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 $$ : \frac{\cos \left( heta-900^{\circ} ight)}{\sin (90+0) an \left(720^{\circ}- heta ight)} $$ 5.3 If $$ \cos 48^{\circ}=t_{1} $$ determine in terms of $$ t $$ $$ 5.3 .1 \cos 228^{\circ} $$ $$ 5.3 .2 \sin \left(-42^{\circ} ight) $$ $$ 5.3 .3 \cos 24^{\circ} $$

Question

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 $$ : \frac{\cos \left(	heta-900^{\circ}ight)}{\sin (90+0) 	an \left(720^{\circ}-	hetaight)} $$ 5.3 If $$ \cos 48^{\circ}=t_{1} $$ determine in terms of $$ t $$ $$ 5.3 .1 \cos 228^{\circ} $$ $$ 5.3 .2 \sin \left(-42^{\circ}ight) $$ $$ 5.3 .3 \cos 24^{\circ} $$

UpStudy AI · Accepted Answer

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
- $$ \cos(228^{\circ}) = -t_1 $$
- $$ \sin(-42^{\circ}) = -t_1 $$
- $$ \cos(24^{\circ}) = t_1 $$
- $$ \cos(228^{\circ}) = -t_1 $$ - $$ \sin(-42^{\circ}) = -t_1 $$ - $$ \cos(24^{\circ}) = t_1 $$

5.I Given that ,
use to prove that
5.2 Simplify
5.3 If determine in terms of

Upstudy AI Solution

Answer

Solution

5.1 Proving

5.2 Simplifying

5.3 Evaluating Trigonometric Functions in Terms of

5.3.1 Finding

5.3.2 Finding

5.3.3 Finding

Summary of Results

Beyond the Answer

Related Questions

Latest Trigonometry Questions

5.I Given that , use to prove that 5.2 Simplify 5.3 If determine in terms of