Please Help: Trig- simplifying, factoring, verifying

[gozilla]

I have a take-home quiz that I need to do well on. I am having trouble with these: I'm really confused on sine and cosine by themselves.

1) Simplify: [1/(1 + sinx)] + [1/(1 - sinx)]

I have no idea where to start.

2) Factor and then simplify: (sec^2x - 1) / (1 + secx)

All I have so far is (secx - 1)(secx + 1)

3) Factor then simplify: cos^4x - 2cos^2 + 1

I have no idea here.

4) Verify: sin^4x - cos^4x = 2sin^2x - 1

I have no idea here either.

Really appreciate it.

 

[Mrspi]

I have a take-home quiz that I need to do well on. I am having trouble with these: I'm really confused on sine and cosine by themselves.

1) Simplify: [1/(1 + sinx)] + [1/(1 - sinx)]

I have no idea where to start.

2) Factor and then simplify: (sec^2x - 1) / (1 + secx)

All I have so far is (secx - 1)(secx + 1)

3) Factor then simplify: cos^4x - 2cos^2 + 1

I have no idea here.

4) Verify: sin^4x - cos^4x = 2sin^2x - 1

I have no idea here either.

Really appreciate it.

Since this is a take-home quiz, I'll just give you a couple of hints:

1) get a common denominator for the fractions and combine them. After you do that, inspect the fundamental identitities (in particular, the Pythagorean identity, sin<SUP>2</SUP>x + cos<SUP>2</SUP> x = 1) to see if you can simplify.

2) Did you try reducing the fraction after you factored the numerator??

3) You might try letting y = cos<SUP>2</SUP> x, and rewriting your expression as

y<SUP>2</SUP> - 2y + 1
Can you factor THAT? After you do, back-substitute cos<SUP>2</SUP> x for the y in the factored expression, and once again study the Pythagorean identity to see if you can simplify.

4) An expression of the form a<SUP>2</SUP> - b<SUP>2</SUP> factors as (a + b)(a - b)

a<SUP>4</SUP> - b<SUP>4</SUP> is the same thing as
(a<SUP>2</SUP>)<SUP>2</SUP> - (b<SUP>2</SUP>)<SUP>2</SUP>
and it factors as (a<SUP>2</SUP> + b<SUP>2</SUP>)(a<SUP>2</SUP> - b<SUP>2</SUP>)

Your expression is of this form.....factor it using the same pattern. Then, it's back to the Pythagorean identity again.....you should be able to simplify.

 

[soroban]

Hello, gozilla!

1) Simplify: \(\displaystyle \,\frac{1}{1\,+\,\sin x}\,+\,\frac{1}{1\,-\,\sin x}\)

As Mrspi recommended, get a common denominator . . .

Multiply top and bottom of the first fraction by \(\displaystyle 1\,-\,\sin x\)
. . \(\displaystyle \frac{1\,-\,\sin x}{1\,-\,\sin x}\cdot\frac{1}{1\,+\,\sin x} \;=\;\frac{1\,-\,\sin x}{1\,-\,\sin^2x}\;=\;\frac{1\,-\,\sin x}{\cos^2x}\)

Multiply top and bottom of the second fraction by \(\displaystyle 1\,+\,\sin x\)
. . \(\displaystyle \frac{1\,+\,\sin x}{1\,+\,\sin x}\cdot\frac{1}{1\,-\,\sin x}\;=\;\frac{1\,+\,\sin x}{1\,-\,\sin^2x} \;= \;\frac{1\,+\,\sin x}{\cos^2x}\)

The problem becomes: \(\displaystyle \,\frac{1\,-\,\sin x}{\cos^2x}\,+\,\frac{1\,+\,\sin x}{\cos^2x}\)

Can you finish it now?

2) Factor and then simplify: \(\displaystyle \,\frac{\sec^2x\,-\,1}{1\,+\,\sec x}\)

Obviously, you didn't do what they suggested . . .

Factor: \(\displaystyle \,\frac{(\sec x\,-\,1)(\sec\,+\,1)}{1\,+\,\sec x}\)

Simplify: \(\displaystyle \,\frac{(\sec x\,-\,1)(\sout{1\,+\,\sec x})}{\sout{1\,+\,\sec x}} \;=\;\sec x\,-\,1\)