Purdue MA 15400 - Trigonometric Function of Angles - D2201384

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Purdue MA 15400 - Trigonometric Function of Angles

School name Purdue University

Course Ma 15400- Algebra and Trigonometry II

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MA 15400 Lesson 3 Section 6.2Trigonometric Function of AnglesThe Fundamental Identities:(1) The reciprocal identities:csc1sinsec1coscot1tan(2) The tangent and cotangent identities:tansincoscotcossin(3) The Pythagorean identities:sin2 cos21 1 tan2sec21 cot2csc2The reciprocal identities are obvious from the definitions of the six trigonometric functions.Take the simple right triangle with sides 3, 4 and 5 with  opposite the side of length 3.To prove the tangent identity, examine the following. The cotangent identity proof is similar.Find sin and cos, now find sin2 cos2sin2 cos21 is a Pythagorean identity since it is derived from the Pythagorean Theorem.2 22 22 22 23 4sin cos5 5(sin ) (cos ) ?sin cos ?3 4?5 59 16?25 2525125sin cos 1q qq qq qq q= =+ =+ =� � � �+ =� � � �� + ==\ + =sintancosoppopp opp hyphypadjadj hyp adjhypqqq= = � = =345θ3 4 3sin cos tan5 5 43sin 3 5 354cos 5 4 45sintancosq q qqqqqq= = == = � =\ =MA 15400 Lesson 3 Section 6.2Trigonometric Function of AnglesDivide both sides by sin2to find another Pythagorean identity.The third Pythagorean identity can by found by dividing by cos2.Each of the three Pythagorean identities creates two more identities by subtracting a term from the left side to the right side. Verify the identity by transforming the left side into the right side.tancot1sin 3 cot 3 cos 3 sectancscsin2 cos2sin21 cot2(1 cos)(1 cos) sin2cos2sec2 1 sin22 2 2 2 2 22 2 2 2 2 22 2 2 2 2 2sin cos 1 1 tan sec 1 cot cscsin 1 cos tan sec 1 cot csc 1cos 1 sin 1 sec tan 1 csc cotq q q q q qq q q q q qq q q q q q+ = + = + == - = - = -= - = - = -2 2 22 22 2 2222 2sin cos 1 Divide each side by sinsin cos 1sin sin sincos1 cscsin1 cot cscq q qq qq q qqqqq q+ =+ =� �+ =� �� \ + =MA 15400 Lesson 3 Section 6.2Trigonometric Function of Angles1 sin2 1 tan2 1cot tancscsecMA 15400 Lesson 3 Section 6.2Trigonometric Function of AnglesUsing the coordinate system, Notice that the adjacent side corresponds to the x-value of the coordinate and the opposite side corresponds the y-value of the coordinate. The idea that the cosine of  corresponds to the x-axis and the sine of  corresponds to the y-axis is one that you need to get used to. This is not saying thatsin θ = the y value nor that cos θ = the x value. Itsimply says there is a correspondence. If  is an angle in standard position on a rectangular coordinate system and if P(-5, 12) is on theterminal side of , find the values of the six trigonometric functions of .If  is an angle in standard position on a rectangular coordinate system and if P(4, 3) is on the terminal side of , find the values of the six trigonometric functions of .(x, y)x = adjy = oppMA 15400 Lesson 3 Section 6.2Trigonometric Function of AnglesFind the exact values of the six trigonometric functions of , if  is in standard position and the terminal side of  is in the specified quadrant and satisfies the given condition.III; on the line 4x – 3y = 0 II; parallel to the line 3x + y – 7 = 0Find the quadrant containing  if the given conditions are true.a) tan  < 0 and cos  > 0b) sec  > 0 and tan  < 0c) csc  > 0 and cot  < 0d) cos  < 0 and csc  < 0e) cos θ < 0 and sec θ > 0All +Icossec +IVsin +cscIItancot +IIIMA 15400 Lesson 3 Section 6.2Trigonometric Function of AnglesUse the fundamental identities to find the values of the trigonometric functions for the given conditions:tan125 and cos 0sec 4 and csc

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