# Thread: Would someone be willing to help me with my Trigonometry makeup work?

1. ## Would someone be willing to help me with my Trigonometry makeup work?

I was pretty sick the past few days and had to miss a few days. Going through my makeup work, I've ran into trouble with the following problems. Out of around 68 problems, these 6 are the only ones that stumped me.

Anyway, the directions are

1&2, Factor the expression and use the fundamental identities to simplify
3&4, perform the multiplication and use the fundamental identities to simplify
5&6, perform the addition or subtraction and use the fundamental identities to simplify

2. ## Re: Would someone be willing to help me with my Trigonometry makeup work?

For number 1 recall $\displaystyle a^2-b^2 = (a-b)(a+b)$

For number 2 make $\displaystyle u = \sin^2 x$ and factor the quadratic in terms of $\displaystyle u$

Post your working for those 2 first after we have those nailed we'll move onto the next ones.

3. ## Re: Would someone be willing to help me with my Trigonometry makeup work?

For 1, would I change the denominator to sinx-1, making the problem (csc^2x-1)(sinx-1)?

As for 2, I have no idea where to even begin.

4. ## Re: Would someone be willing to help me with my Trigonometry makeup work?

No, you would do as Pickslides has advised, and recognise that the numerator is a difference of two squares that can be factorised.

How can you have no idea where to begin for 2? Pickslides has clearly told you to start by making a substitution so that the quadratic can be easily seen and worked with.

5. ## Re: Would someone be willing to help me with my Trigonometry makeup work?

Sorry, incredibly tired. 1 just clicked though.

so it'd be

csc^2(x)-1/csc(x)-1
=(csc(x)-1)(csc(x)+1)/(csc(x)-1)
=csc(x)+1

right?

6. ## Re: Would someone be willing to help me with my Trigonometry makeup work?

Correct. Just make sure to use brackets where they're needed (i.e. the first line).

7. ## Re: Would someone be willing to help me with my Trigonometry makeup work?

So for 2, am I doing this right thus far?

1-sin^2(2x)-sin^4(x)
=(1-sin^2(x))-sin^2(x)-(sin^2(x))(sin^2(x))

8. ## Re: Would someone be willing to help me with my Trigonometry makeup work?

No, try expanding what you wrote, you won't get what you started with.

Do as Pickslides suggested, write it as 1 - sin^2(x) + [sin^2(x)]^2, and replace each sin^2(x) with u... This should simplify the problem greatly.

So, 1-u+u^2?

Yep.

11. ## Re: Would someone be willing to help me with my Trigonometry makeup work?

#2
1-sin^2(2x)-sin^4(x)
=1-(sin 2x)^2-sin^4(x)
=1-(2 sin x cos x)^2-sin^4(x)
=1-4 sin^2 x cos^2 x-sin^4(x)
=1-4 sin^2 x (cos^2 x)-sin^4(x)
=1-4 sin^2 x (1- sin^2 x)-sin^4(x)
=1-4 sin^2 x+ 4 sin^4 x)-sin^4(x)
=1-4 sin^2 x+ 3 sin^4 x
=1-3 sin^2 x-sin^2 x + 3 sin^4 x
=(1-3 sin^2 x)-sin^2 x(1- 3 sin^2 x)
=(1-3 sin^2 x)(1 -sin^2 x)
=(1-3 sin^2 x)cos^2 x

I'm so lost on this one

#3
[cot(x)+csc(x)][cot(x)-csc(x)]
cot^2(x)-csc^2(x)
[csc^2(x)]/[sec^2(x)]-[csc^2(x)]/1
[csc^2(x)]/[sec^2(x)]-[csc^2(x)][sec^2(x)]/[sec^2(x)]

Where to now though?

#4
[3-3sin(x)][3+3sin(x)]
9[1-sin(x)][1+sin(x)]
9[sin^2(x)]
=9cos^2(x)

#5
1/[sec(x)+1]-1/[sec(x)-1]
[sec(x)-1]/[sec^2(x)-1]-[sec(x)+1]/[sec^2(x)-1]

Now what though?

