1. (secx)(cotx)(sinx)
2. (cos^2x)(sin^2x)-(sin^2x)
3. secx(secx - cosx)
4. (cos^2x)(tanx)/(sinx)
Thanks for your help!
This should be clear after you convert to sines and cosines.
Factor out a $\displaystyle \sin^2x.$ What is $\displaystyle \cos^2x-1?$ (Know your identities!)2. (cos^2x)(sin^2x)-(sin^2x)
Convert to cosines, get a common denominator in the second factor, and simplify. (Again, you will need to make use of an important identity)3. secx(secx - cosx)
Convert to sines and cosines, and then reduce.4. (cos^2x)(tanx)/(sinx)
Good!
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?Factor out a $\displaystyle \sin^2x.$ What is $\displaystyle \cos^2x-1?$ (Know your identities!)
I got sin^4x, is this right?
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 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?
You mean $\displaystyle \frac{\cos^2x\tan x}{\sin x},$ right? Or was the cosine not supposed to be squared in the original post?Convert to sines and cosines, and then reduce.
I'm stuck at (cosx)(tanx)/(sinx). Did I do something wrong here?
In either case, convert the tangent to sines and cosines. How? By using an identity that should be second nature by now.
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!
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.
Okay, but what happened to that extra $\displaystyle \sec x$ sitting out front?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?
$\displaystyle \sec x(\sec x-\cos x)$
$\displaystyle =\frac1{\cos x}\left(\frac1{\cos x}-\frac{\cos^2x}{\cos x}\right)$
$\displaystyle =\frac1{\cos x}\cdot\frac{1-\cos^2x}{\cos x}$
$\displaystyle =\frac{\sin^2x}{\cos^2x}.$
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 $\displaystyle \frac{\sin x}{\cos x}?$
Way to go! Hooray!4. (cos^2x)(tanx)/(sinx)
I got cosx. Is this correct?