Involving Diagrams and 1. If $$ \sin heta=\frac{-5}{13} $$ and $$ heta \in\left(90^{\circ} ; 270^{\circ} ight) $$ Calculate without the use of a calculatas and with the aid of the diagram the value of: (a) $$ \sin 2 heta $$ (b) $$ \operatorname{Cos} 2 heta $$ 2. If $$ an \alpha=\frac{12}{5} $$ with $$ \alpha \in\left(90^{\circ} ; 270^{\circ} ight) $$ an $$ \cos \beta=-\frac{8}{17} $$ with $$ \beta \in\left(90^{\circ} ; 270^{\circ} ight) $$. calculate without the use of a calculator with the aid of a diagram the value of

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Involving Diagrams and 1. If $$ \sin 	heta=\frac{-5}{13} $$ and $$ 	heta \in\left(90^{\circ} ; 270^{\circ}ight) $$ Calculate without the use of a calculatas and with the aid of the diagram the value of: (a) $$ \sin 2 	heta $$ (b) $$ \operatorname{Cos} 2 	heta $$ 2. If $$ 	an \alpha=\frac{12}{5} $$ with $$ \alpha \in\left(90^{\circ} ; 270^{\circ}ight) $$ an $$ \cos \beta=-\frac{8}{17} $$ with $$ \beta \in\left(90^{\circ} ; 270^{\circ}ight) $$. calculate without the use of a calculator with the aid of a diagram the value of

UpStudy AI · Accepted Answer

To solve the problems, we will first analyze the given conditions and then use trigonometric identities and properties to find the required values.

### Problem 1

Given:
- $$ \sin \theta = \frac{-5}{13} $$
- $$ \theta \in (90^{\circ}, 270^{\circ}) $$

Since $$ \theta $$ is in the third quadrant, we know that:
- $$ \sin \theta $$ is negative.
- $$ \cos \theta $$ is also negative.

#### Step 1: Find $$ \cos \theta $$

Using the Pythagorean identity:
$$
\sin^2 \theta + \cos^2 \theta = 1
$$
Substituting the value of $$ \sin \theta $$:
$$
\left(\frac{-5}{13}\right)^2 + \cos^2 \theta = 1
$$
$$
\frac{25}{169} + \cos^2 \theta = 1
$$
$$
\cos^2 \theta = 1 - \frac{25}{169} = \frac{169 - 25}{169} = \frac{144}{169}
$$
Taking the square root:
$$
\cos \theta = -\sqrt{\frac{144}{169}} = -\frac{12}{13}
$$

#### Step 2: Calculate $$ \sin 2\theta $$

Using the double angle formula:
$$
\sin 2\theta = 2 \sin \theta \cos \theta
$$
Substituting the values:
$$
\sin 2\theta = 2 \left(\frac{-5}{13}\right) \left(-\frac{12}{13}\right) = 2 \cdot \frac{60}{169} = \frac{120}{169}
$$

#### Step 3: Calculate $$ \cos 2\theta $$

Using the double angle formula:
$$
\cos 2\theta = \cos^2 \theta - \sin^2 \theta
$$
Substituting the values:
$$
\cos 2\theta = \left(-\frac{12}{13}\right)^2 - \left(\frac{-5}{13}\right)^2
$$
$$
= \frac{144}{169} - \frac{25}{169} = \frac{119}{169}
$$

### Summary for Problem 1:
- $$ \sin 2\theta = \frac{120}{169} $$
- $$ \cos 2\theta = \frac{119}{169} $$

---

### Problem 2

Given:
- $$ \tan \alpha = \frac{12}{5} $$ with $$ \alpha \in (90^{\circ}, 270^{\circ}) $$
- $$ \cos \beta = -\frac{8}{17} $$ with $$ \beta \in (90^{\circ}, 270^{\circ}) $$

#### Step 1: Find $$ \sin \alpha $$ and $$ \cos \alpha $$

Since $$ \tan \alpha = \frac{12}{5} $$, we can represent this as:
$$
\sin \alpha = 12k, \quad \cos \alpha = -5k
$$
Using the Pythagorean identity:
$$
\sin^2 \alpha + \cos^2 \alpha = 1
$$
Substituting:
$$
(12k)^2 + (-5k)^2 = 1
$$
$$
144k^2 + 25k^2 = 1
$$
$$
169k^2 = 1 \implies k^2 = \frac{1}{169} \implies k = \frac{1}{13}
$$
Thus:
$$
\sin \alpha = \frac{12}{13}, \quad \cos \alpha = -\frac{5}{13}
$$

#### Step 2: Find $$ \sin \beta $$

Using the Pythagorean identity:
$$
\sin^2 \beta + \cos^2 \beta = 1
$$
Substituting the value of $$ \cos \beta $$:
$$
\sin^2 \beta + \left(-\frac{8}{17}\right)^2 = 1
$$
$$
\sin^2 \beta + \frac{64}{289} = 1
$$
$$
\sin^2 \beta = 1 - \frac{64}{289} = \frac{225}{289}
$$
Taking the square root:
$$
\sin \beta = -\sqrt{\frac{225}{289}} = -\frac{15}{17}
$$

### Summary for Problem 2:
- $$ \sin \alpha = \frac{12}{13} $$
- $$ \cos \alpha = -\frac{5}{13} $$
- $$ \sin \beta = -\frac{15}{17} $$

Thus, the final results are:
1. $$ \sin 2\theta = \frac{120}{169} $$, $$ \cos 2\theta = \frac{119}{169} $$
2. $$ \sin \alpha = \frac{12}{13} $$, $$ \cos \alpha = -\frac{5}{13} $$, $$ \sin \beta = -\frac{15}{17} $$
1. $$ \sin 2\theta = \frac{120}{169} $$, $$ \cos 2\theta = \frac{119}{169} $$ 2. $$ \sin \alpha = \frac{12}{13} $$, $$ \cos \alpha = -\frac{5}{13} $$, $$ \sin \beta = -\frac{15}{17} $$

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