#6
[cos(x)]/[1+sin(x)]+[1+sin(x)]/[cos(x)]
=[cos^2 (x)+ (1+ sin x)^2}/[1+sin(x)][cos(x)]
=[cos^2 (x)+ 1+ 2 sin x+ sin^2 x } / [1+sin(x)][cos(x)]
=[(cos^2 (x)+ sin^2 x) + 1+ 2 sin x} / [1+sin(x)][cos(x)]
=[1+ 1+ 2 sin x} / [1+sin(x)][cos(x)]
=[2+ 2 sin x} / [1+sin(x)][cos(x)]
=2[1+ sin x} / [1+sin(x)] [cos(x)]
=2/ cos(x)
= 2 sec x

12. ## Re: Would someone be willing to help me with my Trigonometry makeup work?

Originally Posted by oddball7465
#2
1-sin^2(2x)-sin^4(x)
=1-(sin 2x)^2-sin^4(x)
=1-(2 sin x cos x)^2-sin^4(x)
=1-4 sin^2 x cos^2 x-sin^4(x)
=1-4 sin^2 x (cos^2 x)-sin^4(x)
=1-4 sin^2 x (1- sin^2 x)-sin^4(x)
=1-4 sin^2 x+ 4 sin^4 x)-sin^4(x)
=1-4 sin^2 x+ 3 sin^4 x
=1-3 sin^2 x-sin^2 x + 3 sin^4 x
=(1-3 sin^2 x)-sin^2 x(1- 3 sin^2 x)
=(1-3 sin^2 x)(1 -sin^2 x)
=(1-3 sin^2 x)cos^2 x

I'm so lost on this one

#3
[cot(x)+csc(x)][cot(x)-csc(x)]
cot^2(x)-csc^2(x)
[csc^2(x)]/[sec^2(x)]-[csc^2(x)]/1
[csc^2(x)]/[sec^2(x)]-[csc^2(x)][sec^2(x)]/[sec^2(x)]

Where to now though?

#4
[3-3sin(x)][3+3sin(x)]
9[1-sin(x)][1+sin(x)]
9[sin^2(x)]
=9cos^2(x)

#5
1/[sec(x)+1]-1/[sec(x)-1]
[sec(x)-1]/[sec^2(x)-1]-[sec(x)+1]/[sec^2(x)-1]

Now what though?

#6
[cos(x)]/[1+sin(x)]+[1+sin(x)]/[cos(x)]
=[cos^2 (x)+ (1+ sin x)^2}/[1+sin(x)][cos(x)]
=[cos^2 (x)+ 1+ 2 sin x+ sin^2 x } / [1+sin(x)][cos(x)]
=[(cos^2 (x)+ sin^2 x) + 1+ 2 sin x} / [1+sin(x)][cos(x)]
=[1+ 1+ 2 sin x} / [1+sin(x)][cos(x)]
=[2+ 2 sin x} / [1+sin(x)][cos(x)]
=2[1+ sin x} / [1+sin(x)] [cos(x)]
=2/ cos(x)
= 2 sec x
Double check your original question 2. The middle term is NOT sin^2(2x), it's sin^2(x).

Do you know how to factorise the quadratic that you found? u^2 - u + 1

13. ## Re: Would someone be willing to help me with my Trigonometry makeup work?

1. csc^2(x-1)/csc(x-1) = (csc(x-1))(csc(x-1))/csc(x-1) = csc(x-1)/1 = 1/sin(x-1)

3. (cot(x) + csc(x))(cot(x)-csc(x)) = cot^2(x)-csc(x)cot(x)+csc(x)cot(x)-csc^2(x) = cot^2(x)-csc^2(x) = 1

4. (3-3sin(x))(3+3sin(x)) = 9+9sin(x)-9sin(x)-9sin^2(x) = 9-9sin^2(x) = 9(1-sin^2(x)) = 9*cos^2(x) = 9cos^2(x)

5. 1/sec(x+1) + 1/sec(x+1) = 2/sec(x+1) = 2*1/sec(x+1) = 2*cos(x+1) = 2cos(x+1)

6. cos(x)/(1+sin(x)) + (1+sin(x))/cos(x) = cos^2(x)/cos(x)(1+sin(x)) + (1+sin(x))(1+sin(x))/cos(x)(1+sin(x)) =
cos^2(x) + 1+2sin(x)+sin^2(x)/cos(x)(1+sin(x)) = 1+1+2sin(x)/cos(x)(1+sin(x)) = 2+2sin(x)/cos(x)(1+sin(x)) =
2(1+sin(x)/cos(x)(1+sin(x)) = 2/cos(x) = 2*1/cos(x) = 2*sec(x) = 2sec(x)

Hope this helps. Good luck!!