Unit 5 Establishing Trigonometric Identities

University:
University of California, Berkeley
Course:
MATH W54 | Linear Algebra and Differential Equations
Academic year:

2021
Views:

267

Pages:

4
Author:

cravelingstnw

Unit 5 Establishing Trigonometric Identities ** These are very crucial for not only this unit, but also in calculus, so make sure you know them very well ○ Trig Identities 𝑠𝑖𝑛𝜃 1 ● 𝑡𝑎𝑛𝜃 = 𝑐𝑜𝑠𝜃 = 𝑐𝑜𝑡𝜃 1 ● 𝑐𝑜𝑠𝜃 = 𝑠𝑒𝑐𝜃 ● 𝑐𝑜𝑡𝜃 = 𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃 1 = 𝑡𝑎𝑛𝜃 1 ● 𝑠𝑒𝑐𝜃 = 𝑠𝑖𝑛𝜃 1 ● 𝑐𝑠𝑐𝜃 = 𝑠𝑖𝑛𝜃 1 ● 𝑠𝑖𝑛𝜃 = 𝑐𝑠𝑐𝜃 ● 𝑠𝑖𝑛2 𝜃 = 1 − 𝑐𝑜𝑠 2 𝜃 ● 𝑐𝑜𝑠 2 𝜃 = 1 − 𝑠𝑖𝑛2 𝜃 ● 𝑠𝑖𝑛2 𝜃 + 𝑐𝑜𝑠 2 𝜃=1 ● 𝑡𝑎𝑛2 𝜃 + 1 = 𝑠𝑒𝑐 2 𝜃 ● 1 + 𝑐𝑜𝑡 2 𝜃 = 𝑐𝑠𝑐 2 𝜃 ● 𝑠𝑖𝑛(−𝜃) = −𝑠𝑖𝑛(𝜃) ● 𝑐𝑜𝑠(−𝜃) = 𝑐𝑜𝑠(𝜃) ● 𝑡𝑎𝑛(−𝜃) = −𝑡𝑎𝑛𝜃 ● 𝑐𝑠𝑐(−𝜃) = −𝑐𝑠𝑐(𝜃) ● 𝑠𝑒𝑐(−𝜃) = 𝑠𝑒𝑐(𝜃) ● 𝑐𝑜𝑡(−𝜃)0 = −𝑐𝑜𝑡(𝜃) ○ Establishing Identity Rules ● Work with only 1 side ● Choose the more complicated side (usually) ● Use simplifying techniques and identities to show both sides are congruent ○ Hints for beginning to establish identities ● Make the form similar for both sides (1 term or 2 terms or 1 fraction...etc) ● Turn your functions into sine and cosine ● Separate fractions with single denominator into multiple fractions ● Find a common denominator to write as 1 fraction ● Multiply or divide each term