For number 1 recall
For number 2 make and factor the quadratic in terms of
Post your working for those 2 first after we have those nailed we'll move onto the next ones.
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
Thanks in advance!
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.
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.
#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
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!!