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Trig identities

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



cos+sintan/sinsec=csc

Homework Equations





The Attempt at a Solution


(cos+sin(sin/cos))/(sin/1/cos)
(cos+sin^2sincos)/(sin/cos)
(cos+sin^2sincos)X(cos/sin)
 

Answers and Replies

  • #2
590
0
How did you go from
(cos+sin(sin/cos))/(sin/1/cos) to
(cos+sin^2sincos)/(sin/cos) ?

The cosine should become the LCM in the numerator.
 
  • #3
4
0
i think i multiplied sin by sin and cos i wouldnt doubt that i did it wrong can u correct me i really would love to fix this !!!
 
  • #4
4
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where were u saying cos shud become the LCM in the numerator?
 
  • #5
33,153
4,838
i think i multiplied sin by sin and cos i wouldnt doubt that i did it wrong can u correct me i really would love to fix this !!!
Have you ever heard of punctuation? The quote above appears to be three different sentences. Using periods at the ends of your sentences would make what you say easier to understand.

A similar problem exists with your first post (in addition to the omission of arguments of the functions shown):
cos+sintan/sinsec=csc
As you have written it, most people in this forum would interpret the above as:
[tex]cos(x) + \frac{sin(x)tan(x)}{sin(x)sec(x)} = csc(x)[/tex]

I suspect that what you really meant, though, was this:
[tex]\frac{cos(x) +sin(x)tan(x)}{sin(x)sec(x)} = csc(x)[/tex]

If you don't know how to use LaTeX, you can write the equation above like so:
(cos(x) +sin(x)tan(x))/(sin(x)sec(x))=csc(x)
 

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