geom 5.1-5.4 ans.pdf - Worksheet Altitude Median Angle bisector ... - Free Printable

To solve the problems related to trigonometric identities and equations, let's go through each part step by step.

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Problem 1:

#### Part (a)
We are given:
$$
\sin \alpha = \frac{3}{5}, \quad \cos \beta = -\frac{4}{5}, \quad \alpha \in \left(0, \frac{\pi}{2}\right), \quad \beta \in \left(\frac{\pi}{2}, \pi\right).
$$
We need to find $\sin(\alpha + \beta)$.

Step 1: Find $\cos \alpha$ and $\sin \beta$.

Using the Pythagorean identity:
$$
\sin^2 \alpha + \cos^2 \alpha = 1.
$$
Substitute $\sin \alpha = \frac{3}{5}$:
$$
\left(\frac{3}{5}\right)^2 + \cos^2 \alpha = 1 \implies \frac{9}{25} + \cos^2 \alpha = 1 \implies \cos^2 \alpha = 1 - \frac{9}{25} = \frac{16}{25}.
$$
Since $\alpha \in \left(0, \frac{\pi}{2}\right)$, $\cos \alpha > 0$:
$$
\cos \alpha = \frac{4}{5}.
$$

Similarly, for $\beta$:
$$
\sin^2 \beta + \cos^2 \beta = 1.
$$
Substitute $\cos \beta = -\frac{4}{5}$:
$$
\sin^2 \beta + \left(-\frac{4}{5}\right)^2 = 1 \implies \sin^2 \beta + \frac{16}{25} = 1 \implies \sin^2 \beta = 1 - \frac{16}{25} = \frac{9}{25}.
$$
Since $\beta \in \left(\frac{\pi}{2}, \pi\right)$, $\sin \beta > 0$:
$$
\sin \beta = \frac{3}{5}.
$$

Step 2: Use the angle addition formula for sine.
The formula is:
$$
\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta.
$$
Substitute the known values:
$$
\sin(\alpha + \beta) = \left(\frac{3}{5}\right)\left(-\frac{4}{5}\right) + \left(\frac{4}{5}\right)\left(\frac{3}{5}\right).
$$
Simplify:
$$
\sin(\alpha + \beta) = -\frac{12}{25} + \frac{12}{25} = 0.
$$

Thus, the answer is:
$$
\boxed{0}
$$

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Problem 2:

#### Part (b)
We are given:
$$
\tan \alpha = \frac{1}{7}, \quad \tan \beta = \frac{1}{3}.
$$
We need to find $\tan(\alpha + \beta)$.

Step 1: Use the tangent addition formula.
The formula is:
$$
\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}.
$$
Substitute the given values:
$$
\tan(\alpha + \beta) = \frac{\frac{1}{7} + \frac{1}{3}}{1 - \left(\frac{1}{7}\right)\left(\frac{1}{3}\right)}.
$$

Step 2: Simplify the numerator and the denominator.

Numerator:
$$
\frac{1}{7} + \frac{1}{3} = \frac{3}{21} + \frac{7}{21} = \frac{10}{21}.
$$

Denominator:
$$
1 - \left(\frac{1}{7}\right)\left(\frac{1}{3}\right) = 1 - \frac{1}{21} = \frac{21}{21} - \frac{1}{21} = \frac{20}{21}.
$$

So:
$$
\tan(\alpha + \beta) = \frac{\frac{10}{21}}{\frac{20}{21}} = \frac{10}{21} \cdot \frac{21}{20} = \frac{10}{20} = \frac{1}{2}.
$$

Thus, the answer is:
$$
\boxed{\frac{1}{2}}
$$

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Problem 3:

#### Part (c)
We are given:
$$
\sin \alpha = \frac{4}{5}, \quad \alpha \in \left(\frac{\pi}{2}, \pi\right).
$$
We need to find $\sin 2\alpha$ and $\cos 2\alpha$.

Step 1: Find $\cos \alpha$.

Using the Pythagorean identity:
$$
\sin^2 \alpha + \cos^2 \alpha = 1.
$$
Substitute $\sin \alpha = \frac{4}{5}$:
$$
\left(\frac{4}{5}\right)^2 + \cos^2 \alpha = 1 \implies \frac{16}{25} + \cos^2 \alpha = 1 \implies \cos^2 \alpha = 1 - \frac{16}{25} = \frac{9}{25}.
$$
Since $\alpha \in \left(\frac{\pi}{2}, \pi\right)$, $\cos \alpha < 0$:
$$
\cos \alpha = -\frac{3}{5}.
$$

Step 2: Use the double-angle formulas.

The double-angle formulas are:
$$
\sin 2\alpha = 2 \sin \alpha \cos \alpha,
$$
$$
\cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha.
$$

Substitute the known values:
$$
\sin 2\alpha = 2 \left(\frac{4}{5}\right)\left(-\frac{3}{5}\right) = 2 \cdot \frac{4}{5} \cdot -\frac{3}{5} = 2 \cdot -\frac{12}{25} = -\frac{24}{25}.
$$

For $\cos 2\alpha$:
$$
\cos 2\alpha = \left(-\frac{3}{5}\right)^2 - \left(\frac{4}{5}\right)^2 = \frac{9}{25} - \frac{16}{25} = -\frac{7}{25}.
$$

Thus, the answers are:
$$
\sin 2\alpha = -\frac{24}{25}, \quad \cos 2\alpha = -\frac{7}{25}.
$$

The final answer is:
$$
\boxed{-\frac{24}{25}, -\frac{7}{25}}
$$

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Final Answers:

1. (a) $\boxed{0}$
2. (b) $\boxed{\frac{1}{2}}$
3. (c) $\boxed{-\frac{24}{25}, -\frac{7}{25}}$