May 14th, 2018, 01:35 PM  #1 
Newbie Joined: Jul 2015 From: N/A Posts: 14 Thanks: 0  Trig Identities (ASAP)
Question: Starting with sin$^2$x + cos$^2$x = 1 and using your knowledge of the quotient and reciprocal identities, derive an equivalent identity in term of cot x and csc x. (show all work) Thank you very much! Last edited by skipjack; May 14th, 2018 at 11:19 PM. 
May 14th, 2018, 01:47 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,629 Thanks: 622 
Is this a homework assignment?

May 14th, 2018, 02:06 PM  #3 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 620 Thanks: 85 
What can you divide each term by to make an equation with cot x and csc x?

May 14th, 2018, 02:44 PM  #4 
Newbie Joined: Jul 2015 From: N/A Posts: 14 Thanks: 0  Thank you!
Thank you!

May 14th, 2018, 11:22 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 19,887 Thanks: 1836 
As cot x and csc x are not defined when x is an integer multiple of $\pi$, there are no precisely equivalent identities.

May 15th, 2018, 02:58 AM  #6 
Newbie Joined: Jul 2015 From: N/A Posts: 14 Thanks: 0 
I thought I was able to figure it out, but I might come up with wrong answers, So that means the answer is not derived from a simple calculation, correct? If the answer for cot X is plus minus square root something, it is wrong, right? Because there are not equivalent identities as cot x and csc x are not defined....Could you please confirm if my understanding is correct? Thank you.

May 15th, 2018, 01:09 PM  #7 
Global Moderator Joined: May 2007 Posts: 6,629 Thanks: 622 
$sin^2x+cos^2x=1$. Divide through by $sin^2x$. Result $1+cot^2x=csc^2x$.

May 15th, 2018, 01:11 PM  #8  
Math Team Joined: Jul 2011 From: Texas Posts: 2,774 Thanks: 1428  Quote:
divide every term in the above equation by $\sin^2{x}$ ... $\dfrac{\sin^2{x}}{\sin^2{x}} + \dfrac{\cos^2{x}}{\sin^2{x}} = \dfrac{1}{\sin^2{x}}$ finish it ...  
May 16th, 2018, 01:58 PM  #9 
Global Moderator Joined: May 2007 Posts: 6,629 Thanks: 622 
You can also get $tan^2x+1=sec^2x$, by dividing by $cos^2x$.

May 16th, 2018, 04:16 PM  #10 
Newbie Joined: Jul 2015 From: N/A Posts: 14 Thanks: 0 
Thank you so much! I divided by sin^2x then got the answer.


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