PARTIAL NOTES for 6.1 Trigonometric Identities

PARTIAL NOTES for 6.1 Trigonometric Identities tanθ = sinθ cosθ cotθ = cosθ sinθ BASIC IDENTITIES cscθ = 1 sinθ secθ = 1 cosθ cotθ = 1 tanθ PYTHAGOREAN IDENTITIES sin θ + cos θ =1 tan θ +1= sec θ 1 + cot θ = csc θ EVEN/ODD IDENTITIES sin( θ) = sinθ cos( θ) = cosθ tan( θ) = tanθ csc( θ) = cscθ sec( θ) = secθ cot( θ) = cotθ sin( π θ) = cosθ cos( π θ) = sinθ tan( π θ) = cotθ sin(π θ) = sinθ cos(π θ) = cosθ tan(π θ) = tanθ IDENTITIES FOR REDUCING FUNCTIONS sin( π +θ) = cosθ sin(π θ) = sinθ cos( π +θ) = sinθ cos(π θ) = cosθ tan( π +θ) = cotθ tan(π θ) = tanθ sin(π +θ) = sinθ sin( π θ) = cosθ cos(π +θ) = cosθ cos( π θ) = sinθ tan(π +θ) = tanθ tan( π θ) = cotθ sin(π +θ) = sinθ cos(π +θ) = cosθ tan(π +θ) = tanθ sin( π +θ) = cosθ cos( π +θ) = sinθ tan( π +θ) = cotθ ** Must also be able to recognize variations, for example: sin θ =1 cos θ sinθ = 1 sin( θ) = sinθ cscθ tan θ sec θ = 1 TO ESTABLISH IDENTITIES: start with one side, usually the one containing the more complicated expression use basic identities and algebraic manipulations to arrive at the other side CAUTION!!! - - - - - - do not treat identities to be established as if they were equations (cannot +, -,, values to each side, etc.) it is sometimes helpful to write one side in terms of sines and cosines add or subtract fractions where appropriate establishing identities occurs through trial & error and LOTS of practice 1

Examples: Establish each identity 1... 4. cosθ(tanθ + cotθ)= cscθ (secθ 1)(secθ +1)= tan θ sin θ + 4cos θ = + cos θ 1 cos θ 1+ sinθ = sinθ 5. Complete the identity 1 secθ tanθ a. cosθ cotθ b. secθ + tanθ c. cotθ + cosθ d. 1 sinθ 6. tan θ tan θ 1 1 1 a. b. -1 c. d. 1 tan θ 1+ cot θ 1 cot θ 7. Is this an identity? sinθ cotθ = secθ

PARTIAL NOTES for 6. Sum & Difference Formulas SUM AND DIFFERENCE FORMULAS cos(α + β) = cosα cosβ sinα sinβ sin(α + β) = sinα cosβ + cosα sinβ tanα + tanβ tan(α + β) = 1 tanα tanβ cos(α β) = cosα cosβ + sinα sinβ sin(α β) = sinα cosβ cosα sinβ tanα tanβ tan(α β) = 1 + tanα tanβ ** Must also be able to recognize variations, reciprocals, etc. Examples: Complete each identity 1. cos( 7π x)=. cos(x π )+ sin(x π )= Write as a single trig function. cos( 1π 0 )cos( π 10 1π π )+ sin( )sin( ) 0 10 Evaluate 4. cos75 o = 5. sin[tan 1 ( 1)+ 6. cos 1 ( )]= csc[6π 5 cos 1 ( 1)]= 7. If cosα = 5 5 ; ; sinβ = 4 5 ; < β < 0 ; find sin(α + β). 8. cot[cos 1 ( )+ 1 sin 1 ( 5 )]= 9 9. ( 1) 1 10. cot[ 5π cos 1 ( 1)]= 11. sin0 o cos65 o sin65 o cos0 o = 1. csc[11π tan 1 ( )]= 5 1. If cscθ =, find cos(θ π ).

Complete each identity 14. tan55 o + tan5 o 1+ tan55 o tan5 o = 15. cot( 4π x)= 16. 8sin( π x)sin( π + x)= 17. 1+ sinθ cosθ + cosθ 1+ sinθ = 18. cosθ secθ cosθ = 19. sec(x π ) cot(x π ) = tanθ 1 cosθ 0. tanθ + sinθ sin θ = 4

PARTIAL NOTES for 6. Double- & Half-Angle Formulas The double-angle formulas can be derived directly from the sum formulas, and the half-angle formulas can be derived from variations of the double-angle formulas. sinθ = sinθ cosθ cosθ = cos θ sin θ DOUBLE-ANGLE FORMULAS cosθ =1 sin θ Variations tanθ = tanθ 1 tan θ ** These are derived directly from the sum formulas. Ex. sinθ = sin(θ + θ) = sinθ cosθ + cosθ sinθ = sinθ cosθ + sinθ cosθ = sinθ cosθ cosθ = cos θ 1 1 cosθ sin θ = cos θ = 1 + cosθ 1 cosθ tan θ = 1 + cosθ sin α = ± cos α = ± tan α = ± 1 cosα 1 + cosα 1 cosα 1 + cosα ** These are derived from the last three variations of the double-angle formulas. Ex. let θ = α, then θ = α, and we have cos α = 1 + cosα cos α = ± 1 + cosα *** The choice of ± depends on the α quadrant of. HALF-ANGLE FORMULAS Variations tan α = 1 cosα sinα tan α = sinα 1 + cosα ** These two do not depend on α finding the quadrant of. 5

