Given: sin(a)=-4/5 and pi_<a_<(3pi)/2; sin(B)=5/13 and 0_<B_<pi/2

Question

( _< means less than or equal to)

find: sin2a

find: cos2B

find: sin(a+B)

find: cos(a-B)

Answer 1

You will need to start by finding cos(a) and cos(b) as you will need these for your calculations.

If sin(a)= -4/5 then cos(a)= -3/5

If sin(b)= (5/13) then cos(b)=(12/13)

Sin(2a)= 2sin(a)Cos(a)= (2)(-4/5)(-3/5)= 24/25

Cos(2a)= Cos^2(a)-Sin^2(a)= (-3/5)^2 -(4/5)^2= 9/25-16/25= -7/25

Sin(a+b)= sin(a)cos(b) + sin(b)cos(a)= (-4/5)(12/13)+ (5/13)(-3/5)= -48/65+ (-15/65)= -63/65

cos(a-b)= cos(a)(cos(b) = sin(a)(sin(b). I leave the calculation for you