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Precalculus, Section 7.6, #14 Double-angle and Half-angle Formulas Use the information given about the angle θ, 0 ≤ θ < 2π, to find the exact value of the following trigonometric functions.1 √ csc (θ) = − 5 cos (θ) < 0 √ We are given csc (θ) = − 5. So, in terms of more familiar trig functions, we have √ csc (θ) = − 5 √ 1 =− 5 sin (θ) √ 1 = − 5 · sin (θ) or 1 sin (θ) = − √ 5 This tells us that θ must be in quadrant III or IV. We are also given cos (θ) < 0, so θ must be in quadrant II or III. Thus we know that θ is in quadrant III. We use this to draw a triangle to help us find the other trig functions of θ. From the diagram, we get √ 2 5 2 cos (θ) = − √ = − 5 5 √ 5 1 sin (θ) = − √ = − 5 5 1 tan (θ) = 2 (a) sin (2θ) sin (2θ) = 2 sin (θ) cos (θ) √ √ 2 5 5 ·− = 2·− 5 5 4 = 5 (b) cos (2θ) cos (2θ) = cos2 (θ) − sin2 (θ) √ !2 √ !2 2 5 5 = − − − 5 5 4·5 5 − 25 25 15 = 25 3 = 5 = 1 Sullivan, Precalculus: Enhanced with Graphing Utilities, p. 495, #14. -2 θ -1 √ 5 Precalculus Double-angle and Half-angle Formulas (c) sin θ 2 r θ 1 − cos (θ) =± sin 2 2 v √ u u1 − −2 5 t 5 =± 2 s √ 2 5 5 5 + 5 =± 2 s √ 5+2 5 1 =± · 5 2 s √ 5+2 5 =± 10 Since we know θ is in quadrant III, we can deduce that 2θ is in quadrant II. (Since 180◦ < θ < 270◦ , it ◦ θ 270◦ must be that 180 or 90◦ < θ2 < 135◦ . Finally, the sine function is positive in quadrant II, 2 < 2 < 2 and thus s √ 5+2 5 θ = sin 2 10 (d) cos θ 2 r θ 1 + cos (θ) =± cos 2 2 v u √ u1 + −2 5 t 5 =± 2 s √ 2 5 5 5 − 5 =± 2 s √ 5−2 5 1 · =± 5 2 s √ 5−2 5 =± 10 From (c) we know cos θ =− 2 θ 2 is in quadrant II, and so cos s √ 5−2 5 10 θ 2 must be negative. Thus, Precalculus Double-angle and Half-angle Formulas (e) tan (2θ) tan (2θ) = 2 tan (θ) 1 − tan2 (θ) = 2· 1− 1 = 1− 4 = 3 (f) tan θ 2 1 4 1 2 1 2 2 s θ 1 − cos (θ) =± tan 2 1 + cos (θ) v √ u u1 − −2 5 5 u √ = ±t 2 5 1+ − 5 v u 5 2√5 u + 5 √ = ±t 5 2 5 5 5 − 5 v u 5+2√5 u = ±t 5√ 5−2 5 5 √ 5+2 5 5 √ =± · 5 5−2 5 s √ 5+2 5 √ =± 5−2 5 s θ 2 is in quadrant II, and so tan s √ θ 5+2 5 √ tan =− 2 5−2 5 From (c) we know Or s √ θ 5+2 5 √ =− tan 2 5−2 5 s √ √ 5+2 5 5+2 5 √ · √ =− 5−2 5 5+2 5 s √ 25 + 20 5 + 20 =− 25 − 20 s √ 5 9+4 5 =− 5 q √ =− 9+4 5 θ 2 must be negative. Thus,