Trig Identities that I can't get a grip on

  • Thread starter Fractal314
  • Start date
  • #1
14
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[tan(pi/4+x)-tan(pi/4-x)]/[tan(pi/4+x)+tan(pi/4-x)]=2sinxcosx

I tried to prove this trig identity but I an really stuck. I think tan of pi/4 is '1', and if I do that then my numerator becomes zero, thus zero=2sinxcosx. But that can't be right, so I don't know what to do now.


LS= (1+tanx-1-tanx)/(1+tanx+1-tanx)

I get 0=2sinxcosx

Any thoughts? Also, I am wondering where this advanced formatting option is or how to do it.
 

Answers and Replies

  • #2
22
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You will need to use the identities

tan(A+B)=(tanA+tanB)/(1-tanA*tanB)

tan(A-B)=(tanA-tanB)/(1+tanA*tanB)
 
  • #3
14
0
Thankyou Overt, I did not see it.

So would I be right in assigning 'y' as pi/4? and let 'x' be 'x'?

For tan(pi/4+x) I will instead get (tanx+tanpi/4)/(1-tanxtanpi/4)?
 
  • #4
22
0
Correct!
 

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