document.write( "Question 825715: If sinθ= 4/5 and θ terminates in Quadrant II, find the exact value of sin2θ.
\n" ); document.write( "F. -24/25
\n" ); document.write( "G. 8/5 (I chose this one and it was wrong)
\n" ); document.write( "H.-12/25
\n" ); document.write( "J. 24/25
\n" ); document.write( "Sorry I found the typo I submitted in one of my previous questions. So sorry about that.
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Algebra.Com's Answer #497536 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
I'm going to use x in place of θ\r
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\n" ); document.write( "\n" ); document.write( "sin(x) = 4/5\r
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\n" ); document.write( "\n" ); document.write( "sin^2(x) = 16/25 ... square both sides\r
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\n" ); document.write( "\n" ); document.write( "sin^2(x) + cos^2(x) = 1\r
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\n" ); document.write( "\n" ); document.write( "16/25 + cos^2(x) = 1\r
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\n" ); document.write( "\n" ); document.write( "cos^2(x) = 1 - 16/25\r
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\n" ); document.write( "\n" ); document.write( "cos^2(x) = 25/25 - 16/25\r
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\n" ); document.write( "\n" ); document.write( "cos^2(x) = (25 - 16)/25\r
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\n" ); document.write( "\n" ); document.write( "cos^2(x) = 9/25\r
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\n" ); document.write( "\n" ); document.write( "cos(x) = -sqrt(9/25) ... we are in quadrant II, so cos(x) < 0\r
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\n" ); document.write( "\n" ); document.write( "cos(x) = -3/5\r
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\n" ); document.write( "\n" ); document.write( "--------------------------------------------------------\r
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\n" ); document.write( "\n" ); document.write( "sin(2x) = 2*sin(x)*cos(x)\r
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\n" ); document.write( "\n" ); document.write( "sin(2x) = 2*(4/5)*(-3/5)\r
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\n" ); document.write( "\n" ); document.write( "sin(2x) = (2/1)*(4/5)*(-3/5)\r
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\n" ); document.write( "\n" ); document.write( "sin(2x) = (2*4*(-3))/(1*5*5)\r
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\n" ); document.write( "\n" ); document.write( "sin(2x) = -24/25\r
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\n" ); document.write( "\n" ); document.write( "Final Answer: sin(2x) = -24/25
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