## Wednesday, January 24, 2007

### Precalculus 5.2 notes: Verifying Trigonometric Identities

5.2 Verifying Trigonometric Identities

Example 1: sin x = 0 is a conditional equation that is true for only some values in the domain.

x = {0, π, 2 π, 3 π, etc..} so x = n π

Example 2: sin2x = 1 – cos2x is an identity and it is true for all real numbers x.

Guidelines:

1. Work with one side of the equation at a time. Usually the more complicated side first.
2. Look for opportunities to factor an expression, add fractions, square a binomial, or create a monomial denominator.
3. Look for opportunities to use the fundamental identities. Note which functions are in the final expression you want. Sines and Cosines pair up well, as do secants and tangents, and cosecants and cotangents.
4. If the preceding guidelines do not help, try converting all terms to sines and cosines.
5. Try something! Even making an attempt that leads to a dead end gives insight.

A). Verifying a Trigonometric Identity

Example 1: cos2β – sin2 β = 2 cos2 β – 1

cos2β –(1 – cos2 β) = 2 cos2 β – 1

cos2β – 1 + cos2 β = 2 cos2 β – 1

2 cos2 β – 1 = 2 cos2 β – 1

Example 2: 2 – csc2x = 1 – cot2x

2 – (1 + cot2x) = 1 – cot2x

2 – 1 – cot2x = 1 – cot2x

1 – cot2x = 1 – cot2x

B). Combining Fractions before using Identities

Example 3: (cot α)/ (csc α – 1) = (csc α + 1)/(cot α)

(cot2α)/ ((csc α – 1)(cot α)) = (csc α + 1)/(cot α)
(csc2α – 1)/ ((csc α – 1)(cot α)) = (csc α + 1)/(cot α)

(csc α + 1) (csc α – 1)/ ((csc α – 1)(cot α)) = (csc α + 1)/(cot α)

(csc α + 1)/ (cot α) = (csc α + 1)/(cot α)

Example 4:

(cos x – cos y)/ (sin x + sin y) + (sin x – sin y)/ (cos x + cos y) = 0

((cos x – cos y)(cos x + cos y) + (sin x – sin y)(sin x + sin y))/ ((sin x + sin y)(cos x + cos y)) = 0

(cos2x – cos2y + sin2x – sin2y)/ ((sin x + sin y)(cos x + cos y)) = 0

(cos2x + sin2x – sin2y– cos2y)/ ((sin x + sin y)(cos x + cos y)) = 0

(cos2x + sin2x – (sin2y+ cos2y))/ ((sin x + sin y)(cos x + cos y)) = 0

(1 – 1)//(sin x + sin y)(cos x + cos y) = 0

0 = 0

C) Dealing with square roots and absolute value

Example 5:
(the square root of ( (1 – cos x)/(1 + cos x))) =
(1 – cos x) / (the absolute value of sin x)

(the square root of ( ((1 – cos x)(1 – cos x))/((1 + cos x)(1 – cos x))) =
(1 – cos x) / (the absolute value of sin x)

(the square root of ( (1 – cos x)2/(1 – cos2x)) =
(1 – cos x) / (the absolute value of sin x)

(the square root of ( (1 – cos x)2/(sin2x))) =
(1 – cos x) / (the absolute value of sin x)

(1 – cos x)/(the absolute value of sin x) =
(1 – cos x) / (the absolute value of sin x)

D). Using properties of logarithms and trigonometric identities to verify the identity.

Example 6: ln (absolute value of(sec x)) = - ln (absolute value of (cos x))

ln (absolute value of (sec x)) = ln (absolute value of (1/ cos x))

ln (absolute value of (sec x)) = ln (absolute value of (sec x))

E). Use the co-function identities to evaluate the expression

Example 7: cos2 14º + cos2 76º

sin2(90º - 14º) + cos2 76º

sin2(76º) + cos2 76º

1

Example 8: sin2 12º + sin2 40º + sin2 50º + sin2 78º

cos2 (90º - 12º) + cos2 (90º - 40º) + sin2 50º + sin2 78º

cos2 78º + cos2 50º + sin2 50º + sin2 78º

1 + 1

2

F). True or False: Is the square root of (tan2x) = tan x ?

Let x = - 45º

The square root of (tan2- 45º) = tan - 45º

The square root of ((-1)2) = -1

1 = -1 is false

Example 9:

tan5x = (tan3x)( sec2x) - tan3x

tan5x = tan3x (sec2x - 1)

tan5x =( tan3x)( tan2x)
tan5x = tan5x

5.2 homework: #44; pg. 389; 3 – 9 odd, 19 – 27 odd, 34, 35, 37, 43, 49, 55, 59, 71 – Quiz on 5.1 and 5.2 next class!!