Factoring and Simplifying Trig Expressions

[burger]

I have a huge packet due and there are some Identity problems I'm having trouble with:

1) Factor and Simplify:

. . .2 Sin^2 X - 2 Sin^4 X

I can't figure out where to go with this one. I tried using Pythagorean identities, but that just turns it into the same thing but with Cosines rather than Sines.

2) Factor and Simplify:

. . .Sec^2 X Csc^2 X - Sec^2 X - Csc^2 X +1

I think I may have solved this one. I turned it into:

. . .2 + cot^2 X + tan^2 X - 1 - tan^2 X - 1 - cot^2 X + 1

...which I think simplifies down to 1.

3) Factor and Simplify:

. . .tan X(Cos X - Csc X)

I worked this one out and got -Cos^2 X Sin X but then I graphed them and thats totally wrong

Sorry if it seems like I'm asking for a lot but I'm really not good at this stuff

 

[stapel]

1) 2 sin²(x) - 2 sin⁴(x)

To factor, try taking the common factor out front:

. . . . .2 sin²(x)[1 - sin²(x)]

Then replace the contents of the brackets by using the appropriate Pythagorean Identity, and then apply a double-angle formula.

2) sec²(x) csc²(x) - sec²(x) - csc²(x) + 1

(It is unfortunate that you didn't show your work, so we cannot check your steps.) Note that sec(ß) = 1/cos(ß) and csc(ß) = 1/sin(ß). Then the above becomes:

. . . . .1 / [sin²(x) cos²(x)] - 1 / cos²(x) - 1 / sin²(x) + 1

Converting the fractions to a common denominator gives

. . . . .[1 - sin²(x) - cos²(x)] / [sin²(x) cos²(x)] + 1

By using a Pythagorean Identity within the brackets, you can arrive at your answer of "1", which is correct.

3) tan(x) [cos(x) - csc(x)]

It is often (usually?) easier to work with sines and cosines, so:

. . . . .[sin(x)/cos(x)][cos(x) - 1/sin(x)]

Multiplying through gives:

. . . . .sin(x) - 1/cos(x)

Converting to a common denominator gives:

. . . . .[sin(x) cos(x) - 1] / cos(x)

I don't know how much more could be done with this. Of course, I could be missing something....

Eliz.

 

[burger]

In the second problem I fail to see how [1 - sin2(x) - cos2(x)] / [sin2(x) cos2(x)] + 1 works out to be 0 + 1




EDIT: wait I see it now thank you

 

[soroban]

Hello, burger!

Ms. Stapel gave you some very good advice . . .

1) Factor and simplify: \(\displaystyle \:2\cot\sin^2x\,-\,2\cdot\sin^4x\)

You seem to be familiar with the Pythagorean Identities . . . good!

Did you follow their directions? . Factor and simplify.

We have:\(\displaystyle \:2\cdot\sin^2x\,-\,2\cdot\sin^4x\)

Factor: \(\displaystyle \;2\cdot\sin^2x\underbrace{\left(1 \,-\,\sin^2x)}\)
We have: \(\displaystyle \;2\sin^2x\,\cdot\,\cos^2x\)

2) Factor and simplify: \(\displaystyle \:\sec^2x\cdot\csc^2x\,-\,\sec^2x\,-\,\csc^2x\,+\,1\)

Same advice: factor.

We have: \(\displaystyle \:\underbrace{\sec^2x\cdot\csc^2x\,-\,\sec^2x}\,-\underbrace{\csc^2x\,+\,1}\)

. . .Factor: \(\displaystyle \;\sec^2x\left(\csc^2x\,-\,1\right)\,-\,\left(\csc^2x\,-\,1)\)

. . .Factor: . . . . \(\displaystyle \underbrace{(\csc^2x\,-\,1)}\underbrace{(\sec^2x\,-\,1)}\)

Therefore: . . . . . . . \(\displaystyle \cot^2x\;\cdot\;\tan^2x\;\;=\;\;1\)

3) Factor and simplify: \(\displaystyle \:\tan x(\cos x\,-\,\csc x)\)

Change everything to sines and cosines . . .

We have: \(\displaystyle \L\:\frac{\sin x}{\cos x}\left(\cos x \,-\,\frac{1}{\sin x}\right) \;=\;\frac{\sin x}{\sout{\cos x}}\cdot\frac{\sout{\cos x}}{1}\,-\,\frac{\sout{\sin x}}{\cos x}\cdot\frac{1}{\sout{\sin x}}\)

. . \(\displaystyle \L=\;\sin x \,- \,\frac{1}{\cos x}\;=\;\sin x\,-\,\sec x\)