Fill in the missing value on the unit circle: What is the value of $sin(θ)$ when $cos(θ) = -frac{1}{2}$?

Given the unit circle equation:

$\cos^2(θ) + \sin^2(θ) = 1$

Substitute $\cos(θ) = -\frac{1}{2}$:

$\left(-\frac{1}{2}\right)^2 + \sin^2(θ) = 1$

Simplify:

$\frac{1}{4} + \sin^2(θ) = 1$

Subtract $\frac{1}{4}$ from both sides:

$\sin^2(θ) = 1 – \frac{1}{4}$

$\sin^2(θ) = \frac{3}{4}$

Taking the square root of both sides:

$\sin(θ) = \pm \sqrt{\frac{3}{4}}$

$\sin(θ) = \pm \frac{\sqrt{3}}{2}$

So, the value of $\sin(θ)$ when $\cos(θ) = -\frac{1}{2}$ is $\pm \frac{\sqrt{3}}{2}$.

Starting with the unit circle equation $cos^2(θ) + sin^2(θ) = 1$:

Substitute $cos(θ) = -frac{1}{2}$:

$left(-frac{1}{2}
ight)^2 + sin^2(θ) = 1$

Calculate $left(-frac{1}{2}
ight)^2$:

$frac{1}{4} + sin^2(θ) = 1$

Therefore:

$sin^2(θ) = 1 – frac{1}{4}$

$sin^2(θ) = frac{3}{4}$

Take the square root:

$sin(θ) = pm frac{sqrt{3}}{2}$

The values of $sin(θ)$ are $pm frac{sqrt{3}}{2}$.

From $cos^2(θ) + sin^2(θ) = 1$ and $cos(θ) = -frac{1}{2}$:

$left(-frac{1}{2}
ight)^2 + sin^2(θ) = 1$

$frac{1}{4} + sin^2(θ) = 1$

$sin^2(θ) = frac{3}{4}$

$sin(θ) = pm frac{sqrt{3}}{2}$

The solutions are $pm frac{sqrt{3}}{2}$.

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