Thread: Solving simultaneous trig and linear equations

Dear , the solution is SO easy : first : alfa-beta = 2pi/3 => beta =alfa - 2pi/3 , cos ( beta ) = cos ( alfa - 2pi/3 )

as we know cos ( a - b ) = cos a cos b + sin a sin b

=> cos ( alfa - 2pi/3 ) = cos (alpha ) cos ( 2pi/3 ) + sin ( alfa ) sin ( 2pi/3 )

=> you can calculate cos alpha wich is square root ( 1 - sin ( alpha ) ^ 2 ) becarefull you have to take the negative value becoz alpha betwiin pi and 3 pi / 2 and in this region the cos in negative cos alpha = - ( square root 7 ) / 7

the final answer is : cos ( beta ) = (sqr root 3 ) / 14 - 6/7

note : check again i solved it very fast

The answer isn't correct, doesn't matter, cause I figured it out by myself
here 's correct answer ( Thanks anyway)

Originally Posted by livmed

The answer isn't correct, doesn't matter, cause I figured it out by myself
here 's correct answer ( Thanks anyway)

it is about the method not about the answer