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May 14th, 2018, 12:35 PM   #1
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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 10:19 PM.
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May 14th, 2018, 12:47 PM   #2
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Is this a homework assignment?
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May 14th, 2018, 01:06 PM   #3
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What can you divide each term by to make an equation with cot x and csc x?
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May 14th, 2018, 01:44 PM   #4
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Thank you!

Thank you!
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May 14th, 2018, 10:22 PM   #5
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As cot x and csc x are not defined when x is an integer multiple of $\pi$, there are no precisely equivalent identities.
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May 15th, 2018, 01:58 AM   #6
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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.
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May 15th, 2018, 12:09 PM   #7
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$sin^2x+cos^2x=1$. Divide through by $sin^2x$. Result $1+cot^2x=csc^2x$.
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May 15th, 2018, 12:11 PM   #8
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Quote:
Originally Posted by nyuviolets View Post
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 ...
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May 16th, 2018, 12:58 PM   #9
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You can also get $tan^2x+1=sec^2x$, by dividing by $cos^2x$.
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May 16th, 2018, 03:16 PM   #10
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Thank you so much! I divided by sin^2x then got the answer.
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