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    • Thread Starter
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    Hey Guys.

    I'm stuck on this higher question..

    A is the point (8,4). The line OA is inclined at an angle of p radians to the x axis.

    Find the exact values od :

    (i) sin(2p) ;
    (ii)cos(2p).

    I know i have to use the equations sin2A = 2sinAcosA and cos2A = 1-2cos^2 A - 1.

    I also know that the opposite is 4 and the adjacent is 8. It is a right angled triangle so that should make the hypotenuse route 48. Therefore you can use SOHCAHTOA to find sinA and cosA. But this is in a non calculator paper therefore making it harder.

    Can someone tell me where i have went wrong.

    Thanks
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    (Original post by PuRe-TaCtIcZ)
    Hey Guys.

    I'm stuck on this higher question..

    A is the point (8,4). The line OA is inclined at an angle of p radians to the x axis.

    Find the exact values od :

    (i) sin(2p) ;
    (ii)cos(2p).

    I know i have to use the equations sin2A = 2sinAcosA and cos2A = 1-2cos^2 A - 1.

    I also know that the opposite is 4 and the adjacent is 8. It is a right angled triangle so that should make the hypotenuse route 48. Therefore you can use SOHCAHTOA to find sinA and cosA. But this is in a non calculator paper therefore making it harder.

    Can someone tell me where i have went wrong.

    Thanks
    The hypoteneuse is not \sqrt{48}, so check/post your working for that.

    Your formula for cos2A isn't right, but that may just be typos.

    Aside from those, what's your actual problem?

    Post some working if you'd like people to check it.
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    I don't know how you got 48 for the hypoteneuse. It's \sqrt{4^2 + 8^2} = \sqrt{80} = 4\sqrt{5}.

    Not sure how you'd work out the sin and cos values though.
 
 
 
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