Question - Solving Trigonometric Expressions on the Unit Circle
Solution:
The image shows four trigonometric expressions for which the values need to be found over the interval 0 ≤ θ < 360° using the unit circle. Let's solve each one:1. sec^(-1)(-√2) = 135°The secant function is the reciprocal of the cosine function. sec(θ) = 1/cos(θ). Since sec(θ) is -√2, we are looking for when cos(θ) = -1/√2. This occurs at an angle of 135°, assuming the principal value (the smallest positive angle).2. csc^(-1)(2√3/3) = 40°The cosecant function is the reciprocal of the sine function. csc(θ) = 1/sin(θ). Since csc(θ) is 2√3/3, we are looking for when sin(θ) = 3/2√3, which simplifies to 1/√3. This occurs at an angle of 30°. Since cosecant is positive in the first and second quadrants, and the principal value requires the smallest positive angle, the answer would be 30°.3. cos^(-1)(-√3/2) = 150°The inverse cosine of -√3/2 implies that we are looking for an angle where the cosine value is -√3/2. This happens in the second quadrant, where the cosine is negative, and the corresponding angle is 150°, which is the principal value.4. sin^(-1)(-1/2) = 210° or 330°The inverse sine of -1/2 implies we are seeking an angle where the sine value is -1/2. This occurs in the third and fourth quadrants where the sine is negative. The corresponding principal angles are 210° and 330°, both of which are solutions over the interval 0 ≤ θ < 360°. However, since we typically take the principal value for the inverse sine, the solution would be the smallest positive angle - 210°.These angles correspond to the values of the angles where the trigonometric functions of the given values are attained on the unit circle within the specified interval.
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