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Re: Trig Identities

  • #1

Homework Statement



1. cos θ - 1 / cos θ = sec θ - sec^2 θ
2. (sin θ + cos θ)^2 = 1 + 2 sin θ cos θ
3. (1 / 1 + cos θ) + (1 / 1 - cos θ) = 2csc^2θ

Homework Equations



1 / cos θ = sec θ
1 / sin θ = csc θ
sin^2 θ + cos^2 θ = 1
cos^2 θ = 1 - sin^2 θ
sin^2 θ = 1 - cos^2 θ

The Attempt at a Solution



1. cos θ - 1 / 1 - sin^2 θ
cos θ / (1 + sin^2 θ) (1 - sin^2 θ)

2. sin^2 θ + cos^2 θ = 1

3. LCD : (1 + cos θ) (1 - cos θ)

1 / 1 + cos θ (1 - cos θ)

1 - cos θ / (1 + cos θ) (1 - cos θ)


1 / 1 - cos θ (1 + cos θ)

1 + cos θ / (1 - cos θ) (1 + cos θ)

I'm really confused with these problems. I appreciate the help anyway, thanks. Happy holidays.
 

Answers and Replies

  • #2
33,170
4,857

Homework Statement



1. cos θ - 1 / cos θ = sec θ - sec^2 θ
2. (sin θ + cos θ)^2 = 1 + 2 sin θ cos θ
3. (1 / 1 + cos θ) + (1 / 1 - cos θ) = 2csc^2θ

Homework Equations



1 / cos θ = sec θ
1 / sin θ = csc θ
sin^2 θ + cos^2 θ = 1
cos^2 θ = 1 - sin^2 θ
sin^2 θ = 1 - cos^2 θ

The Attempt at a Solution



1. cos θ - 1 / 1 - sin^2 θ
cos θ / (1 + sin^2 θ) (1 - sin^2 θ)
In this problem you have apparently replaced cosθ with 1 - sin^2(θ), which is incorrect.

In problems 1 and 3, you have fractional expressions with more than term in numerator or denominator. Use parentheses to show what's in the numerator or denominator.

To prove identities, the typical approach is to start with one side of the original equation, and work with it until you arrive at the other side. When you have two expressions that are equal, use = to show that.

For 1, you are trying to prove that (cos θ - 1) / cos θ = sec θ - sec^2 θ is an identity.
Note the parentheses I added. These are necessary.

I would start with the right side, and write the secant terms in their equivalent cosine forms.

2. sin^2 θ + cos^2 θ = 1
For 2, start with the left side, and expand and simplify it.

(sin θ + cos θ)^2 = ?

You should have an unbroken chain of equalities that end with 1 + 2 sin θ cos θ.

3. LCD : (1 + cos θ) (1 - cos θ)

1 / 1 + cos θ (1 - cos θ)

1 - cos θ / (1 + cos θ) (1 - cos θ)


1 / 1 - cos θ (1 + cos θ)

1 + cos θ / (1 - cos θ) (1 + cos θ)

I'm really confused with these problems. I appreciate the help anyway, thanks. Happy holidays.
 
  • #3
Okay, for 1, by starting with the right side

(1 / cos θ) - (1 / cos^2 θ), therefore that shows sec θ - sec^2 θ

- But do you multiply the numerator and demonator by the lcd which is cos^2 θ

For 2, (sin θ + cos θ)^2 = ?
- by the 2 as an exponent, would you multiply 2 by sin and cos, therefore it would be like this - sin^2 θ + cos^2 θ = ?

For 3, (1 / 1 + cos θ) + (1 / 1 - cos θ) = 2csc^2 θ
I know 1 / sin θ = csc θ and that sin^2 θ + cos^2 θ = 1 but how do i get from cos to sin
 
  • #4
For 2, i just realize what i did wrong.

(sinθ + cosθ)^2 = sin^2θ + 2sinθcosθ + cos^2θ
(sin^2θ + cos^2θ= 1), you get 1 + 2 sin θ cos θ.

But for 1, i have a question when you simplify the secant in terms of cosine. Would you cross multiply but then again there's the subtraction sign there instead of a multiplication sign? then would you multiply by the lcd to get the demoniator to be the same for both sides.

for 3, im confused on this problem.
 
  • #5
33,170
4,857
Okay, for 1, by starting with the right side

(1 / cos θ) - (1 / cos^2 θ), therefore that shows sec θ - sec^2 θ
No, start with the right side. You don't need all the conversation. Use = !
sec θ = sec2 θ = 1/(cos θ) - 1/(cos2 θ)

Yes, you can multiply by cos2 θ over itself.

- But do you multiply the numerator and demonator by the lcd which is cos^2 θ

For 2, (sin θ + cos θ)^2 = ?
- by the 2 as an exponent, would you multiply 2 by sin and cos, therefore it would be like this - sin^2 θ + cos^2 θ = ?

For 3, (1 / 1 + cos θ) + (1 / 1 - cos θ) = 2csc^2 θ
I know 1 / sin θ = csc θ and that sin^2 θ + cos^2 θ = 1 but how do i get from cos to sin
You added parentheses, but you added them in the wrong place. The left side should look like this: 1/(1 + cos θ) + 1/(1 - cos θ)

What you have would be properly interpreted as 1 + cos θ + 1 - cos θ, which is 2. I'm sure that's not what you meant.
 
  • #6
33,170
4,857
For 2, i just realize what i did wrong.

(sinθ + cosθ)^2 = sin^2θ + 2sinθcosθ + cos^2θ
(sin^2θ + cos^2θ= 1), you get 1 + 2 sin θ cos θ.
Leave out this part: (sin^2θ + cos^2θ= 1), and leave out the "you get." Use = !!!!

It should look like this:
(sinθ + cosθ)2 = sin2 θ + 2sin θ cos θ + cos2 θ = 1 + 2 sin θ cos θ

If you need to include an explanation for one step (such as replacing sin2 θ + cos2 θ by 1, put it off to the side, not in the flow of equal expressions.
But for 1, i have a question when you simplify the secant in terms of cosine. Would you cross multiply but then again there's the subtraction sign there instead of a multiplication sign? then would you multiply by the lcd to get the demoniator to be the same for both sides.

for 3, im confused on this problem.
What is the cross multiplying you're talking about? If you have an equation such as x/2 = 3/4, you can cross multiply to get 4x = 6, or x = 3/2, but you don't have an equation. What you have is an expression that you are trying to simplify or expand to make look like the other side of the equation.

Since you are not dealing with an equation, you are very limited in the things you can do, simplify an expression, expand an expression, add 0, or multiply by 1.

In problem 1, you have sec θ - sec2 θ = 1/cos θ - 1/cos2 θ. You can multiply by 1 (i.e., cos2 θ/cos2 θ), and that might be helpful.
 

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