how would you do this one: show that: tan(pi/4 + A) - tan(pi/4 - A) = 2 tan 2A thanks
ezzy85 hmm...yeah..... Joined Nov 4, 2002 Messages 556 Gender Undisclosed HSC N/A Mar 22, 2003 #1 how would you do this one: show that: tan(pi/4 + A) - tan(pi/4 - A) = 2 tan 2A thanks
S spice girl magic mirror Joined Aug 10, 2002 Messages 785 Mar 22, 2003 #2 Originally posted by ezzy85 show that: tan(pi/4 + A) - tan(pi/4 - A) = 2 tan 2A Click to expand... use compound angles for both trig expressions on the LHS and see wot we get: tan(pi/4 + A) = [tan(pi/4) + tan(A)] / [1 - tan(pi/4)tanA] = (1 + tanA)/(1-tanA) similarly tan(pi/4 - A) = (1-tanA)/(1+tanA) subtract one from the other with common denominator: LHS = [ (1+tanA)^2 + (1-tanA)^2 ]/(1-tan^2(A)) = 4tanA / (1-tan^2(A)) = 2(2tanA)/(1-tan^2(A)) = 2tan2A
Originally posted by ezzy85 show that: tan(pi/4 + A) - tan(pi/4 - A) = 2 tan 2A Click to expand... use compound angles for both trig expressions on the LHS and see wot we get: tan(pi/4 + A) = [tan(pi/4) + tan(A)] / [1 - tan(pi/4)tanA] = (1 + tanA)/(1-tanA) similarly tan(pi/4 - A) = (1-tanA)/(1+tanA) subtract one from the other with common denominator: LHS = [ (1+tanA)^2 + (1-tanA)^2 ]/(1-tan^2(A)) = 4tanA / (1-tan^2(A)) = 2(2tanA)/(1-tan^2(A)) = 2tan2A
