Question

The angle \theta_1θ
1
​
theta, start subscript, 1, end subscript is located in Quadrant \text{II}IIstart text, I, I, end text, and \cos(\theta_1)=-\dfrac{12}{19}cos(θ
1
​
)=−
19
12
​
cosine, left parenthesis, theta, start subscript, 1, end subscript, right parenthesis, equals, minus, start fraction, 12, divided by, 19, end fraction .
What is the value of \sin(\theta_1)sin(θ
1
​
)sine, left parenthesis, theta, start subscript, 1, end subscript, right parenthesis?

Answer 1

Answer:

$sin\theta_1 = \dfrac{\sqrt{217}}{19}$

Step-by-step explanation:

It is given that:

$cos\theta_1 = -\dfrac{12}{19}$

And we have to find the value of $sin\theta_1 = ?$

As per trigonometric identities, the relation between $sin\theta\ and \ cos\theta$ can represented as:

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

Putting $\theta_1$ in place of $\theta$ Because we are given

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

Putting value of cosine:

$cos\theta_1 = -\dfrac{12}{19}$

$sin^2\theta_1 + (\dfrac{12}{19})^2 = 1\\\Rightarrow sin^2\theta_1 + \dfrac{144}{361} = 1\\\Rightarrow sin^2\theta_1 = 1-\dfrac{144}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{361-144}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{217}{361}\\\Rightarrow sin\theta_1 = +\sqrt{\dfrac{217}{361}}, -\sqrt{\dfrac{217}{361}}\\\Rightarrow sin\theta_1 = +\dfrac{\sqrt{217}}{19}, -\dfrac{\sqrt{217}}{19}$

It is given that $\theta_1$ is in 2nd quadrant and value of sine is always positive in 2nd quadrant. So, the answer is.

$\Rightarrow sin\theta_1 = \dfrac{\sqrt{217}}{19}$

Answer 2

Answer:

−  square root 15/4

Step-by-step explanation: