-90^(@)

Question

-90^(@) <= theta <= 90^(@). Find the value of theta in degrees. {:[sin(theta)=-(sqrt3)/(2)],[theta=◻^(@)]:}

Bytelearn · Accepted Answer

Identify Theta Values: Identify the values of theta for which the sine function equals negative square root of 3 divided by 2. The sine function equals negative square root of 3 divided by 2 at specific reference angles in the unit circle. Since the value is negative, we are looking for angles in the third or fourth quadrant. However, since theta is restricted to the range [-90 degrees, 90 degrees], we are only interested in the fourth quadrant where sine is negative.Determine Reference Angle: Determine the reference angle whose sine is square root of 3 divided by 2. The reference angle for which the sine function equals square root of 3 divided by 2 is 60 degrees. This is because sin(60 degrees) = sqrt(3)/2.Find Fourth Quadrant Angle: Find the angle in the fourth quadrant that corresponds to the reference angle. Since the sine is negative and we are looking for an angle in the fourth quadrant, the angle theta will be -60 degrees because it is the negative of the reference angle.Check Validity: Check if the found angle is within the given range. The angle -60 degrees is within the range [-90 degrees, 90 degrees], so it is a valid solution.

$-90^{\circ} \leq \theta \leq 90^{\circ}$ . Find the value of $\theta$ in degrees. $\newline$ $\begin{array}{l} \sin (\theta)=-\frac{\sqrt{3}}{2} \\ \theta=\square^{\circ} \end{array}$

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−90∘≤θ≤90∘ -90^{\circ} \leq \theta \leq 90^{\circ} −90∘≤θ≤90∘. Find the value of θ \theta θ in degrees.\newlinesin⁡(θ)=−32θ=□∘ \begin{array}{l} \sin (\theta)=-\frac{\sqrt{3}}{2} \\ \theta=\square^{\circ} \end{array} sin(θ)=−23​​θ=□∘​

$-90^{\circ} \leq \theta \leq 90^{\circ}$ . Find the value of $\theta$ in degrees. $\newline$ $\begin{array}{l} \sin (\theta)=-\frac{\sqrt{3}}{2} \\ \theta=\square^{\circ} \end{array}$