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## Trigonometry Exam 3 Study Guide

by: Joseph Notetaker

29

0

8

# Trigonometry Exam 3 Study Guide MTH 181

Marketplace > Missouri State University > Math > MTH 181 > Trigonometry Exam 3 Study Guide
Joseph Notetaker
MSU

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This is a study guide for our exam next wednesday. Be sure to prepare
COURSE
Trigonometry
PROF.
Micheal E. Cagle
TYPE
Study Guide
PAGES
8
WORDS
CONCEPTS
Math
KARMA
50 ?

## Popular in Math

This 8 page Study Guide was uploaded by Joseph Notetaker on Friday April 1, 2016. The Study Guide belongs to MTH 181 at Missouri State University taught by Micheal E. Cagle in Spring 2016. Since its upload, it has received 29 views. For similar materials see Trigonometry in Math at Missouri State University.

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Date Created: 04/01/16
Trigonometry exam 3 study guide Pythagorean Identities: 2 2 Sin θ+Cos θ=1 2 2 Tan θ+1=Sec θ 2 2 1+Cot θ=Csc θ X 0 π π π π Conjugates: 6 4 3 2 A+B--------A-B so (a+b)*(a-b)=a -b Sin 0 1 √ 2 √3 1 2 2 2 sin α−sin α=cos α−cos α 4 Co 1 3 2 1 0 √ √ s 2 2 2 sin α+cos α ≠ 1 Ta 0 √ 3 1 √ 3 DN 2 2 Instead: sin α=(1−cos α)  and n 3 E sin α=(1−cos α)(1−cos α)  or 1−2cos α+cos α4 Sin + All + Csc + Tan + Cos + Cot + Sec + 1 cosθ= 2                       Cosine is positive in quadrants 1 and 4 and has two solutions 1 π cosr= 2sr= 3         Use the table to find the value of the reference angles Q1: π Q4: 5π 3 3                  Assign the values from the unit circle Q1:θ= +k∗2π 3              Add two rotations to equal θ Q4:θ= 5π +k∗2π 3             Answer for sin and cos problems must have two values Let’s try a tangent problem: tanθ= √3 3   tanrisr= π 6 Q1:θ= +k∗π 6     tangent problems only require one value because tangent problems create  a                                    straight diagonal line                                    Now for a more interesting problem: −2sin3θ+1=0 1 sin3θ=                         Use algebra to create a rational equation (note: sin3θ ≠ 3sinθ) 2 sinr= isr= π 2 6        Q1:3θ= +k∗2π 6        Write answers in terms of 3θ Q2:3θ= 5π +k∗2π 6   Q1:θ= π + k∗2π 18 3           Use division   Q2:θ= 5π +∗2π   18 3 Next something even harder: 2 4 csc θ= 3 2 cscθ=+¿−            Find the square roots √3 √3 sinθ=+¿−            Reciprocate for a sin problem 2 √3 π sinr= isr=          2 3 π Q1:θ= +k∗2π        There should be an answer in every quadrant    3 2π Q2:θ= 3 +k∗2π       because the reference angle is positive and negative 4π Q3:θ= 3 +k∗2π   5π Q4:θ= +k∗2π   3 Now for Quadratic equations: 2 cos θ−cosθ=0 cosθ(cosθ−1 )=0         Factor cosθ=0∨cosθ=1 cosr=1isr=0   π cosr=0isr= 2 π π θ= +k∗π 2                                Quadrants are irrel2van  and 0 represent a  right θ=k∗2π                                     angle Last one: cosθ=1−sin 2 1−sinθ¿ cosθ¿ =¿                      Square both sides ¿ cos θ=1−2sinθ+sin θ 2 1−sin θ=1−2sinθ+si n θ 2       Use Pythagorean identities 2sin θ−2sinθ=0                           Move terms to one side 2sinθ(sinθ−1=0                          Factor 2sinr=0issinr=0isr=0 π sinr=1isr= 2   θ= +k∗π                                      Remember to check for extraneous solutions 2 θ=k∗2 π Cos(a-b) = cosacosb + sinasinb Cos(a+b) = cosacosb – sinasinb Sin(a-b) = sinacosb - cosasinb Sin(a+b) = sinacosb + cosasinb tana−tanb Tan(a-b) =1+tanatanb tana+tanb Tan(a+b) = 1−tanatanb Here is an example problem: 7π 7π 2π π Find the Exact value ofn12 (hint: 12 = 3 − 4 ) sin 2π− π =sin2π cosπ −cos2π sin (3 4 ) 3 4 3 4 Plug in the values ¿ √3 √2 − −1 √2 Derive reference ( 2 ( 2 (2) (2) angles 6 − 2 ¿ √ − √ Multiply 4 4 6+ 2 √ √ Combine (this is the 4 finished form of the problem Double angle formulas: Half angle formulas(+ and – is either/ or not both): Sin: sin2x=2sinxcosx Sin: x 1−cosx sin2=± √ 2 2 2 Cos: cos2x=cos x−si n x Cos: x 1+cosx cos2=± √ 2 2 cos2x=2cos x−1 Tan: tan = 1−cosx 2 sinx cos2x=1−2si n x tan2 x=2tanx Tan: 1−tan x −29 3π −21 −20 Example: secθ= π<θ< this means sin= cos= 20 2 29 29 tan=21 20 θ = Q3 −21 −20 840 Sin2θ = 2 = ( 29 ( 29 841 −20 −41 29¿ −1=841 Cos2θ = 2¿ 21¿ 2¿ 1−¿ Tan2θ = 21 2(20) ¿ π θ 3π 2 2 <4 θ = Q2 1+20 sin= 29 2 √ 2 29 20 29 29 ¿√ 2 49 = 29 √ 2 ¿ 49 √58 7√58 ¿ 58 20 θ 1−29 cos2− √ 2 29−20 ¿− 29 29 √ 2 9 29 ¿−√ 2 ¿− 9 √58 ¿−3√58 58 −1+20 29 −21 ∗29 θ 29 ta2 = 29 49 −7 ¿−21= 3

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