Oh yeh sure sorryFor future reference, can you please post subject specific content content in subject specific forums?
?How can you use the identity of Cos2x here? Or am i just being dumb
I don't know what the hell to do about the 1/2How can you use the identity of Cos2x here? Or am i just being dumb
Thank but I'm a bit confused about what happened from here 2+2(2cos^2((A-B)/2)-1) --> 4cos^2((A-B)/2)for L you expand and get sin^2A+cos^2A+sin^2B+cos^2B+2sinAsinB+2cosAcosB then u use sin^2x+cos^2x=1 and
cos(A-B)=cosAcosB+sinAsinB to get 2+2(cos(A-B))then you use cos2x=2cos^2(x)-1to get 2+2(2cos^2((A-B)/2)-1)=4cos^2((A-B)/2)
Can someone also explain how to do 14 a. ThanksSomeone please do L (identity)View attachment 29432
y = cos2a = cos^a - sin^a = cos^a - (1-cos^a) = 2cos^a -1Can someone also explain how to do 14 a. Thanks
Ahh Shit I'm stupidy = cos2a = cos^a - sin^a = cos^a - (1-cos^a) = 2cos^a -1
express costheta in terms of y, sub into x equation.