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: 6,436 Thanks: 562  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  
Newbie Joined: Dec 2011 Posts: 12 Thanks: 0  Re: Trig Identities Quote:
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: 464 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  
Global Moderator Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4  Re: Trig Identities Quote:
The second term already had sin(x) in the denominator. 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: 464 Math Focus: Calculus/ODEs  Re: Trig Identities
1.) Another approach: 2.) Another approach: 3.) 

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