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by: Joseph Notetaker

Triginometry MTH 181

Joseph Notetaker
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About this Document

Notes for the week of March 14-18
Micheal E. Cagle
Class Notes




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This 5 page Class Notes was uploaded by Joseph Notetaker on Thursday March 17, 2016. The Class Notes belongs to MTH 181 at Missouri State University taught by Micheal E. Cagle in Spring 2016. Since its upload, it has received 34 views. For similar materials see Trigonometry in Math at Missouri State University.


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Date Created: 03/17/16
Trigonometry Notes Chapter 2 Section 1&2 Pythagorean Identities: Sin θ+Cos θ=1 Tan θ+1=Sec θ 2 1+Cot θ=Csc θ 2 Conjugates: 2 2 A+B--------A-B so (a+b)*(a-b)=a -b Verify the Identity: sinθ 1+cosθ = Binomial in the denominator is not good 1−cosθ sinθ sinθ ∗(1+cosθ) 1−cosθ Multiply by conjugate (1+cosθ) sinθ(1+cosθ) 2 (1-cosθ)(1+cosθ)=sin2θ because 1-cosθ=sinθ sin θ 1+cosθ 1+cosθ = Factor out sinθ sinθ sinθ Replace with single term: Tan θ+1=Sec θ 2 Sec θ+1=N/A Verify the Identity: tanθ secθ−1 =cscθ+cotθ Left side tanθ ∗(secθ+1) secθ−1 Multiply by the conjugate (secθ+1) tanθ(secθ+1) 2 (secθ­1)(secθ+1)=tan θ tan θ secθ+1 tanθ  Factor out tanθ 1 cosθ +1 Express as sin and cos sin cos 1 +1 ∗ cosθ Multiply (cosθ )( sinθ 1 cosθ 1 + cosθ over cosθ cancels leaving sinθ sinθ sinθ cscθ+cotθ=cscθ+cotθ Simplify Verify the identity: 2 tanθ+cotθ= tan θ+1 tanθ Move to Left side sinθ+cosθ cosθ sinθ Express as sin and cos sinθ sinθ cosθ cosθ ∗( ) + ∗( ) Multiply by reciprocals cosθ sinθ sinθ cosθ 2 2 sin θ + cos θ iθ oθ siθ cθ Simplify 2 1 = tan θ+1 2 2 iθ oθ tanθ sin θ+cos θ=1 tan θ+1 tanθ Move to Right side sec θ tan θ+1=sec θ tanθ 1 2 cos θ sinθ Express as sin and cos cosθ 1 2 ∗cosθ cos θ Multiply sinθ 1 1 cosθ∗sinθ = sinθ∗cosθ Section 2 sin α−sin α=cos α−cos α 4 4 4 sin α+cos α ≠ 1 2 2 4 2 2 2 4 Instead: sin α=(1−cos α)  and  sin α=(1−cos α)(1−cos α)  or  1−2cos α+cos α Here are some example problems: 1 1 + =−2cscxcotx            Move to left side cosx+1 cosx−1 1 cosx−1 1 cosx+1 ∗( )+ ∗ ( )   Multiply by conjugates cosx+1 cosx−1 cosx−1 cosx+1 cosx−1 cosx+1 2 + 2                              Now that the denominator matches add the fractions −sin x −sin x 2cos 2 −sin x                                                   (cosx+1) + (cosx­1) = 2cos  −2∗1 ∗cosx sinx                               Express as the product of fractions sinx −2cscxcotx                                            Simplify 1+secθ sinθ+tanθ =csc                                  Move to left side 1 1+ cosθ                                              Express as sin and cos sinθ+ sinθ cosθ cosθ+1 niθ s+sinθ                                     Multiply all terms by cos cosθ+1 sinθ(cosθ+1)                                        Factor sinθ out of the denominator 1 sinθ                                                         Factor out cosθ+1 csc                                                           Simplify


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