1. ## Simplifying trig functions

1. (secx)(cotx)(sinx)

2. (cos^2x)(sin^2x)-(sin^2x)

3. secx(secx - cosx)

4. (cos^2x)(tanx)/(sinx)

2. Originally Posted by live_laugh_luv27
1. (secx)(cotx)(sinx)
This should be clear after you convert to sines and cosines.

2. (cos^2x)(sin^2x)-(sin^2x)
Factor out a $\displaystyle \sin^2x.$ What is $\displaystyle \cos^2x-1?$ (Know your identities!)

3. secx(secx - cosx)
Convert to cosines, get a common denominator in the second factor, and simplify. (Again, you will need to make use of an important identity)

4. (cos^2x)(tanx)/(sinx)
Convert to sines and cosines, and then reduce.

3. Originally Posted by Reckoner
This should be clear after you convert to sines and cosines.

I came out with 1. Is this correct?

Factor out a $\displaystyle \sin^2x.$ What is $\displaystyle \cos^2x-1?$ (Know your identities!)

I got sin^4x, is this right?

Convert to cosines, get a common denominator in the second factor, and simplify. (Again, you will need to make use of an important identity)

I'm stuck factoring inside the parentheses. I have (1/cosx) - (cos^2/cosx), = (1/cosx) - cosx. Did I do something wrong?

Convert to sines and cosines, and then reduce.
I'm stuck at (cosx)(tanx)/(sinx). Did I do something wrong here?

Thank you so much for your help!

4. Originally Posted by live_laugh_luv27
This should be clear after you convert to sines and cosines.

I came out with 1. Is this correct?
Good!

Factor out a $\displaystyle \sin^2x.$ What is $\displaystyle \cos^2x-1?$ (Know your identities!)

I got sin^4x, is this right?
Not quite. The identity is $\displaystyle \sin^2x+\cos^2x=1,$ which can be written as $\displaystyle 1-\cos^2x=\sin^2x.$ But you have $\displaystyle \cos^2x-1.$ What can you do to flip these terms around?

Convert to cosines, get a common denominator in the second factor, and simplify. (Again, you will need to make use of an important identity)

I'm stuck factoring inside the parentheses. I have (1/cosx) - (cos^2/cosx), = (1/cosx) - cosx. Did I do something wrong?
Your goal in getting a common denominator is to combine the fractions. Do that, and you should notice something special about the numerator.

Convert to sines and cosines, and then reduce.

I'm stuck at (cosx)(tanx)/(sinx). Did I do something wrong here?
You mean $\displaystyle \frac{\cos^2x\tan x}{\sin x},$ right? Or was the cosine not supposed to be squared in the original post?

In either case, convert the tangent to sines and cosines. How? By using an identity that should be second nature by now.

5. 2. (cos^2x)(sin^2x)-(sin^2x)
Would you have to make the sin^2x negative? Would the answer then be -sin^4x?

3. secx(secx - cosx)
I factored out the 1/cosx, and made the inside of the parantheses equal sin^2x. Now, when I multiplied 1/cosx and sin^2x, I got sin^2x/cosx. How would I reduce this?

4. (cos^2x)(tanx)/(sinx)
I got cosx. Is this correct?

thanks!

6. Originally Posted by live_laugh_luv27
2. (cos^2x)(sin^2x)-(sin^2x)
Would you have to make the sin^2x negative? Would the answer then be -sin^4x?
Yes. What you are doing is factoring out a -1: $\displaystyle (a-b)=-(b-a).$ So

$\displaystyle \cos^2x\sin^2x-\sin^2x=\sin^2x\left(\cos^2x-1\right)$

$\displaystyle =\sin^2x\left[-\left(1-\cos^2x\right)\right]$

$\displaystyle =-\sin^2x\sin^2x=-\sin^4x.$

I kept the negative sign with the $\displaystyle 1-\cos^2x$ factor only to make sure you understood what was happening. In your own work, bringing the negative all the way out in front is fine.

3. secx(secx - cosx)
I factored out the 1/cosx, and made the inside of the parantheses equal sin^2x. Now, when I multiplied 1/cosx and sin^2x, I got sin^2x/cosx. How would I reduce this?
Okay, but what happened to that extra $\displaystyle \sec x$ sitting out front?

$\displaystyle \sec x(\sec x-\cos x)$

$\displaystyle =\frac1{\cos x}\left(\frac1{\cos x}-\frac{\cos^2x}{\cos x}\right)$

$\displaystyle =\frac1{\cos x}\cdot\frac{1-\cos^2x}{\cos x}$

$\displaystyle =\frac{\sin^2x}{\cos^2x}.$

Can you see how to simplify this? No? Let me make it clearer:

$\displaystyle \frac{\sin^2x}{\cos^2x}$

$\displaystyle =\left(\frac{\sin x}{\cos x}\right)^2.$

Now what is $\displaystyle \frac{\sin x}{\cos x}?$

4. (cos^2x)(tanx)/(sinx)
I got cosx. Is this correct?
Way to go! Hooray!

7. For #3, it would be tanx^2, then? Or just tanx?

Thank you so much for all of your help...I really appreciate it!

8. Originally Posted by live_laugh_luv27
For #3, it would be tanx^2, then?
Yes!

...Or just tanx?
No! You cannot just make the exponent disappear. You are not a magician!

9. Originally Posted by Reckoner
Yes!

No! You cannot just make the exponent disappear. You are not a magician!
Ok, just checking. Thanks again!