By William J. Bruce, W. J. Langford, E. A. Maxwell and I. N. Sneddon (Auth.)
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19. i F> [y \5 4 ^N? 0 -3 FIG. 19 ly -78 78 X -1 3 \ . ^ ^ 3 -1 3 P' Fio. 20 TRIGONOMETRIC FUNCTIONS OF ANGLES 37 From the reference triangle, we get sin Θ = -f-, tan Θ = — f, cot Θ = — Î, sec Θ = — | , and csc 0 = f. Example 2. Find the other functions of Θ if sin 0 = — %. Since the quadrant of Θ is not specified, we must consider Θ in quadrants HI and IV because sin Θ is negative. But sin θ = y r and, since r is positive, we choose r = 3 and get >> = — 1. It follows that x = ± V8 and two reference triangles are needed V8 (Fig.
The right triangle formed by drawing a perpendicular line from any point P on the terminal side of an angle to the xaxis is called the reference triangle. We consider now, functions of angles whose reference angles are 30°, 45°, and 60°. For a right triangle with an angle of 45° choose the legs to be of unit length and the hypotenuse is Vl (Fig. 15). Λ/2 1 The functions of 45° are, thus, sin 45° = —- = _ s e c 450 Vl 2 ' = — =» V2, etc. 34 ANALYTIC FIG. 15 FIG. FIG. 17 If the angle is 135° (reference angle 45°) the reference triangle may be labelled as in Fig.
These two sets of values of 0 form together the solutions of 2 sin2 0 - sin 0 - 1 = 0 Example 4. Solve the equation cos2 0 + 2 sin2 0 = f. Since sin2 0=1— cos2 0, this equation may be written 5 cos2 0 + 2 (1 - cos2 0) = - . Simplifying, we get 3 cos2 0 = — Thus, cos 0 = ± - V3~ 2 Since cos 0 is both positive and negative, 0 can be in any quadrant. The reference angle is 30° ( — ). Therefore, 0 = . . , - 30°, 30°, 150°, 210°, 330° Just as in solving algebraic equations, a word of caution is necessary in the solutions of trigonometric equations.
Analytic Trigonometry by William J. Bruce, W. J. Langford, E. A. Maxwell and I. N. Sneddon (Auth.)
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