If #Sin A =2/3# where #0°< A < 90°#, what is the value of #Sin 2A?# Please help me find the value?

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Answer 1 · 2018-05-21T17:46:41

Using a couple of identities, I found that #Sin(2A)=(4sqrt(5))/9#

First, I wrote out the double angle identity for sine:

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

We know the solution for #sin(a)# so we can substitute that in:

#sin(2A)=2color(blue)((2/3))cos(A)#

#sin(2A)=4/3cos(A)#

Next, I used the sine and cosine sum-of-squares identity:

#sin^2(A)+cos^2(A)=1#

#cos^2(A)=1-sin^2(A)#

#cos(A)=sqrt(1-sin^2(A))#

#cos(A)=sqrt(1-color(blue)((2/3))^2)#

#cos(A)=sqrt(1-4/9)#

#cos(A)=sqrt(5/9)#

#color(red)(cos(A)=(sqrt(5))/3#

Finally, I substituted the solution for #cos(A)# into the solution for #sin(2A)#:

#sin(2A)=4/3color(red)((sqrt(5)/3))#

#sin(2A)=(4xxcolor(red)(sqrt(5)))/(3xxcolor(red)(3)#

#color(green)(sin(2A)=(4sqrt(5))/9)#