Trigonometry by Heidi A. Howard

By Heidi A. Howard

Instructor's recommendations handbook to accompany Trigtonometry, eighth version via Lial, Hornsby, Schneider

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2  3   3 1 1 ( x )  x  =  x 2    2  2   2  2  3 x 3 3 in the area of a triangle, which gives Since sin θ = 2 = , we can substitute sin θ for 2 x 2 1 2 1 x sin θ . Thus, the area of the hexagon is 6 x 2 sin θ = 3x 2 sin θ . 2 2 FG H IJ K 62. (a) From the figure in the text and the definition of tan θ , we can see that tan θ = (b) Solving for x we have, tan θ = y . x y y ⇒ x tan θ = y ⇒ x = . 4: Using the Definitions of the Trigonometric Functions 1. 2. 1 is its own reciprocal.

D; 750° − 2 ⋅ 360° = 30° (30° is in quadrant I) 6. B; 480° − 360° = 120° and 180° − 120° = 60° (120° is in quadrant II) 2 is a good choice for r because in a 30° − 60° right triangle, the hypotenuse is twice the length of the shorter side (the side opposite to the 30° angle). By choosing 2, one avoids introducing a fraction (or decimal) when determining the length of the shorter side. Choosing any even positive integer for r would have this result; however, 2 is the most convenient value. 8. – 9.

Sin 2 180° + cos 2 180° sin180° = y 0 x −1 = = 0 and cos180° = = = −1 r 1 r 1 sin 2 180° + cos 2 180° = 02 + ( −1) = 0 + 1 = 1 2 40. sin 2 360° + cos 2 360° sin 360° = sin 0° = y 0 x 1 = = 0 and cos 360° = cos 0° = = = 1 r 1 r 1 sin 2 360° + cos 2 360° = 0 2 + 12 = 0 + 1 = 1 41. sec 2 180° − 3sin 2 360° + 2 cos180° sec180° = r 1 y 0 x −1 = = −1, sin 360° = sin 0° = = = 0, and cos180° = = = −1 x −1 r 1 r 1 sec 2 180° − 3sin 2 360° + 2 cos180° = ( −1) − 3 ( 0 ) + 2 ( −1) = 1 − 0 − 2 = −1 2 42. 3: Trigonometric Functions 25 43.

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