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TRIGONOMETRY. Sign for sin , cos  and tan  Quadrant I 0° <  < 90° Quadrant II 90 ° <  < 180° Quadrant III 180° <  < 270° Quadrant IV 270 ° <  <

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Presentation on theme: "TRIGONOMETRY. Sign for sin , cos  and tan  Quadrant I 0° <  < 90° Quadrant II 90 ° <  < 180° Quadrant III 180° <  < 270° Quadrant IV 270 ° <  <"— Presentation transcript:

1 TRIGONOMETRY

2 Sign for sin , cos  and tan  Quadrant I 0° <  < 90° Quadrant II 90 ° <  < 180° Quadrant III 180° <  < 270° Quadrant IV 270 ° <  < 360° ALL (+) SIN  (+) TAN  (+) COS  (+)  =  Let  = acute angle  = 180 °−   = 180 °+   = 360 °−     

3 Finding angle  when given sin  Given that 0 °    360°, find  when sin  = 0.7660 sin  = −0.5736 Quadrant II 90 ° <  < 180° SIN  (+)  = 180 °−  Quadrant III 180 ° <  < 270° TAN  (+)  = 180 °+  Quadrant IV 270 ° <  < 360° COS  (+)  = 360 °−  sign (+) == sin -1 0.7660  = 50° (acute angle)   = 50°, 130° Quad I & Quad II Quadrant I 0 ° <  < 90°  =  sign ( − ) Quad III & Quad IV == sin -1 0.5736  = 35°   = 180° + 35°, 360°−35° = 215°, 325°

4 Finding angle  when given cos  Given that 0 °    360°, find  when (a) (a) cos  = 0.7660 (b) (b) cos  = −0.5736 Quadrant 2 90 ° <  < 180° SIN  (+)  = 180 °−  Quadrant 3 180 ° <  < 270° TAN  (+)  = 180 °+  Quadrant 4 270 ° <  < 360° COS  (+)  = 360 °−  sign(+) == cos -1 0.7660  = 40°   = 40°, 360 − 40° = 40°, 320° Quad I & Quad IV Quadrant I 0 ° <  < 90°  =  sign ( − ) Quad II & Quad III == cos -1 0.5736  = 55°   = 180° −55°, 180°+35° = 125°, 235°

5 Find angle  when given tan  Given that 0 °    360°, find  when (a) (a) tan  = 1.7660 (b) (b) tan  = −2.5 Quadrant 2 90 ° <  < 180° SIN  (+)  = 180 °−  Quadrant 3 180 ° <  < 270° TAN  (+)  = 180 °+  Quadrant 4 270 ° <  < 360° KOS  (+)  = 360 °−  sign (+) == tan -1 1.7660  = 60°29’ Hence  = 60°29’, 180° + 60°29’ = 60°29’, 240° 29’ Quadrant I and Quadrant 3 Quadrant 1 0 ° <  < 90°  =  sign ( − ) Quadrant 2 and Quadrant 4 == tan -1 2.5  = 68°12’ Hence  = 180° − 68°12’, 360°−68°12’ = 111°48’, 291°48’

6 Practice makes perfect!!! 1. Given sin x° =0.7547 and 90°  x  180°, find x. 2. Given cos x = cos 34° and 270°  x  360°, find x. 3. Given cos x = − 0.6926 and 90°  x  180°, find x. 4. Given tan x = 0.8 and 180°  x  360°, find x. 5. Given tan x = −0.8098 and 270°  x  360°, find x. Answer: (1) 131° (2)326° (3)133°50’ (4)218°40’ (5)321°


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