Question

find the value of sin 2θ given θ is in the first quadrant and tan θ = 8/9. double-angle formulas sin(2a) = 2sinacosa cos(2a) = cos²a - sin²a = 1 - 2sin²a = 2cos²a - 1 tan(2a) = 2tana / (1 - tan²a) sin 2θ = ?

Explanation:

Step1: Recall the double - angle formula for sine

The double - angle formula for sine is $\sin(2\theta)=2\sin\theta\cos\theta$. 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

\[

$$\begin{align*} \sin(2\theta)&=2\sin\theta\cos\theta\\ &=2\times\frac{8}{\sqrt{145}}\times\frac{9}{\sqrt{145}}\\ &=2\times\frac{72}{145}\\ &=\frac{144}{145} \end{align*}$$

\]

Answer:

$\frac{144}{145}$