Trig ientities
3 answers
http://www.sosmath.com/trig/Trig5/trig5/trig5.html
Trig identites
(1/1+tanx)+(1/1+cotx)=1
For this one, I wanted to know if 1+tanx is the same as 1+tan^2x, because 1+tan^2x=sec^2x
These are the ones that I have no idea how to do:
((cosx-cosy)/(sinx+siny))+((sinx-siny)/(cosx+cosy))=0
(sin^3x+cos^3x)/(sinx+cosx)=1-sinxcosx
csc2x=cscx/2cosx
cos3x/cosx=1-4sin^2x
**sorry I didn't type it the first time, my computer froze and posted it like that**
(1/1+tanx)+(1/1+cotx)=1
For this one, I wanted to know if 1+tanx is the same as 1+tan^2x, because 1+tan^2x=sec^2x
These are the ones that I have no idea how to do:
((cosx-cosy)/(sinx+siny))+((sinx-siny)/(cosx+cosy))=0
(sin^3x+cos^3x)/(sinx+cosx)=1-sinxcosx
csc2x=cscx/2cosx
cos3x/cosx=1-4sin^2x
**sorry I didn't type it the first time, my computer froze and posted it like that**
no, 1+tanx is not the same as 1+tan^2x
use your sum-to-product formulas:
cosx-cosy = -2 sin(x+y)/2 sin(x-y)/2
sinx+siny = 2 sin(x+y)/2 cos(x-y)/2
so the quotient is just -tan(x-y)/2
the 2nd term is similar
recall that (a+b)^3 = (a+b)(a^2-ab+b^2)
csc2x = 1/sin2x = 1/(2sinx cosx)
cos3x = cos(2x+x)
= cos2x cosx - sin2x sinx
and so on
use your sum-to-product formulas:
cosx-cosy = -2 sin(x+y)/2 sin(x-y)/2
sinx+siny = 2 sin(x+y)/2 cos(x-y)/2
so the quotient is just -tan(x-y)/2
the 2nd term is similar
recall that (a+b)^3 = (a+b)(a^2-ab+b^2)
csc2x = 1/sin2x = 1/(2sinx cosx)
cos3x = cos(2x+x)
= cos2x cosx - sin2x sinx
and so on