sinA =4/5 , then sin2A = ?

Question

I don't know Trigonometry as better ,,because i didn't learn

Answer 1

You need to apply the double angle formula for the sine:

 

sin(2A) = 2sin(A)cos(A)

 

Since sin(A) = 4/5:

 

sin(2A) = 2*(4/5)*cos(A) = (8/5)cos(A)

 

To find cos(A), use the Pythagorean identity:

 

sin²(A) + cos²(A) = 1

cos²(A) = 1 - sin²(A)

cos²(A) = 1 - (4/5)²

cos²(a) = 9/25

√cos²(A) = √(9/25)

 

cos(A) = ±3/5

 

So:

 

sin(2A) = (8/5)*(±3/5) = ±24/25

 

Answer 2

This is where you use the double angle identity in which:

 

sin2A=2sinA*cosA

 

The next step is to draw a right triangle in which the sinA is 4/5.

 

From geometry, this turns out to be a 3-4-5 right triangle, hence cosA=3/5

 

Next substitute the numbers to determine sin2A in which is:

 

sin2A=2*4/5*3/5=24/25.

 

Hope this helps.