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Another trig expression

  • Thread starter mileena
  • Start date
  • #1
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Homework Statement



Simplify:

(sin t + cos t)2
/
sin t cos t

Homework Equations



tan t = sin t/cos t
cot t = cos t/sin t
cot t = 1/tan t

The Attempt at a Solution



(sin t + cos t)2
/
sin t cos t


sin2 t + 2sin t cos t + cos2 t
/
sin t cos t


(sin t/cos t) + 2 + (cos t/sin t)

tan t + cot t + 2

tan t + (1/tan t) + 2


(tan2 t + 2tan t + 1)
/
tan t


(tan t + 1)2
/
tan t

The above answer look wrong. I don't know what else to do though.
 
Last edited:

Answers and Replies

  • #2
Notice you have: sin2t + cos2t......
That will simplify to 1 + 2sintcost / sintcost

Try going from there..

Are you converting to tan because the books answer is in tan form?? Swapping to Tan isnt necessary. If anything 1/sintcost can become csctsect
 
  • #3
haruspex
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It's often not clear what should be regarded as the simplest form. If you define at as minimising the number of references to trig functions, you can get this down to one reference. Are you familiar with a formula involving sin(2x)?
 
  • #4
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Notice you have: sin2t + cos2t......
That will simplify to 1 + 2sintcost / sintcost

Try going from there..

Are you converting to tan because the books answer is in tan form?? Swapping to Tan isnt necessary. If anything 1/sintcost can become csctsect
Oh duh! We didn't review any of the Pythagorean Identities in class, so I just assumed they wouldn't be used here. That, and I suck at trigonometry!

Here it is again:

(sin t + cos t)2
/
sin t cos t =


1/(sin t cos t) =

csc t sec t
 
  • #5
129
0
It's often not clear what should be regarded as the simplest form. If you define at as minimising the number of references to trig functions, you can get this down to one reference. Are you familiar with a formula involving sin(2x)?
Well, probably not. I just know the Law of Sines, Law of Cosines, and Pythagorean Identities.

But I do admit this forum is giving me a great "crash course" in trigonometry.
 
  • #6
Oh duh! We didn't review any of the Pythagorean Identities in class, so I just assumed they wouldn't be used here. That, and I suck at trigonometry!

Here it is again:

(sin t + cos t)2
/
sin t cos t =


1/(sin t cos t) =

csc t sec t
You have to FOIL out the parenthesis. (sint + cost)2 does not equal 1
You will have:

sin2t + cos2t + 2sintcost / sintcost

which gives you when simplified:

1+ 2sintcost / sintcost

Now, simplify.....
 
Last edited:
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  • #7
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Ok, I see what I did wrong (I took sin x + cos x = 1 when in reality it is sin2 x + cos2 x =1).

I think I got it now (or I hope):

1+ 2sin t cos t / sin t cos t

[1/sin t cos t] + [ (2sin t cos t) / (sin t cos t) ]

csc t sec t + 2
 
  • #8
haruspex
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Ok, I see what I did wrong (I took sin x + cos x = 1 when in reality it is sin2 x + cos2 x =1).

I think I got it now (or I hope):

1+ 2sin t cos t / sin t cos t

[1/sin t cos t] + [ (2sin t cos t) / (sin t cos t) ]

csc t sec t + 2
That's a right answer. To get it down to one trig reference you can use sin(2x) = 2 sin(x)cos(x).
 
  • #9
129
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Thank you! It's a miracle I did it correctly, with the forum's help, of course. A lot of help, actually :)
 

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