Question

The angle θ 1 is located in Quadrant IV, and cos ⁡ ( θ 1 ) = 9/ 19 , theta, start subscript, 1, end subscript, right parenthesis, equals, start fraction, 9, divided by, 19, end fraction . What is the value of sin ⁡ ( θ 1 ) sin(θ 1 ​ )sine, left parenthesis, theta, start subscript, 1, end subscript, right parenthesis? Express your answer exactly.

Answer 1

Answer:

$sin\theta_1 = - \frac{2\sqrt{70}}{19}$

Step-by-step explanation:

We are given that $\theta_1$ is in fourth quadrant.

$cos\theta_1$ is always positive in 4th quadrant and  

$sin\theta_1$ is always negative in 4th quadrant.

Also, we know the following identity about $sin\theta$ and $cos\theta$:

$sin^2\theta + cos^2\theta = 1$

Using \theta_1 in place of \theta:

$sin^2\theta_1 + cos^2\theta_1 = 1$

We are given that $cos\theta_1 = \frac{9}{19}$

$\Rightarrow sin^2\theta_1 + \dfrac{9^2}{19^2} = 1\\\Rightarrow sin^2\theta_1 = 1 - \dfrac{81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{361-81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{280}{361}\\\Rightarrow sin\theta_1 = \sqrt{\dfrac{280}{361}}\\\Rightarrow sin\theta_1 = +\dfrac{2\sqrt{70}}{19}, -\dfrac{2\sqrt{70}}{19}$

$\theta_1$ is in 4th quadrant so $sin\theta_1$ is negative.

So, value of $sin\theta_1 = - \frac{2\sqrt{70}}{19}$