Question

double-angle formulas
find the value of \\( \sin 2\theta \\) given \\( \theta \\) is in the first quadrant and \\( \tan \theta = \frac{8}{9} \\).
double-angle formulas
\\( \sin(2a) = 2\sin a\cos a \\)
\\( \cos(2a) = \cos^2 a - \sin^2 a \\)
\\( \quad \quad \quad = 1 - 2\sin^2 a \\)
\\( \quad \quad \quad = 2\cos^2 a - 1 \\)
\\( \tan(2a) = \frac{2\tan a}{1 - \tan^2 a} \\)
\\( \sin 2\theta = \frac{144}{?} \\)

Explanation:

Step1: Recall the double - angle formula for sine

The double - angle formula for sine is $\sin(2\theta)=2\sin\theta\cos\theta$. Also, we know that $\tan\theta = \frac{\sin\theta}{\cos\theta}=\frac{8}{9}$, so we can let $\sin\theta = 8k$ and $\cos\theta=9k$ for some positive real number $k$ (since $\theta$ is in the first quadrant, both $\sin\theta$ and $\cos\theta$ are positive).

Step2: Use the Pythagorean identity $\sin^{2}\theta+\cos^{2}\theta = 1$

Substitute $\sin\theta = 8k$ and $\cos\theta = 9k$ into the Pythagorean identity:
\[

$$\begin{align*} (8k)^{2}+(9k)^{2}&=1\\ 64k^{2}+81k^{2}&=1\\ 145k^{2}&=1\\ k^{2}&=\frac{1}{145}\\ k&=\frac{1}{\sqrt{145}} \quad (\text{since }k>0\text{ as }\theta\text{ is in the first quadrant}) \end{align*}$$

\]

Step3: Calculate $\sin\theta$ and $\cos\theta$

$\sin\theta=8k = \frac{8}{\sqrt{145}}$ and $\cos\theta = 9k=\frac{9}{\sqrt{145}}$

Step4: Calculate $\sin(2\theta)$ using the double - angle formula

$\sin(2\theta)=2\sin\theta\cos\theta=2\times\frac{8}{\sqrt{145}}\times\frac{9}{\sqrt{145}}$
\[

$$\begin{align*} \sin(2\theta)&=2\times\frac{72}{145}\\ &=\frac{144}{145} \end{align*}$$

\]

Answer:

145