in a fraction by the same trig function ○ Examples: Sum and Difference ● 𝑐𝑜𝑠(𝛼 + 𝛽) = 𝑐𝑜𝑠𝛼𝑐𝑜𝑠𝛽 − 𝑠𝑖𝑛𝛼𝑠𝑖𝑛𝛽 ● 𝑐𝑜𝑠(𝛼 − 𝛽) = 𝑐𝑜𝑠𝛼𝑐𝑜𝑠𝛽 + 𝑠𝑖𝑛𝛼𝑠𝑖𝑛𝛽 ● 𝑠𝑖𝑛(𝛼 + 𝛽) = 𝑠𝑖𝑛𝛼𝑐𝑜𝑠𝛽 + 𝑐𝑜𝑠𝛼𝑠𝑖𝑛𝛽 ● 𝑠𝑖𝑛(𝛼 − 𝛽) = 𝑠𝑖𝑛𝛼𝑐𝑜𝑠𝛽 − 𝑐𝑜𝑠𝛼𝑠𝑖𝑛𝛽 𝑡𝑎𝑛𝛼+𝑡𝑎𝑛𝛽 ● 𝑡𝑎𝑛(𝛼 + 𝛽) = 1−𝑡𝑎𝑛𝛼𝑡𝑎𝑛𝛽 𝑡𝑎𝑛𝛼−𝑡𝑎𝑛𝛽 ● 𝑡𝑎𝑛(𝛼 − 𝛽) = 1+𝑡𝑎𝑛𝛼𝑡𝑎𝑛𝛽 ● Example 1: ○ 𝑐𝑜𝑠75° = 𝑐𝑜𝑠(30° + 45°) = 𝑐𝑜𝑠30° ⋅ 𝑐𝑜𝑠45° − 𝑠𝑖𝑛30° ⋅ 𝑠𝑖𝑛45° √2 ■ = √3 2 ⋅ ■ = √6 4 − 2 1 −2⋅ √2 2 √2 4 ■ Answer: √6+√2 4 7𝜋 ● Example 2: 𝑠𝑖𝑛 12 4𝜋 3𝜋 ○ 𝑠𝑖𝑛( 12 + 12 ) 𝜋 𝜋 𝜋 𝜋 ○ 𝑠𝑖𝑛 3 ⋅ 𝑐𝑜𝑠 4 + 𝑐𝑜𝑠 3 ⋅ 𝑠𝑖𝑛 4 √2 ○ √3 2 ⋅ ○ √6 4 + 2 1 +2⋅ √2 2 √2 4 ○ Answer: √6+√2 4 ● Half-Angle Formulas 𝜃 1−𝑐𝑜𝑠𝜃 ○ 𝑠𝑖𝑛 2 = ±√ 2 𝜃 1+𝑐𝑜𝑠𝜃 ○ 𝑐𝑜𝑠 2 = ±√ 𝜃 ○ 𝑡𝑎𝑛 2 = 1−𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃 2 𝑠𝑖𝑛𝜃 = 1+𝑐𝑜𝑠𝜃 3 𝜋 5 2 ● Example: If 𝑠𝑖𝑛𝜃 = 𝑎𝑛𝑑 <𝜃<𝜋 ○ 𝑠𝑖𝑛(2𝜃) = 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 3 ○ 2⋅5⋅ −4 5 = −24 35 ● Double Angle Formulas ○ 𝑠𝑖𝑛(2𝜃) = 2𝑠𝑖𝑛θcosθ ○ 𝑐𝑜𝑠(2𝜃) = 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 = 2𝑐𝑜𝑠 2 𝜃 − 1 = 1 − 2𝑠𝑖𝑛2 2𝑡𝑎𝑛𝜃 ○ 𝑡𝑎𝑛(2𝜃) = 1−𝑡𝑎𝑛2 𝜃 ● Example: 𝑐𝑜𝑠15° √2+√3 ○ 2 Law of Sines ● 𝑠𝑖𝑛𝐴 𝑎 = 𝑠𝑖𝑛𝐵 𝑏 = 𝑠𝑖𝑛𝐶 𝑐 Law of Cosines ● Finding Sides ○ 𝑐 2 = 𝑎2 + 𝑏 2 − 2𝑎𝑏𝑐𝑜𝑠𝐶 ○ 𝑏 2 = 𝑎2 + 𝑐 2 − 2𝑎𝑐𝑐𝑜𝑠𝐵 ○ 𝑎2 = 𝑏 2 + 𝑐 2 − 2𝑏𝑐𝑐𝑜𝑠𝐴 ● Finding Angles ○ 𝑐𝑜𝑠𝐴 = ○ 𝑐𝑜𝑠𝐵 = ○ 𝑐𝑜𝑠𝐶 = 𝑏 2 +𝑐 2 −𝑎2 2𝑏𝑐 𝑎2 +𝑐 2 −𝑏2 2𝑎𝑐 𝑎2 +𝑏 2 −𝑐 2 2𝑎𝑏