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## Trigonometry Section 2.3

by: Joseph Notetaker

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# Trigonometry Section 2.3 MTH 181

Marketplace > Missouri State University > Math > MTH 181 > Trigonometry Section 2 3
Joseph Notetaker
MSU
Trigonometry
Micheal E. Cagle

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These notes give a basic outline of section 2.3
COURSE
Trigonometry
PROF.
Micheal E. Cagle
TYPE
Class Notes
PAGES
0
WORDS
CONCEPTS
Trigonometry
KARMA
25 ?

## Popular in Math

This 0 page Class Notes was uploaded by Joseph Notetaker on Tuesday March 22, 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 11 views. For similar materials see Trigonometry in Math at Missouri State University.

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Date Created: 03/22/16
Trigonometry Section 23 Here are some things that will be helpful All The unit circle will also be helpful X 0 E E E E for this section 6 4 3 2 Sin 0 1 1 2 2 2 CO 1 E Q l 0 Sin s 2 2 2 Csc Ta 0 B 1 x3 DN n 7 E Tan Cot e1 COS E Cosine is positive in 1 and 4 and has two solutions 1 H cosr zsr 2 3 H 5H 1 4 Q 3 Q 3 Q19k2 Add two rotations to equal 9 Q4ze57Hkss2n Let s try a tangent problem DEC Assign the values from the unit circle Use the table to find the value of the reference angles Answer for sin and cos problems must have two values quadrants tan 9 3 H tan r lS r 6 1 e H k Q 39 g 1 tangent problems only require one value because tangent problems create straight diagonal line Now for a more interesting problem 2 n39l0 e 1 5m 3 5 Use algebra to create a rational equation note sin39 i 3sin9 sin r is r E 2 6 H Q133 92Ek2H Write answers in terms of 39 Q2395Hk2H H kgtllt2H Q1 921 8 3 Use diVision 5H kgtIlt2H 29 Q 18 3 Next something even harder 4 2 92 CSC 3 cscG HZ i E Find the square roots Sine 6 7 Reciprocate for a sin problem 3 H s1nr lsr 2 H Q 1 3 e 2 g k 2 H There should be an answer in every quadrant 2 H Q 2 3 e 7 k 2 H because the reference angle is positive and negative Q394THk2H Q4ze57Hk2n Now for Quadratic equations co 529 6059 0 cosGcosG l0 Factor case 0 Vcosel cosrlisr0 H cosr015r 2 H H e k H Quadrants are irrelevant because 3 and 0 represent a right 9k2H angle Last one case 1 sin 1 sin 9 2 case 32 3 Square both sides Z co 5291 25in9sin29 1 Si n29 1 2 Sine Si quot2 9 Use Pythagorean identities 2 si 1129 2 Sine 0 Move terms to one side 2 sine sine 1 0 Factor 2sinr0issinr0isr0 H s1nr115r 2 H e E k H Remember to check for extraneous solutions 9k2H

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