If $$ 6 \sin heta=5 \operatorname{sind} \cos heta

Question

If $$ 6 \sin 	heta=5 \operatorname{sind} \cos 	heta<0 $$, find, without using a calculator, the value of $$ 31 \quad 1-2 \cos ^{2} 	heta $$ $$ 32 \quad \cos 	heta 	an 	heta $$. 4 If $$ 4 \cos 	heta=3 $$ and $$ 	an \beta=5,90^{\circ}<	heta<360^{\circ} $$ and $$ \beta>90^{\circ} $$, prove without using a calculator that $$ \frac{7}{	an ^{2} 	heta}+74 \sin ^{2} \beta=34 $$. 5 If $$ \cos A=3 $$ and $$ A+B=90^{\circ} $$, use a sketch to find the value of $$ \frac{	an A \sin A}{\cos B} $$.

UpStudy AI · Accepted Answer

Let's break down the problem step by step.

### Part 1: Solve $$ 6 \sin \theta = 5 \operatorname{sind} \cos \theta < 0 $$

1. **Understanding the equation**: The equation $$ 6 \sin \theta = 5 \operatorname{sind} \cos \theta $$ can be interpreted as $$ 6 \sin \theta = 5 \sin \theta \cos \theta $$. The term $$ \operatorname{sind} $$ is often used to denote $$ \sin \theta $$ in degrees, but here we will treat it as $$ \sin \theta $$.

2. **Rearranging the equation**:
   $$
   6 \sin \theta - 5 \sin \theta \cos \theta = 0
   $$
   Factoring out $$ \sin \theta $$:
   $$
   \sin \theta (6 - 5 \cos \theta) = 0
   $$
   This gives us two cases:
   - $$ \sin \theta = 0 $$
   - $$ 6 - 5 \cos \theta = 0 $$

3. **Solving for $$ \theta $$**:
   - If $$ \sin \theta = 0 $$, then $$ \theta = 0^\circ, 180^\circ, 360^\circ $$.
   - If $$ 6 - 5 \cos \theta = 0 $$, then $$ \cos \theta = \frac{6}{5} $$ which is not possible since $$ \cos \theta $$ must be between -1 and 1.

4. **Considering the condition $$ \cos \theta < 0 $$**: This means $$ \theta $$ must be in the second or third quadrant. Therefore, the only valid solution is $$ \theta = 180^\circ $$.

### Part 2: Calculate $$ 31 \quad 1 - 2 \cos^2 \theta $$

1. **Finding $$ \cos \theta $$**:
   Since $$ \theta = 180^\circ $$, we have:
   $$
   \cos 180^\circ = -1
   $$

2. **Calculating $$ 1 - 2 \cos^2 \theta $$**:
   $$
   1 - 2 \cos^2 180^\circ = 1 - 2(-1)^2 = 1 - 2(1) = 1 - 2 = -1
   $$

### Part 3: Calculate $$ 32 \quad \cos \theta \tan \theta $$

1. **Finding $$ \tan \theta $$**:
   Since $$ \theta = 180^\circ $$:
   $$
   \tan 180^\circ = 0
   $$

2. **Calculating $$ \cos \theta \tan \theta $$**:
   $$
   \cos 180^\circ \tan 180^\circ = (-1)(0) = 0
   $$

### Part 4: Given $$ 4 \cos \theta = 3 $$ and $$ \tan \beta = 5 $$

1. **Finding $$ \cos \theta $$**:
   $$
   \cos \theta = \frac{3}{4}
   $$

2. **Finding $$ \sin \theta $$** using $$ \sin^2 \theta + \cos^2 \theta = 1 $$:
   $$
   \sin^2 \theta = 1 - \left(\frac{3}{4}\right)^2 = 1 - \frac{9}{16} = \frac{7}{16}
   $$
   Thus, $$ \sin \theta = \frac{\sqrt{7}}{4} $$ (since $$ \theta $$ is in the second quadrant).

3. **Finding $$ \sin \beta $$**:
   Since $$ \tan \beta = 5 $$, we can use the identity:
   $$
   \tan \beta = \frac{\sin \beta}{\cos \beta} \implies \sin \beta = 5 \cos \beta
   $$
   Using $$ \sin^2 \beta + \cos^2 \beta = 1 $$:
   $$
   (5 \cos \beta)^2 + \cos^2 \beta = 1 \implies 25 \cos^2 \beta + \cos^2 \beta = 1 \implies 26 \cos^2 \beta = 1 \implies \cos^2 \beta = \frac{1}{26}
   $$
   Thus, $$ \sin^2 \beta = 1 - \frac{1}{26} = \frac{25}{26} $$.

4. **Proving $$ \frac{7}{\tan^2 \theta} + 74 \sin^2 \beta = 34 $$**:
   $$
   \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{\frac{7}{16}}{\left(\frac{3}{4}\right)^2} = \frac{\frac{7}{16}}{\frac{9}{16}} = \frac{7}{9}
   $$
   Thus, $$ \frac{7}{\tan^2 \theta} = \frac{7}{\frac{7}{9}} = 9 $$.

Now, calculate $$ 74 \sin^2 \beta $$:
   $$
   74 \sin^2 \beta = 74 \cdot \frac{25}{26} = \frac{1850}{26} = 71.1538
   $$

Adding these:
   $$
   9 + 71.1538 \approx 80.1538 \neq 34
   $$
   Therefore, the equation does not hold.

### Part 5: Given $$ \cos A = 3 $$ and $$ A + B = 90^\circ $$

1. **Understanding the relationship**: Since $$ A + B = 90^\circ $$, we have $$ B = 90^\circ - A $$.

2. **Finding $$ \cos B $$**:
   $$
   \cos B = \sin A
   $$

3. **Finding $$ \tan A $$**:
   Since $$ \cos A = 3 $$ is not possible (as $$ \cos A $$ must be between -1 and 1), we cannot proceed with this part.

### Summary of Results:
- $$ 31: -1 $$
- $$ 32: 0 $$
- Part 4 does not hold true.
- Part 5 is invalid due to the impossibility of $$ \cos A = 3 $$.
- For $$ 6 \sin \theta = 5 \sin \theta \cos \theta $$, the solution is $$ \theta = 180^\circ $$. - Calculating $$ 1 - 2 \cos^2 \theta $$ at $$ \theta = 180^\circ $$ gives $$ -1 $$. - Calculating $$ \cos \theta \tan \theta $$ at $$ \theta = 180^\circ $$ gives $$ 0 $$. - Given $$ 4 \cos \theta = 3 $$ and $$ \tan \beta = 5 $$, the equation $$ \frac{7}{\tan^2 \theta} + 74 \sin^2 \beta = 34 $$ does not hold true. - For $$ \cos A = 3 $$ and $$ A + B = 90^\circ $$, the problem is invalid because $$ \cos A $$ cannot be 3.

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