document.write( "Question 737089: Use the given information to find cos 2x, sin 2x, and tan 2x.
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document.write( "cosx=sqrt3/7 and 3π/2 < x < 2π\r
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document.write( "since I have cos, I created a triangle using sqrt3 as the opposite side. and 7 as the adjacent side. Then I used the Pythagorean theorem to find the hypotenuse.
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document.write( "hypotenuse^2=(sqrt3)^2+(7)^2
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document.write( "hypotenuse^2=3+49
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document.write( "hypotenuse^2=52
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document.write( "hypotenuse=sqrt52\r
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document.write( "Then to find sin2x, I used sin2x=2sin(xcos(x), so...
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document.write( "sin2x=2sin(xcos(x)
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document.write( "= 2(sqrt3/sqrt52)(sqrt3/7)
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document.write( "= 2 (sqrt9/7sqrt52)
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document.write( "= 3/7sqrt13\r
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document.write( "but that was incorrect and I can't go any further in the problem with an incorrect sin2x. Please assist me with where I went wrong or how to go about finding this answer. thanks!
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Algebra.Com's Answer #450182 by mananth(16946) You can put this solution on YOUR website! cos x = \n" ); document.write( "\n" ); document.write( "square both sides\r \n" ); document.write( "\n" ); document.write( "cos^2 x=3/49\r \n" ); document.write( "\n" ); document.write( "1-cos^2 x=1-3/49\r \n" ); document.write( "\n" ); document.write( "sin^2 x = 46/49\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "cos 2x = 2 cos^2 x -1\r \n" ); document.write( "\n" ); document.write( "=2*3/49-1 \n" ); document.write( "=6/49-1\r \n" ); document.write( "\n" ); document.write( "=-43/49\r \n" ); document.write( "\n" ); document.write( "Cos 2x= -43/49\r \n" ); document.write( "\n" ); document.write( "Sin 2x = 2sin x . cos x\r \n" ); document.write( "\n" ); document.write( "= \n" ); document.write( "\n" ); document.write( "= \n" ); document.write( "\n" ); document.write( "Tan (2x) = sin (2x)/cos (2x)\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |