mathopressor said:
these are a couple more math problems I am having trouble with. if you can help me i would appreciate it. if you could just put the number of the problem your helping me with that would be great
1)simplify -3sin^(5)x - 3sin^(3)x cos^(2)x
What is the highest common factor in this expression?
Also, notice that almost everything is in terms of sin but we have a [itex]\cos^2{x}[/itex] term. Remember this useful formula
[tex]\sin^2{x}+\cos^2{x}=1[/tex]
so
[tex]\cos^2{x}=1-\sin^2{x}[/tex]
if you apply this substitution, then you'll have everything in terms of sin and will probably be able to simplify even further.
mathopressor said:
2.)simplify 2cot^(2)x -3cotx - 9 / cot^(2)x - 9
You should put brackets around the numerator and denominator as so
(2cot^(2)x -3cotx - 9) / (cot^(2)x - 9)
else it could be misinterpreted as being
[tex]2\cot^2{x}-3\cot{x}-\frac{9}{\cot^2{x}}-9[/tex]
Begin by letting [itex]y=\cot{x}[/itex] so you'll have a quadratic in y in the numerator and denominator, then factorize those quadratics and you should be able to cancel out a common factor, and then convert back to cot(x) at the end.
mathopressor said:
3.)simplify 5cos^(4)x - 5sin^(4)x and write in terms of cos x.
Can you factorize [itex]x^4-y^4[/itex] as a difference of two squares?
mathopressor said:
4.)simplify tan^(2)x + (1 + sec x)^2 write in terms of sec x.
That important formula from earlier
[tex]\sin^2{x}+\cos^2{x}=1[/tex]
if we divide through by [itex]\cos^2{x}[/itex] then we get
[tex]\frac{\sin^2{x}}{\cos^2{x}}+\frac{\cos^2{x}}{\cos^2{x}}=\frac{1}{\cos^2{x}}[/tex]
[tex]\tan^2{x}+1=\sec^2{x}[/tex]
So now you can use this to get rid of the tan(x) term in your expression.