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 March 1st, 2012, 03:51 PM #1 Newbie   Joined: Dec 2011 Posts: 12 Thanks: 0 Trig Identities I am so lost. How do I solve these three problems, please use descriptive explanation. I missed the day we did these in class. 1. sin x + cos x cot x = csc x 2. sin^4 x + 2sin^2 x cos^2 x + cos^4 x = 1 (I realize this is some kind of perfect square, but where do I began with this? I'm confused) 3. csc^4 x - cot^4 x = cot^2 x + csc^2 x Thank you!
 March 1st, 2012, 04:05 PM #2 Global Moderator   Joined: May 2007 Posts: 6,379 Thanks: 542 Re: Trig Identities All three of them are based on using sin^2 x + cos^2 x = 1. For example - the second problem is simply squaring this expression. The others will become clearer if you convert everything to sin or cos, using cot = cos/sin and csc = 1/sin. Note that problem 3 can be simplified by observing that the right side is a factor in the left side.
March 1st, 2012, 04:30 PM   #3
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Re: Trig Identities

Quote:
 Originally Posted by mathman All three of them are based on using sin^2 x + cos^2 x = 1. For example - the second problem is simply squaring this expression. The others will become clearer if you convert everything to sin or cos, using cot = cos/sin and csc = 1/sin. Note that problem 3 can be simplified by observing that the right side is a factor in the left side.
Ok so for 1. are you telling me that it should look like this:

sin x + cos x cos/sin = 1/sin ? (I changed cot= cos/sin and csc x = 1/sin) Did I do this correctly?

 March 1st, 2012, 04:36 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs Re: Trig Identities Yes (although don't forget the arguments for the trig functions...cot(x) = cos(x)/sin(x) etc.), now combine the terms with a common denominator, and the resulting numerator will be a familiar expression.
 March 1st, 2012, 04:49 PM #5 Newbie   Joined: Dec 2011 Posts: 12 Thanks: 0 Re: Trig Identities Oh my gosh, I'm getting ready to bang my head against my keyboard. So this is what I got. What am I doing wrong? Original problem: sin x + cos x cot x = csc x I changed the problem to: sin x + cos x cosx/sin x = 1/sinx Then I got a common denominator (which was obviously sin x) : sinx(sinx)/sinx + cosx(sinx)/sinx cosx/sinx = 1/sinx Then I got: sin^2 x/sinx + cosx sin x cos x / sinx = 1/sinx How do I get rid of the mess in the middle? am I changing cosx to 1/sec? I am so confused.
March 1st, 2012, 04:56 PM   #6
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Re: Trig Identities

Quote:
 Originally Posted by swish ... Then I got a common denominator ( which was obviously sin x) : sinx(sinx)/sinx +[color=#BF0000] cosx(sinx)/sinx cosx/sinx[/color] = 1/sinx ...
No.

You should have sin^2(x)/sin(x) + cos^2(x)/sin(x)
on the left hand side.

i.e.

[sin^2(x) + cos^2(x)]/sin(x)

or
1/sin(x)

 March 1st, 2012, 06:18 PM #7 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs Re: Trig Identities 1.) Another approach: $\sin(x)+\cos(x)\cot(x)=\sin(x)+\frac{\cos^2(x)}{\s in(x)}=\sin(x)+\frac{1-\sin^2(x)}{\sin(x)}=\sin(x)+\frac{1}{\sin(x)}-\sin(x)=\csc(x)$ 2.) $\sin^4(x)+2\sin^2(x)\cos^2(x)+\cos^4(x)=$$\sin^2(x )+\cos^2(x)$$^2\)=1^2=1$ Another approach: $\sin^4(x)+2\sin^2(x)\cos^2(x)+\cos^4(x)=\sin^4(x)-2\sin^2(x)\cos^2(x)+\cos^4(x)+4\sin^2(x)\cos^2(x)=$ $$$\cos^2(x)-\sin^2(x)$$^2+4\sin^2(x)\cos^2(x)=\cos^2(2x)+\sin^ 2(2x)=1$ 3.) $\csc^4(x)-\cot^4(x)=$$\csc^2(x)+\cot^2(x)$$$$\csc^2(x)-\cot^2(x)$$=$ $$$\csc^2(x)+\cot^2(x)$$\csc^2(x)$$1-\cos^2(x)$$=$$\csc^2(x)+\cot^2(x)$$\frac{\sin^2(x) }{\sin^2(x)}=\csc^2(x)+\cot^2(x)$

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