Solve Trig Identities Homework Equations

But you also have sec θ - sec2 θ, and sec θ = 1/cos θ, so you can substitute 1/cos θ for sec θ.
  • #1
priscilla98
93
0

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.
 
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  • #2
priscilla98 said:

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.

priscilla98 said:
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.

priscilla98 said:
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, I am confused on this problem.
 
  • #5
priscilla98 said:
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.

priscilla98 said:
- 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
priscilla98 said:
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.
priscilla98 said:
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, I am 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.
 

What is the purpose of solving trigonometric identities?

The purpose of solving trigonometric identities is to simplify and manipulate complex trigonometric expressions into simpler forms that are easier to work with and understand. This can also help in solving trigonometric equations and proving trigonometric identities.

What are the basic trigonometric identities?

The basic trigonometric identities include the Pythagorean identities, sum and difference identities, double angle identities, and half angle identities. These identities are used to express trigonometric functions in terms of other trigonometric functions, making it easier to manipulate and solve equations.

How do you prove a trigonometric identity?

To prove a trigonometric identity, you need to manipulate one side of the equation using algebraic and trigonometric properties until it is equal to the other side of the equation. This can involve using basic trigonometric identities, substituting values, and simplifying expressions. If both sides of the equation are equal, then the identity is proven.

What are some tips for solving trigonometric identities?

Some tips for solving trigonometric identities include using basic identities, simplifying expressions, substituting values, and being familiar with common trigonometric values and formulas. It is also helpful to use graphing calculators or trigonometric tables to check your work and verify solutions.

How can solving trigonometric identities be applied in real life?

Solving trigonometric identities can be applied in various fields such as engineering, physics, and astronomy. It can be used to solve problems involving angles and distances, such as calculating the height of a building or the trajectory of a projectile. Trigonometric identities are also used in designing and analyzing structures and circuits.

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