2. 1+ cotx tany. Solution: LS 1+cotx tan y sin(x+y) sin.xcos y Therefore since LS-RS, 1+cotx tany. RS sin(x+y) sin .xcos y sin x cos y+sin y cos.x sin.xcos y sin x cos y sin x cos y sin y cos .x sin.x cos y sin(x+y) sin .x cos y (sin y\cos x) cos y sin x -1+tan y cot.x =1+cotxtan y

Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE: 1. Give the measures of the complement and the supplement of an angle measuring 35°.
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Part A:  Select TWO examples of solutions to trig identities from the Content section (Examples or Check your Understanding)  or from Assignment 1 Solutions. post on the image pick two question.

  • Annotate the solution for the example explaining the steps. 
  • Refer to the Tips for Solving Trigonometric Identities from the Content section. 

Part B: Make up your OWN identity.  Start with a trigonometric expression, and apply substitutions and algebraic processes to create equivalent expressions. Note that this is NOT the same as proving an existing trig identity. 

Justify each step.  Verify your identity graphically and algebraically.

 

 

Checklist   

 

Annotations

  • Selects two example(s) from class content or assignment
  • Accurately identifies the techniques at each step
  • Some of the techniques are from  “Tips for Solving Trigonometric Identities”
  • Includes the use of compound angle formulae 
  • Examples show a range of techniques
  • Description is clear, demonstrating understanding and reasoning
  • Uses appropriate mathematical terminology

 

Own Solution    

 

  • Selects and sequences effective strategies to create an identity
  • Demonstrates an understanding of equivalence 
    • uses other identities 
    • uses algebraic techniques
  • Accurately identifies the techniques at each step
  • Verifies identity numerically
  • Verifies identity graphically (Desmos)
  • Uses appropriate mathematical notation and conventions
2. 1+cotxtanym
Solution:
LS
1+cot.x tan y
sin(x + y)
sin x cos y
Therefore since LS-RS, 1+cotx tan y
RS
sin(x + y)
sin.xcos y
sin x cos y + sin y cos.x
sin x cos y
sin x cos y
sin.xcos y
= 1+
sin y cos.x
sin.x cos y
sin(x + y)
sin.x cos y
sin y\cos x)
cos y sin x
- 1+tan y cot.x
=1+cotxtan y
Transcribed Image Text:2. 1+cotxtanym Solution: LS 1+cot.x tan y sin(x + y) sin x cos y Therefore since LS-RS, 1+cotx tan y RS sin(x + y) sin.xcos y sin x cos y + sin y cos.x sin x cos y sin x cos y sin.xcos y = 1+ sin y cos.x sin.x cos y sin(x + y) sin.x cos y sin y\cos x) cos y sin x - 1+tan y cot.x =1+cotxtan y
1. sec²x-2secxcos x + cos²x = sin² x tan² x
Solution:
LS
sec² x-2 secx cos x + cos²x
=
= (secx-cosx)²
COS X
1-cos²x
COS X
COS X
²x+cos²x-cos²x
COS X
(sin²x+
sin² x
COS X
sin x
cos²x
RS
sin² x tan² x
(sin² x sin²x
cos²x
²+)(
= sin² x tan² x
Therefore since LS = RS, sec² x-2 secx cos x + cos²x = sin² x tan²x
Transcribed Image Text:1. sec²x-2secxcos x + cos²x = sin² x tan² x Solution: LS sec² x-2 secx cos x + cos²x = = (secx-cosx)² COS X 1-cos²x COS X COS X ²x+cos²x-cos²x COS X (sin²x+ sin² x COS X sin x cos²x RS sin² x tan² x (sin² x sin²x cos²x ²+)( = sin² x tan² x Therefore since LS = RS, sec² x-2 secx cos x + cos²x = sin² x tan²x
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