Question

SinA=3/5 find sin2A help me find the ans

Answer 1

Answer:

Step-by-step explanation:

Given,

sinA = 3 / 5

Thus, by using cos^2 B = 1 - sin^2 B :

= > cos^2 A = 1 - ( 3 / 5 )^2

= > cos^2 A = ( 25 - 9 ) / 25

= > cos^2 A = 16 / 25

= > cosA = 4 / 5 { A is an acute angle, cosB can't be -ve }

From the properties of trigonometric functions :

sin2A = 2sinAcosA

Thus,

= > sin2A

= > sinAcosA

= > ( 3 / 5 )( 4 / 5 )

= > 12 / 25

Hence the required value of sin2A is 12 / 25.

Answer 2

Answer:

sinA = 3/5

we know that Sin²A + Cos²A = 1

so, cos²A = 1-sin²A

==> cos²A = 1 - (3/5)²

==> cos²A = 1 - 9/25

==> cos²A = (25-9)/25

==> cos²A = 16/25

==> cosA = √16/25

==> cosA = 4/5 (it will be positive becausw it lies in Ist quadrant}

we know that

Sin2A = 2SinAcosA

sin2A = 2(3/5)(4/5)

sin2A = 24/25