Question

If sinA=3/5 and A is acute , then find sin2A

Answer 1

        Using cos²θ= 1 - sin²Aθ

=> cos²A = 1 - (3/5)^2

=> cos²A =  (25 - 9)/25

=> cos²A = 16/25

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

From the properties of trigonometric functions :

• sin2A = 2sinAcosA

=> sin2A

=> 2sinAcosA

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

=> 24/25

Hence the required value of sin2A is 24/25.

Answer 2

Question :--

• If sinA=3/5 and A is acute , then find sin2A = ?

Formula used :--

→ SinA = Perpendicular / Hypotenuse = P/H

→ CosA = Base / Hypotenuse = B/H .

→ Pythagoras theoram , P² + B² = H² .

→ Sin2A = 2*SinA*CosA

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Solution :---

Given , SinA = 3/5

→ SinA = 3/5 = P/H

we get,

→ P = 3

→ H = 5

Now, putting value in Pythagoras theoram, we get,

→ (3)² + (B)² = 5²

→ 25 - 9 = B²

→ 16 = B²

Square root both both sides we get,

→ B = 4 .

Putting this value now , we get,

→ sin@ = P/H = 3/5 .

→ cos@ = B/H = 4/5 .

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Now, putting these values in Question we get,

→ Sin2A

→ 2*SinA*CosA

→ 2*(3/5)*(4/5)

→ 24/25

Hence, value of Sin2A is 24/25.....