PARTIAL NOTES for 6.1 Trigonometric Identities

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1 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

2 Examples: Establish each identity 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 a. b. -1 c. d. 1 tan θ 1+ cot θ 1 cot θ 7. Is this an identity? sinθ cotθ = secθ

3 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) 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 ).

4 Complete each identity 14. tan55 o + tan5 o 1+ tan55 o tan5 o = 15. cot( 4π x)= 16. 8sin( π x)sin( π + x)= 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

5 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

6 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 If cosθ = 6 π, <θ < π ; find tanθ. 7. If cosθ =, 0 <θ < π ; find tan θ cos 7π 1 = Complete each identity 9. sin θ + cos θ = cot θ = 11. cos θ sin θ sinθ = 1. tan θ = 1 tan θ 6

7 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

8 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

9 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 )= 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 ))]= 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

10 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

11 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 ! 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 = (!cosθ = sinθ ) (!cosθ +5sinθ = 0)! cosθ! sinθ!1 = tanθ!cotθ +5 = 0! cotθ = 5 11

12 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