Y yasaminG New member Joined Nov 6, 2005 Messages 27 Nov 19, 2005 #1 cos ( A+B) cos B+ sin (A+B) sin B=cos A and how do we know that Tan 2@=2tan@/1-tan^2@ i really appreciate your help :!: thank you
cos ( A+B) cos B+ sin (A+B) sin B=cos A and how do we know that Tan 2@=2tan@/1-tan^2@ i really appreciate your help :!: thank you
G Gene Senior Member Joined Oct 8, 2003 Messages 1,904 Nov 19, 2005 #2 I assume you have the sum of angle formulas cos(a+b)=cos(a)cos(b)-sin(a)sin(b) sin(a+b)=sin(a)cos(b)+cos(a)sin(b) *************************** cos(a+b)cos(b)+sin(a+b)sin(b) = cos(a)cos²(b)-sin(a)sin(b)cos(b) + sin(a)sin(b)cos(b)+cos(a)sin²(b) = cos(a){cos²(b)+sin²(b)} = cos(a) *************************** tan(2@)=sin(@+@)/cos(@+@) = 2sin(@)cos(@)/(cos²(@)-sin²(@)) = 2sin(@)cos(@)/(cos²@)(1-sin²(@)/cos²(@)) = 2(sin(@)/cos(@))/(1-sin²(@)/cos²(@)) = 2tan(@)/(1-tan²(@)) Watch your ()s. 2tan(@)/1-tan²(@) is NOT the same as 2tan(@)/(1-tan²(@)) What you typed means (2tan(@)/1)-tan²(@) = 2tan(@)-tan²(@)
I assume you have the sum of angle formulas cos(a+b)=cos(a)cos(b)-sin(a)sin(b) sin(a+b)=sin(a)cos(b)+cos(a)sin(b) *************************** cos(a+b)cos(b)+sin(a+b)sin(b) = cos(a)cos²(b)-sin(a)sin(b)cos(b) + sin(a)sin(b)cos(b)+cos(a)sin²(b) = cos(a){cos²(b)+sin²(b)} = cos(a) *************************** tan(2@)=sin(@+@)/cos(@+@) = 2sin(@)cos(@)/(cos²(@)-sin²(@)) = 2sin(@)cos(@)/(cos²@)(1-sin²(@)/cos²(@)) = 2(sin(@)/cos(@))/(1-sin²(@)/cos²(@)) = 2tan(@)/(1-tan²(@)) Watch your ()s. 2tan(@)/1-tan²(@) is NOT the same as 2tan(@)/(1-tan²(@)) What you typed means (2tan(@)/1)-tan²(@) = 2tan(@)-tan²(@)
