How Can I Simplify These Trigonometric Expressions?

[mathopressor]

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

2.)simplify 2cot^(2)x -3cotx - 9 / cot^(2)x - 9

3.)simplify 5cos^(4)x - 5sin^(4)x and write in terms of cos x.

4.)simplify tan^(2)x + (1 + sec x)^2 write in terms of sec x.

 

[Mentallic]

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 $\cos^2{x}$ term. Remember this useful formula

$$\sin^2{x}+\cos^2{x}=1$$

so

$$\cos^2{x}=1-\sin^2{x}$$

if you apply this substitution, then you'll have everything in terms of sin and will probably be able to simplify even further.

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

$$2\cot^2{x}-3\cot{x}-\frac{9}{\cot^2{x}}-9$$

Begin by letting $y=\cot{x}$ 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.

3.)simplify 5cos^(4)x - 5sin^(4)x and write in terms of cos x.

Can you factorize $x^4-y^4$ as a difference of two squares?

4.)simplify tan^(2)x + (1 + sec x)^2 write in terms of sec x.

That important formula from earlier

$$\sin^2{x}+\cos^2{x}=1$$

if we divide through by $\cos^2{x}$ then we get

$$\frac{\sin^2{x}}{\cos^2{x}}+\frac{\cos^2{x}}{\cos^2{x}}=\frac{1}{\cos^2{x}}$$

$$\tan^2{x}+1=\sec^2{x}$$

So now you can use this to get rid of the tan(x) term in your expression.