Simplify

[coder797]

Please help me...
Simplify:
sinx+sin2x+sin3x / cosx+cos2x+cos3x = <----this is one fraction
solution is tg2x

 

[stapel]

sinx+sin2x+sin3x / cosx+cos2x+cos3x = <----this is one fraction
solution is tg2x

When you say "this is one fraction", are you indicating that you left off the grouping symbols? Also, are the numbers after the "sin" and before the "x" meant to be part of the argument, or are they meant to be exponents? Does "tg" mean "tan" (for "tangent")? Does "2x" indicate "tan^2(x)" or "tan(2x)"?

What have you tried? How far have you gotten? Where are you stuck? Showing your work will help us figure out what you mean, and will enable us to help you get un-stuck, so please be complete. Thank you!

 

[Deleted member 4993]

Please help me...
Simplify:
(sinx+sin2x+sin3x) / (cosx+cos2x+cos3x) = <----this is one fraction
solution is tg2x

Use

sin(A) + sin(B) = 2*sin[(A+B)/2] * cos[(A-B)/2]

cos(A) + cos(B) = 2*cos[(A+B)/2] * cos[(A-B)/2]

Apply these to 'x' and '3x' ................... and show us what you get.

 

[coder797]

Use

sin(A) + sin(B) = 2*sin[(A+B)/2] * cos[(A-B)/2]

cos(A) + cos(B) = 2*cos[(A+B)/2] * cos[(A-B)/2]

Apply these to 'x' and '3x' ................... and show us what you get.

2*sinxcosx + 2*sin2xcosx / cos^2 x-sin^2 x + 2+cos2xcosx

im stuck here

 

[Deleted member 4993]

(2*sinxcosx + 2*sin2xcosx) / (cos^2 x-sin^2 x + 2+cos2xcosx)

im stuck here

First - you need to use grouping symbols () as shown to indicate proper order of operation.

Please show us how did you arrive at the numerator and the denominator.