#### Read section5.2precalc.cwk text version

Section 5.2 Verifying Trigonometric Identities Notes.

There are Guidelines for Verifying Trigonometric Identities on page 360. You might find these guidelines helpful, please take some time to read them.

Example Problems.

#22 page 366 Verify the identity.

sec 6 x (sec x tan x ) - sec 4 x (sec x tan x ) = sec 5 x tan3 x Remember that there are multiple ways to do this problem. When I look at this problem, I notice that there is a subtraction sign on the left side of the equation but not on the right side. I'm going to factor the left side to see if I can get rid of the subtraction sign. sec 6 x (sec x tan x ) - sec 4 x (sec x tan x ) = sec 5 x tan3 x sec 4 x (sec x tan x )(sec 2 x - 1) = sec 5 x tan3 x tan2 x + 1= sec 2 x Using one of our Pythagorean Identities 2 2 tan x = sec x - 1 sec 4 x (sec x tan x )(sec 2 x - 1) = sec 5 x tan3 x sec 4 x (sec x tan x ) tan2 x = sec 5 x tan3 x sec 5 x tan3 x = sec 5 x tan3 x

#26 page 366 Verify the identity.

(1+ sin y )[1+ sin(-y )] = cos 2 y

I notice two things in this problem: (1) The angle measurement (-y ) must be changed. We can do this because the sine function is an odd function and sin(-y ) = -sin y . (2) We need to simplify the left side so there are no plus or minus signs.

(1+ sin y )[1+ sin(-y )] = cos 2 y (1+ sin y )[1- sin y ] = cos 2 y

1- siny + sin y - sin2 y = cos 2 y 1- sin2 y = cos 2 y sin2 y + cos2 y = 1 Using one of our Pythagorean Identities cos2 y = 1- sin2 y 1- sin2 y = cos2 y cos2 y = cos2 y

#34 page 366 Verify the identity algebraically.

1- cos 1- cos = 1+ cos sin I think it is obvious that we need to get rid of the radical symbol. We can do this by squaring both sides. 1- cos 1- cos = 1+ cos sin 1- cos 2 1- cos 2 = 1+ cos sin 1- cos (1- cos ) = 2 1+ cos ( sin ) 1- cos (1- cos ) = 1+ cos sin2

2

2

When you square the absolute value you get a positive number and the absolute value symbol is no longer necessary. sin2 + cos 2 = 1 Also from one of our Pythagorean Identities sin 2 = 1- cos2

1- cos (1- cos ) = 1+ cos sin2 2 1- cos (1- cos ) = 1+ cos 1- cos 2 Now factor 1- cos2 1- cos (1- cos ) = 1+ cos 1- cos 2 1- cos (1- cos )(1- cos ) = 1+ cos (1- cos )(1+ cos ) 1- cos (1- cos ) = 1+ cos (1+ cos ) #38 page 366 Verify the identity algebraically. sec 2 - x -1 = cot 2 x 2 We can't verify a trig identity unless the angle measurements are the same. Using a Cofunction Identity sec 2 - x = csc 2 x 2 sec 2 - x -1 = cot 2 x 2 csc 2 x -1 = cot 2 x 1+ cot 2 x = csc 2 x And once again from one of our Pythagorean Identities 2 2 cot x = csc x - 1 sec 2 - x -1 = cot 2 x 2 csc 2 x -1 = cot 2 x cot 2 x = cot 2 x

2

2

#44 page 366 Verify the identity algebraically.

sin x (1- 2cos 2 x + cos4 x ) = sin5 x When I look at this problem I notice that we have two different trig functions on the left side of the equation and only one on the right side. I'm going to try to factor the left side. sin x (1- 2cos2 x + cos4 x ) = sin5 x sin x (1- cos 2 x )(1- cos 2 x ) = sin5 x sin2 x + cos2 x = 1 Using one of our Pythagorean Identities sin2 x = 1- cos2 x sin x (1- 2cos2 x + cos4 x ) = sin5 x sin x (1- cos 2 x )(1- cos 2 x ) = sin5 x sin x sin2 x sin 2 x = sin5 x sin 5 x = sin5 x #48 page 366 Verify the identity algebraically. cot csc + 1 = csc - 1 cot I have to admit that when I look at this one I'm not sure what to do. But based on all the problems that we have completed so far, I'm willing to bet that this involves a Pythagorean Identity 1+ cot 2 = csc 2 . I'm going to multiply the right hand side of the equation by (csc - 1) . (csc - 1) cot (csc + 1) (csc -1) = csc - 1 cot (csc -1) cot csc 2 - csc + csc - 1 = csc - 1 cot (csc - 1) cot csc 2 -1 = csc - 1 cot (csc -1) cot cot2 = csc - 1 cot (csc -1) cot cot = csc - 1 csc -1

I want you to remember that there are multiple ways to do these problems. If you get stuck please try to substitute some identity so I can give you some partial credit.

Homework.

page 366/ 21-49 odd

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