This preview shows page 1. Sign up to view the full content.
Unformatted text preview: ce, all solutions here are
of the form θ = π − α + 2πk = π − arcsin 3 + 2πk , for integers k . 2. Even though we are told t represents a real number, it we can visualize this problem in terms
of angles on the Unit Circle, so at least mentally,11 we cosmetically change the equation to
We could solve x2 = 4 using square roots as well to get x = ± 4, but, we would simplify the answers to x = ±2.
In practice, this is done mentally, or in a classroom setting, verbally. Carl’s penchant for pedantry wins out here.
10 718 Foundations of Trigonometry
tan(θ) = −2. Tangent is negative in two places: in Quadrant II and Quadrant IV. If we
proceed as above using a reference angle approach, then the reference angle α must satisfy
0 < α < π and tan(α) = 2. Such an angle is α = arctan(2) radians. A Quadrant II
angle with reference angle α is π − α. Hence, the Quadrant II solutions to the equation are
θ = π − α + 2πk = π − arctan(2) + 2πk for integers k . A Quadrant IV angle with reference
angle α is 2π...
View Full Document