+B)}{\cos(A-B)}=\frac{\cos(A) \cos(B)-\sin(A) \sin(B)}{\cos(A) \cos(B)+\sin(A) \sin(B)} \) is always true. Verify the identity by following steps: - step0: Verify: \(\frac{\cos\left(A+B\right)}{\cos\left(A-B\right)}=\frac{\cos\left(A\right)\cos\left(B\right)-\sin\left(A\right)\sin\left(B\right)}{\cos\left(A\right)\cos\left(B\right)+\sin\left(A\right)\sin\left(B\right)}\) - step1: Choose a side to work on: \(\frac{\cos\left(A+B\right)}{\cos\left(B-A\right)}=\frac{\cos\left(A\right)\cos\left(B\right)-\sin\left(A\right)\sin\left(B\right)}{\cos\left(A\right)\cos\left(B\right)+\sin\left(A\right)\sin\left(B\right)}\) - step2: Choose the other side to work on: \(\frac{\cos\left(A+B\right)}{\cos\left(B-A\right)}=\frac{\cos\left(B+A\right)}{\cos\left(A-B\right)}\) - step3: Verify the identity: \(\textrm{false}\) Let's solve the problems step by step. ### 5.1 Proving \( \cos (A+B) = \cos A \cdot \cos B - \sin A \cdot \sin B \) We start with the known identity: \[ \cos (A-B) = \cos A \cdot \cos B + \sin A \cdot \sin B \] To derive \( \cos (A+B) \), we can use the identity for \( \cos (A-B) \) by substituting \( B \) with \( -B \): \[ \cos (A+B) = \cos (A - (-B)) = \cos A \cdot \cos (-B) + \sin A \cdot \sin (-B) \] Using the properties of cosine and sine: \[ \cos (-B) = \cos B \quad \text{and} \quad \sin (-B) = -\sin B \] Substituting these into the equation gives: \[ \cos (A+B) = \cos A \cdot \cos B - \sin A \cdot \sin B \] Thus, we have proven that: \[ \cos (A+B) = \cos A \cdot \cos B - \sin A \cdot \sin B \] ### 5.2 Simplifying \( \frac{\cos \left(\theta-900^{\circ}\right)}{\sin (90+0) \tan \left(720^{\circ}-\theta\right)} \) First, we simplify the expression: - \( \sin(90 + 0) = \sin(90) = 1 \) - \( \tan(720^{\circ} - \theta) = \tan(-\theta) = -\tan(\theta) \) Thus, the expression simplifies to: \[ \frac{\cos(\theta - 900^{\circ})}{1 \cdot (-\tan(\theta))} = -\frac{\cos(\theta - 900^{\circ})}{\tan(\theta)} \] Next, we can simplify \( \cos(\theta - 900^{\circ}) \): \[ \cos(\theta - 900^{\circ}) = \cos(\theta - 2 \cdot 360^{\circ} - 180^{\circ}) = \cos(\theta - 180^{\circ}) = -\cos(\theta) \] Putting it all together: \[ -\frac{-\cos(\theta)}{\tan(\theta)} = \frac{\cos(\theta)}{\tan(\theta)} = \frac{\cos^2(\theta)}{\sin(\theta)} \] So the final simplified expression is: \[ \frac{\cos^2(\theta)}{\sin(\theta)} \] ### 5.3 Evaluating Trigonometric Functions in Terms of \( t_1 \) Given \( \cos 48^{\circ} = t_1 \): #### 5.3.1 Finding \( \cos 228^{\circ} \) Using the cosine identity: \[ \cos(228^{\circ}) = \cos(180^{\circ} + 48^{\circ}) = -\cos(48^{\circ}) = -t_1 \] #### 5.3.2 Finding \( \sin(-42^{\circ}) \) Using the sine identity: \[ \sin(-42^{\circ}) = -\sin(42^{\circ}) = -\sin(90^{\circ} - 48^{\circ}) = -\cos(48^{\circ}) = -t_1 \] #### 5.3.3 Finding \( \cos 24^{\circ} \) Using the cosine identity: \[ \cos(24^{\circ}) = \sin(90^{\circ} - 24^{\circ}) = \sin(66^{\circ}) = \sin(90^{\circ} - 48^{\circ}) = \cos(48^{\circ}) = t_1 \] ### Summary of Results - \( \cos(228^{\circ}) = -t_1 \) - \( \sin(-42^{\circ}) = -t_1 \) - \( \cos(24^{\circ}) = t_1 \)