Examples: Locate the quadrant π 1. If, then is in quadrant. 4 <θ < π θ θ. If π <θ < π, then is in quadrant. θ. If 0 <θ < π, cscθ = 5, cotθ > 0, then is in quadrant. θ 4. If π <θ < 0, tanθ =, cscθ < 0, then is in quadrant. Evaluate 5. If cosθ = π, <θ < π ; find sinθ. 10 6. If cosθ = 6 π, <θ < π ; find tanθ. 7. If cosθ =, 0 <θ < π ; find tan θ. 5 8. cos 7π 1 = Complete each identity 9. sin θ + cos θ = 10. 1+ cot θ = 11. cos θ sin θ sinθ = 1. tan θ = 1 tan θ 6

1. Evaluate sec[ 1 ( tan 1 ) ]= 14. cot[cos 1 ( )]= 15. sin[ 1 tan 1 ( 4)]= π 16. If cosx = 7, ; find. 18 < x < π cot x 17. tan 19π 1 = 18. If sin x = 5π 4, ; find 5 < x < π sec x 19. If sinx = 8, π < x < π ; find csc x. 17 Complete each identity 0. sin θ = tan θ 1. sin θ cos θ = 7

PARTIAL NOTES for 6.5 Inverse Trig Functions Recall: For a function to have an inverse, it must be one-to-one. The trigonometric functions are not one-to-one. Therefore, for them to have an inverse, we must restrict the domain so they will be one-to-one. NOTATION for Inverses:! y = sin 1 x! y = cos 1 x! y = tan 1 x OR OR OR! y = arcsin x! y = arccosx! y = arctan x which we read as: y is an angle whose y is an angle whose y is an angle whose sine is x cosine is x tangent is x 8

Examples: Evaluate 1. arcsin( 1 )=. cos 1 =. cos 1 ( 1)= 4. ( 1)]= 5. tan[arcsin(1)]= 6. cos[arctan( 1)]= 7. sin[sin 1 (.7)]= 8. tan[tan 1 ( )]= 9. cos 1 (cos π )= 10. 11. 5 sin 1 (sin 5π )= 6 tan 1 (tan 5π 1. cos 1 (cos( 1π ))= 1. 5 sin 1 (sin( 11π ))= 1 18 )= 14. If 0 < b <1, then cos(cos 1 b)= 15. If a >1, then sec[sin 1 ( 1 )]= a Evaluate 16. cot[ (cos 1 0)+(sin 1 ( 1 ))]= 17. 4 tan 1 ( ) cos 1 ( 1)= 18. sin[ 9π tan 1 ( 1 )]= Solve 19. sin 1 x = π 0. tan ( 1 x x) = 7π 4 9

PARTIAL NOTES for 6.6 Solving Trig Equations I To solve a trigonometric equation means to find all possible values of the angle θ on a given interval. To solve for θ, when the given angle is a variation on θ : 1. find values, from 0 to π, for the given angle. use the values of the given angle to solve for values of θ ; include the period [these will be the general solutions of the equation]. to the values of θ, add/subtract the period to find all values for θ that are in the given interval [these will be the specific solutions of the equation] Examples: Solve 1. cosθ = on. cosθ = on 0 θ π [ π,0]. tanθ = on [ π,π ] 4. cosθ = 1 on 0 θ < π 5. tanθ = on [ π,0] ( ) = [ π,π ] 6. csc θ π on 6 7. Given tan( θ + π )=, find all possible general solutions that satisfy the equation. 4 4 π 8. General solutions for a certain trigonometric equation are given by + nπ and π + nπ, where 6 n is an integer. Find all particular solutions over [ π, π ]. 9. Find the number of solutions for cosθ = 4 over the interval π θ 5π. 5 10

PARTIAL NOTES for 6.7 Solving Trig Equations II (just a few algebra hints) I. set to zero & factor, then find # solutions by same procedure as in section 6.6 1.! cos θ + cosθ 1 = 0.!sinθ secθ = secθ ( cosθ 1) cosθ +1! secθ sinθ 1! ( ) = 0!sinθ secθ secθ = 0 ( ) = 0 II. use an identity first, so the equation can be factored, then continue 1.!cosθ = cosθ.! csc θ = cotθ +1.! tan θ = secθ! cos θ 1 cosθ = 0! 1+ cot θ = cotθ +1! sec θ 1 secθ = 0! cos θ cosθ 1 = 0! cot θ cotθ = 0! sec θ secθ = 0 cotθ ( cotθ 1) = 0!! sec θ secθ = 0 ( secθ +1) secθ! ( ) = 0 III. if in terms of reciprocal functions, multiply both sides by either of the functions, then continue 1.! cotθ = tanθ.! cscθ +1 = 4sinθ (! cotθ = tanθ )! cotθ (! cscθ +1 = 4sinθ )! sinθ! cot θ =! + sinθ = 4sin θ! cotθ = ±! 4sin θ sinθ = 0 ( 4sinθ +) sinθ 1! ( ) = 0 IV. if in terms of sine & cosine only, multiply both sides by reciprocal of one of the functions, then continue 1.!cosθ = sinθ.!cosθ +5sinθ = 0 1 1 (!cosθ = sinθ ) (!cosθ +5sinθ = 0)! cosθ! sinθ!1 = tanθ!cotθ +5 = 0! cotθ = 5 11

Examples: Find the number of solutions over the given interval. 1. sin θ sinθ +1 = 0 on 0 θ < π. cos θ = 0 on π θ 5π 4. cosθ += sin θ on 0 θ < π 4. sinθ = cosθ on π θ < 5π 5. tan θ 5tanθ += 0 on π θ π 6. cosθ 1 = secθ on π θ < π 7. cosθ +5sinθ = 0 on 5π <θ < π 1