Trigonometric Identities

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  • #1
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Homework Statement


Express (1+cot^2 x) / (cot^2 x) in terms of sinx and/or cosx

Homework Equations


cot(x) = 1/tan(x)
sin^2(x) + cos^2(x) = 1

The Attempt at a Solution


I do not know if I am solving this problem correctly. Is there an easier route than the way I have solved it, if it is solved correctly?

= (1+cot^2x) / (cot^2x)
= 1+ [ (cos^2x) / (sin^2x) ] ÷ [ (cos^2x) / (sin^2x) ]
= 1 + [ (cos^2x) / (sin^2x) ] x [ (sin^2x / cos^2x) ]
= [ (sin^2x / sin^2x) + (cos^2x / sin^2x) ] x [ (sin^2x) / (cos^2x) ]
= [ (sin^2x + cos^2x) / (sin^2x) ] x [ (sin^2x )/ (cos^2x) ]
= [ 1 / sin^2x ] x [ sin^2x / cos^2x]
= sin^2x / (sin^2x)(cos^2x)
= 1 / cos^2x
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Homework Statement


Express (1+cot^2 x) / (cot^2 x) in terms of sinx and/or cosx

Homework Equations


cot(x) = 1/tan(x)
sin^2(x) + cos^2(x) = 1

The Attempt at a Solution


I do not know if I am solving this problem correctly. Is there an easier route than the way I have solved it, if it is solved correctly?

= (1+cot^2x) / (cot^2x)
= 1+ [ (cos^2x) / (sin^2x) ] ÷ [ (cos^2x) / (sin^2x) ]
= 1 + [ (cos^2x) / (sin^2x) ] x [ (sin^2x / cos^2x) ]
= [ (sin^2x / sin^2x) + (cos^2x / sin^2x) ] x [ (sin^2x) / (cos^2x) ]
= [ (sin^2x + cos^2x) / (sin^2x) ] x [ (sin^2x )/ (cos^2x) ]
= [ 1 / sin^2x ] x [ sin^2x / cos^2x]
= sin^2x / (sin^2x)(cos^2x)
= 1 / cos^2x

Looks just fine to me.
 
  • #3
Ray Vickson
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Homework Statement


Express (1+cot^2 x) / (cot^2 x) in terms of sinx and/or cosx

Homework Equations


cot(x) = 1/tan(x)
sin^2(x) + cos^2(x) = 1

The Attempt at a Solution


I do not know if I am solving this problem correctly. Is there an easier route than the way I have solved it, if it is solved correctly?

= (1+cot^2x) / (cot^2x)
= 1+ [ (cos^2x) / (sin^2x) ] ÷ [ (cos^2x) / (sin^2x) ]
= 1 + [ (cos^2x) / (sin^2x) ] x [ (sin^2x / cos^2x) ]
= [ (sin^2x / sin^2x) + (cos^2x / sin^2x) ] x [ (sin^2x) / (cos^2x) ]
= [ (sin^2x + cos^2x) / (sin^2x) ] x [ (sin^2x )/ (cos^2x) ]
= [ 1 / sin^2x ] x [ sin^2x / cos^2x]
= sin^2x / (sin^2x)(cos^2x)
= 1 / cos^2x

Why don't you use the fact that
$$
\frac{1 +\cot^2 x}{\cot^2 x} = \frac{1}{\cot^2 x} + 1 ?
$$
Then you can finish off the whole thing in one more line of simple algebra (plus the identity ##\cos^2 x + \sin^2 x = 1##).
 
  • #4
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Why don't you use the fact that
$$
\frac{1 +\cot^2 x}{\cot^2 x} = \frac{1}{\cot^2 x} + 1 ?
$$
Then you can finish off the whole thing in one more line of simple algebra (plus the identity ##\cos^2 x + \sin^2 x = 1##).

so if 1 / cot^2x = tan^2x = sin^2x / cos^2x

Then all we have to do is: = (sin^2x/cos^2x ) + (cos^2x/cos^2x)
=( sin^2x + cos^2x)/ cos^2x
= 1 / cos^2x

Thanks so much haha, I didn't even notice that. Also another question for proofs. We know that sin^2x + cos^2x = 1 (pythagorean identity). Can we say the same for sinx + cosx= 1? I am pretty sure not but just double checking.
 
  • #5
Dick
Science Advisor
Homework Helper
26,263
619
so if 1 / cot^2x = tan^2x = sin^2x / cos^2x

Then all we have to do is: = (sin^2x/cos^2x ) + (cos^2x/cos^2x)
=( sin^2x + cos^2x)/ cos^2x
= 1 / cos^2x

Thanks so much haha, I didn't even notice that. Also another question for proofs. We know that sin^2x + cos^2x = 1 (pythagorean identity). Can we say the same for sinx + cosx= 1? I am pretty sure not but just double checking.

Absolutely not. Try it if ##x=\pi##.
 
  • #6
jimkris69
The brackets [ ] on your second line are incorrect
 

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