determine the value of theta, an angle in standard position, where $\sin$sin theta= -1/2 and tan theta = $\sqrt{3}\div3$√3÷3 , $0^{\circ}\le theta<360^{\circ}$0?≤theta<360?
6 years ago
Answered By Leonardo F
This is a question involving the ability to identify the positive and negative signs of trigonometric functions in the unit circle. Since the question said that the value of $\sin\left(\theta\right)=-\frac{1}{2}$sin(θ)=−12 , we can disregard the negative sign for now and only consider the absolute value of the sin function of theta. The smallest possible angle for theta is 30o, because sin(300)=1/2. The sin function is positive in the first and second quadrants, and negative in the third and fourth. Hence, theta can only be in the third or fourth quadrants, because the sin is actually -1/2. The corresponding angle of 30o in the second quadrant is 150o, in the third, 210o and, in the fourth, 330o. So, our angle theta can be either 210o or 330o.
Let's consider now the tangent funtion. We know that: $\tan\left(\theta\right)=\frac{\sqrt{3}}{3}$tan(θ)=√33 . The smallest possible angle for theta is also 30o, because $\tan\left(30^o\right)=\frac{\sqrt{3}}{3}$tan(30o)=√33 . The corresponding angle of 30o in the second quadrant is 150o, in the third, 210o and, in the fourth, 330o. The tangent function is positive in the first and third quadrants, and negative in the second and fourth. Since the tangent of theta must be positive and the sin of theta must be negative, the only quadrant possible is the third one, with an angle of 210o.
6 years ago
Answered By Leonardo F
This is a question involving the ability to identify the positive and negative signs of trigonometric functions in the unit circle. Since the question said that the value of $\sin\left(\theta\right)=-\frac{1}{2}$sin(θ)=−12 , we can disregard the negative sign for now and only consider the absolute value of the sin function of theta. The smallest possible angle for theta is 30o, because sin(300)=1/2. The sin function is positive in the first and second quadrants, and negative in the third and fourth. Hence, theta can only be in the third or fourth quadrants, because the sin is actually -1/2. The corresponding angle of 30o in the second quadrant is 150o, in the third, 210o and, in the fourth, 330o. So, our angle theta can be either 210o or 330o.
Let's consider now the tangent funtion. We know that: $\tan\left(\theta\right)=\frac{\sqrt{3}}{3}$tan(θ)=√33 . The smallest possible angle for theta is also 30o, because $\tan\left(30^o\right)=\frac{\sqrt{3}}{3}$tan(30o)=√33 . The corresponding angle of 30o in the second quadrant is 150o, in the third, 210o and, in the fourth, 330o. The tangent function is positive in the first and third quadrants, and negative in the second and fourth. Since the tangent of theta must be positive and the sin of theta must be negative, the only quadrant possible is the third one, with an angle of 210o.
Hence, the value of theta is 210o.