Trigonometry: prove:
tan3a = cosa - cos5a/sin5a-sina
It throws me for a loop when proofs have integers in the functions. Studying for a test and got this own wrong on my mid term, please explain thanks!
by 
liquidkool
ANSWERS: 2
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Should it read: tan3a = (cosa - cos5a) / (sin5a - sina)
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Trig formula for subtraction of cos and sin: cos(A) - cos(B) = - 2 sin((A+B)/2) sin((A-B)/2) sin(A) - sin(B) = 2 cos((A+B)/2) sin((A-B)/2)) Using these: tan 3a = (cos a - cos 5a) / (sin 5a - sin a) = (-2 sin(3a)sin(-2a))/(2cos(3a)sin(2a)) = (sin(3a)sin(2a))/(cos(3a)sin(2a)) = sin(3a)/cos(3a) = tan(3a) It is possible to work out the formula as long as you remember just these ones (the only addition ones I could remember): sin(A+B) = sin(A)cos(B) + cos(A)sin(B) cos(A+B) = cos(A)cos(B) - sin(A)sin(B) and remember that sin is odd and cos is even, so sin(-X)=-sin(X) but cos(-X)=cos(X) If you want sin(X)-sin(Y) you make A+B=X and A-B=Y and use the sin formula twice.
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