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MCRFlyleaf6
Nov 23, 2008, 12:53 PM
I really need to get these identities right so I can make up for the test I just faile on this stuff. I have 3 questions I am stuck on:

simplify: cot(x) + tan(x)


verify: sec^4(x) -1= tan^4(x) + 2tan^2(x)


simplify: 1-cos(x)
sec(x)-1


If you can help I really appreciate it!! I can use all the help I can get

MCRFlyleaf6
Nov 23, 2008, 01:03 PM
I think I may have the 1st one up there but Im not quite sure if it's the simplest form or even if I'm approaching it right, I have:

cotx + tanx substitute cot and tan for sin and cos
cosx/sinx + sinx/cosx then make a common denominator

(cosx/cosx)x(cosx/sinx) + (sinx/cosx)x(sinx/sinx) then combine everything

(cos^2x)/(cosxsinx) + (sin^2x)/(cosxsinx)
simplify

(cos^2x + sin^2s)/cosxsinx then you can use a pythagorean identity

and I end up with:
1/ cosxsinx

galactus
Nov 23, 2008, 01:12 PM
verify: sec^{4}(x) -1= tan^{4}(x) + 2tan^{2}(x)


This one isn't bad if you use the identity sec^{2}(x)=tan^{2}(x)+1

(sec^{2}(x))^{2}-1

(tan^{2}(x)+1)^{2}-1

Now, expand out and see what you get. I bet it's the right side.:)

galactus
Nov 23, 2008, 01:13 PM
I think I may have the 1st one up there but Im not quite sure if its the simplest form or even if im approaching it right, I have:

cotx + tanx substitute cot and tan for sin and cos
cosx/sinx + sinx/cosx then make a common denominator

(cosx/cosx)x(cosx/sinx) + (sinx/cosx)x(sinx/sinx) then combine everything

(cos^2x)/(cosxsinx) + (sin^2x)/(cosxsinx)
simplify

(cos^2x + sin^2s)/cosxsinx then you can use a pythagorean identity

and I end up with:
1/ cosxsinx

That looks good to me. :D

You could also note that cos(x)sin(x)=\frac{sin(2x)}{2}

That gives \frac{2}{sin(2x)}

galactus
Nov 23, 2008, 01:17 PM
simplify: \frac{1-cos(x)}{sec(x)-1}


If you can help I really appreciate it!! I can use all the help I can get

For this one, convert the sec to 1/cos and simplify. \frac{1-cos(x)}{\frac{1}{cos(x)}-1}.

Do the algebra and it whittles down to something very simple.

Let me know if you remain stuck.

Unknown008
Nov 26, 2008, 02:35 AM
That looks good to me. :D

You could also note that cos(x)sin(x)=\frac{sin(2x)}{2}

That gives \frac{2}{sin(2x)}

Hey! But if I was given that question... could I instead put in cosec(x)sec(x) ?
That seems easier for my level. I've not yet worked with coefficients in front of the x in trigonometry...