Recent content by JohanX

Thanks for the help👍

Hi I'm Johan, from South Africa, and I'm doing my first year at Intec College(distance) studying Diploma Analytical Chemistry.

Im using the double angles, half angles: sin^2x + cos ^2x = 1 sin(a+/-b)= sinAcosB +/- cosAsinB cos(a+/b)= cosAcosB -/+ sinAsinB tan(a+/-b)= tanA+/-tanB / 1-/+ tanAtanB sin2A = 2sinAcosA cos2A= cos^2A - sin^2A cos2A= 1 - 2sin^2A tan2A= 2tanA / 1 - tan^2A sin A/2 = √1/(1-cosA) cosA/2=...

2tanx= 1/2tan2x 4tanx=2tanx / 1 - tan^2x 1- tan^2x = 2tanx / 4tanx tan^2x = 1/2 tanx = √1/2 Textbook answer 0 ; +/- 0.615 rad Not sure where I am going wrong.