If sin A = 4/5, where A is an acute angle, determine sin 2A and cos 2A, using 2A = (A + A).
please EXPLAIN how you got the answer, thanks in advance!
please EXPLAIN how you got the answer, thanks in advance!
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sin(2A) = 2sin(A)*cos(A)= 2*(4/5)(3/5) =24/25 ......Ans
cos(2A) = 1 - 2 sin^2(A)=1- 2(4/5)^2 = 1- 32/25 = -7/25 Ans
Second Method
2A = A+A
Taking sin both side
sin(2A) =sin(A)*cos(A)+ cos(A)*sin(A)
= (4/5)*(3/5) +(3/5)(4/5) = 12/25 + 12/25 =24/25 .....Ans
Taking cos both side
cos(2A) =cos(A)*cos(A)- sin(A)*sin(A)
= (3/5)*(3/5) -(4/5)(4/5) = 9/25 - 16/25 =-7/25 .....Ans
cos(2A) = 1 - 2 sin^2(A)=1- 2(4/5)^2 = 1- 32/25 = -7/25 Ans
Second Method
2A = A+A
Taking sin both side
sin(2A) =sin(A)*cos(A)+ cos(A)*sin(A)
= (4/5)*(3/5) +(3/5)(4/5) = 12/25 + 12/25 =24/25 .....Ans
Taking cos both side
cos(2A) =cos(A)*cos(A)- sin(A)*sin(A)
= (3/5)*(3/5) -(4/5)(4/5) = 9/25 - 16/25 =-7/25 .....Ans