Question

how do you solve cosx+1/sin^3=cscx/1-cosx

Answer 1

Kinda tricky, but I'll try and hint you in the right direction and see where we can get. Start by making sin^3(x) into sinx(sin^2(x)). Then change sin^2(x) into cosines using this identity.
\[\sin ^{2}x + \cos ^{2}x + 1\]
If you are able to correctly replace sin^2(x), then you'll be able to do some factoring in the denominator with difference of squares to get things to cancel.

Answer 2

ok you lost me wen you wrote that sin^2x changes to what you wrote after

Answer 3

\[(a ^{2}-b ^{2}) = (a-b)(a+b)\]

Answer 4

how do I use the identity

Answer 5

Well, the identity above lets you do a substitution. Since we know that
\[\sin ^{2}x + \cos ^{2}x = 1\], we can solvefor sin^2(x) and then replace it. So when you isolate sin^2(x), what do you have on the other side of the equal sign?

Answer 6

cos^2x???

Answer 7

1-cos^2(x)

Answer 8

oh yah

Answer 9

Subtract cos^2(x) from both sides and you see that sin^2(x) is the exact same thing as 1-cos^2(x). Because of that I can rewrite what we have to become:
\[\frac{ cosx+1 }{ sinx(1-\cos ^{2}x) }\]

Now the cosine part on bottom is a difference of squares. Use the fact that
\[(a ^{2}-b ^{2}) = (a-b)(a+b)\] to help you finish.

Answer 10

so would I hav to use that on the cos^2

Answer 11

Correct.

Answer 12

what would I hav to do after I have factored cos^2

Answer 13

What'd you factor 1-cos^2(x) into?

Answer 14

idk

Answer 15

Well 1-cos^2(x) fits the form above of a^2-b^2. So that means you can turn it into (a-b)(a+b). So what is a?

Answer 16

1???

Answer 17

Yep. So sqrt of 1 is 1, lol. So b would be cosx, right?

Answer 18

yes

Answer 19

a^2 = 1
b^2 = cos^2(x)
a = 1
b = cosx
(a^2-b^2) = (a-b)(a+b) therefore
(1^2 - cos^2(x)) = (1+cosx)(1-cosx)
Kinda see what I did?

Answer 20

yah I see

Answer 21

Awesome. SO that leaves us with:
\[\frac{ cosx+1 }{ (sinx)(cosx+1)(cosx-1) }\]

Answer 22

yes/ok

Answer 23

cosx + 1 cancels and we have.
\[\frac{ 1 }{ sinx }*\frac{ 1 }{ 1-cosx }\]
I wrote it backwards last time, my bad. This is the way it should be, lol.

Answer 24

Now just change 1/sin into csc :P

Answer 25

so wat would happen to the last cosx-1

Answer 26

never mind I know it would be csx / 1-cosx

Answer 27

Right xD

Answer 28

ok thank you

Answer 29

=D

Answer 30

would I be able to ask you one last question???

Answer 31

go for it.

Answer 32

but its a different problem would that be ok

Answer 33

Yeah, Ill see if I can do it.

Answer 34

ok well its this 1-tan^2x / 1+tan^2x=1-2sin^2x

Answer 35

This problem uses two identities. One of them is just a transformation of one we used before and the other one is a more obscure one. If you have to do a lot ofthese identity problerms, I suggest having formulas handy.

Answer 36

I got some

Answer 37

*formulas

Answer 38

Alright, cool. Well the first identity we're going to use is a transformation of:
\[\sin ^{2}x + \cos ^{2}x = 1 \]What I'm going to do is divide everything by cos^2(x). This will just modify the identity into different trig functions, but is still valid:
\[\frac{ \sin ^{2}x }{ \cos ^{2}x }+\frac{ \cos ^{2}x }{ \cos ^{2}x }= \frac{ 1 }{ \cos ^{2}x }\]this becomes
\[\tan ^{2}x + 1 = \sec ^{2}x\] Now as you should notice, you have something like this result directly in your problem :P

Answer 39

do you think you could draw it

Answer 40

Yeah, I can. This is just showing where the first identity we use comes from, though.

Answer 41

ok

Answer 42

mhm. So from there I'm going to split our problem into 2 fractions like this:
\[\frac{ 1 }{ \sec ^{2}x }-\frac{ \tan ^{2}x }{ \sec ^{2}x }\]
Now from here I can change everything into sines and cosines. Would you know how to do that?

Answer 43

sec would change to 1/cosx and the tan to sinx/cosx

Answer 44

right

Answer 45

is this the problem or is this the way it would sort of be??

Answer 46

This is me going through the problem. Im just showing you step by step. The first step was changing the bottom into sec^2(x), then we separated the fractions. Now we're turning everything into sines and cosines. So if we do what you said, we end up with:
\[\cos ^{2}x-\frac{ \sin ^{2}xcos ^{2}x }{ \cos ^{2}x }\]
Remember, secx is 1/cosx, but that also means 1/secx is cosx. That is why we have the cos^2(x) in the numerator of both fractions, we had to flip them up. Can you see how that happened?

Answer 47

umm I cant tell how you set up the equation once you turned everything into sines and cosines

Answer 48

after you turned 1+tan^x into sec^2x did you keep the top which is 1-tan^2x??

Answer 49

Ah. Well, for every term you have in the numerator, you can make itinto it's own fraction. So I took the term of 1 and made it the numerator of 1 fraction and the -tan^2(x) and made it the numerator of a 2nd fraction. For example:
\[\frac{ 1+x }{ 3+y } = \frac{ 1 }{ 3+y }+\frac{ x }{ 3+y }\]
This is exactly what I did with our equation. We had:
\[\frac{ 1-\tan ^{2}x }{ 1+\tan ^{2}x }\]I turned this into two fractions, one for each term in the numerator:
\[\frac{ 1 }{ 1+\tan ^{2}x }- \frac{ \tan ^{2}x }{ 1+\tan ^{2}x }\]Using the identity we opened with, I turned the 1+tan^2(x) into sec^2(x)
\[\frac{ 1 }{ \sec ^{2}x }-\frac{ \tan ^{2}x }{ \sec ^{2}x }\]
From there, we know that secx is the same as 1/cosx. But since we have 1/secx instead, we say that 1/secx is just cosx. Doing that allows me to rewrite the sec^2(x) parts like so:
\[\cos ^{2}x-\tan ^{2}xcos ^{2}x\]Finally, we know tanx is the same as sinx/cosx, so rewriting that I have:
\[\cos ^{2}x-\frac{ \sin ^{2}xcos ^{2}x }{ \cos ^{2}x }\]

Answer 50

I cant tell how you set this up \[\cos ^{2}x-\frac{ \sin ^{2}xcos ^{2}x }{ \cos ^{2}x }\]