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 May 14th, 2018, 01:35 PM #1 Newbie   Joined: Jul 2015 From: N/A Posts: 16 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: 6,683 Thanks: 658 Is this a homework assignment?
 May 14th, 2018, 02:06 PM #3 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 632 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: 16 Thanks: 0 Thank you! Thank you!
 May 14th, 2018, 11:22 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,274 Thanks: 1959 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: 16 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: 6,683 Thanks: 658 $sin^2x+cos^2x=1$. Divide through by $sin^2x$. Result $1+cot^2x=csc^2x$.
May 15th, 2018, 01:11 PM   #8
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Quote:
 Originally Posted by nyuviolets 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)
$\sin^2{x} + \cos^2{x} = 1$

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: 6,683 Thanks: 658 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: 16 Thanks: 0 Thank you so much! I divided by sin^2x then got the answer.

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