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Other Trigonometry Functions Tangent: tan x sin x cos x 2 n 1 Undefined if cos x 0 , or x 2 Cotangent: cot x cos x sin x Undefined if sin x 0 , or x n Secant: sec x 1 cos x 2 n 1 Like Tangent, undefined for x 2 Cosecant: csc x 1 sin x Like Cotangent, undefined for x n In all the above expressions, n 0, 1, 2, In terms of the sides of a right triangle, tan x OPP ADJ MATH 1310 Ronald Brent © 2016 All rights reserved. cot x ADJ OPP sec x Lecture 4B HYP ADJ csc x HYP OPP 1 of 12 Graphs of Other Trigonometry Functions y y = tan x 5 4 3 2 1 0 5 3 5 3 2 2 2 2 2 2 x -1 -2 -3 -4 -5 MATH 1310 Ronald Brent © 2016 All rights reserved. Lecture 4B 2 of 12 y y = sec x 5 4 3 2 1 0 5 3 5 3 2 2 2 2 2 2 x -1 -2 -3 -4 -5 MATH 1310 Ronald Brent © 2016 All rights reserved. Lecture 4B 3 of 12 y y = cot x 5 4 3 2 1 0 x -1 -2 -3 -4 -5 MATH 1310 Ronald Brent © 2016 All rights reserved. Lecture 4B 4 of 12 y = csc x y 5 4 3 2 1 0 x -1 -2 -3 -4 -5 MATH 1310 Ronald Brent © 2016 All rights reserved. Lecture 4B 5 of 12 Solving Right Triangles: Example: Solve: 32 o 8 We know that the side adjacent to the 32 angle is ADJ = 8. To get OPP, the side opposite the angle, consider that in this case OPP tan 32 , ADJ and so OPP ADJ tan32 (8) (.624869352) 4.998954815 . Given the opposite side, OPP, you can get the hypotenuse, HYP, from the Pythagorean theorem, or by using HYP sec32 , ADJ to obtain HYP = ADJ sec 32 (8) (1.179178403) 9.433427227 . Of course, the last angle is 58 . MATH 1310 Ronald Brent © 2016 All rights reserved. Lecture 4B 6 of 12 Non-Right Angle Triangles Given the triangle show below with sides of length a, b, and c, and opposite angles A, B, and C, we have: a b C B A Law of Sines c sin A sin B sin C a b c Law of Cosines c 2 a 2 b2 2 a b cos C also b2 a 2 c 2 2 a c cos B and a 2 b2 c 2 2 b c cos A MATH 1310 Ronald Brent © 2016 All rights reserved. Lecture 4B 7 of 12 Example: Solve: 60 o 25o 15 Using the Law of Sines c A B = 60 o b a = 15 C = 25 o Notice that A = 95 , so sin 95 sin 60 sin 25 . 15 b c You can solve for b and c as 15 sin 60 b sin 95 15 sin 25 and c . sin 95 Using a calculator, you can get decimal equivalents: b 13.04 and c 6.36 MATH 1310 Ronald Brent © 2016 All rights reserved. Lecture 4B 8 of 12 Example: Solve 4 75 o 5 First, label the triangle: A b c = 4 B = 75 o C a = 5 b2 a 2 c 2 2ac cosB 25 16 40 cos 75 38.32 b 6.19 . Now you can find the angle C from the Law of Sines: sin B sin C sin 75 sin C , so and 6.19 4 b c sin 75 sin C 4 0.624 6.19 C = angle, between 0 and 90, whose sine is .624. Using a calculator we find out that C 38.6 A 66.4 . MATH 1310 Ronald Brent © 2016 All rights reserved. Lecture 4B 9 of 12 Trigonometric Identities y (cos , sin ) x (1,0) (cos , sin ) sin( ) sin cos( ) cos sin cos 2 MATH 1310 Ronald Brent © 2016 All rights reserved. tan( ) tan cos sin 2 Lecture 4B 10 of 12 Given sin 2 cos 2 1 , 2 if we divide both sides of this equation by cos we get sin 2 cos 2 1 cos 2 cos 2 sin 2 cos 2 1 cos 2 cos 2 cos 2 , tan 2 1 sec 2 . 2 Similarly if we divide by sin , we can derive 1 cot 2 csc 2 . MATH 1310 Ronald Brent © 2016 All rights reserved. Lecture 4B 11 of 12 Sum and Difference Formulas: sin( A B ) sin A cos B sin B cos A sin( A B ) sin A cos B sin B cos A cos( A B ) cos A cos B sin A sin B cos( A B ) cos A cos B sin A sin B Double Angle Formulas sin 2 2 sin cos cos 2 cos2 sin 2 Half Angle Formulas 1 cos 2 cos 2 12 1 cos ( 2 ) 2 cos 2 sin 2 12 sin 2 MATH 1310 Ronald Brent © 2016 All rights reserved. Lecture 4B 1 cos 2 1 cos ( 2 ) 2 12 of 12