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    How dfq does(1-sin^2x)+2sinx equal 2-(sin^2x-2sinx+1)?

    It doesn't make any sense at all. Can anyone give me a method I could use? thx :d
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    (Original post by CookieHero)
    How dfq does(1-sin^2x)+2sinx equal 2-(sin^2x-2sinx+1)?

    It doesn't make any sense at all. Can anyone give me a method I could use? thx :d
    2-(sin^2x-2sinx+1) = 2 - sin^2x + 2 sinx - 1 = 1 - sin^2x + 2 sin x, as required.
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    (Original post by CookieHero)
    How dfq does(1-sin^2x)+2sinx equal 2-(sin^2x-2sinx+1)?

    It doesn't make any sense at all. Can anyone give me a method I could use? thx :d
    It's a simple manipulation, they've simply added one, then subtracted it again. So:
    (1-\sin^2(x))+2\sin(x)=[1-(\sin^2(x)-2\sin(x))]+1-1=(2-\sin^2(x)+2\sin(x)+1).
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    (Original post by joostan)
    It's a simple manipulation, they've simply added one, then subtracted it again. So:
    (1-\sin^2(x))+2\sin(x)=[1-(\sin^2(x)-2\sin(x))]+1-1=(2-\sin^2(x)+2\sin(x)+1).
    My method of starting from the RHS is far simpler as it doesn't require seeing the "trick" of adding 1 etc.
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    (Original post by ConstatSententia)
    My method of starting from the RHS is far simpler as it doesn't require seeing the "trick" of adding 1 etc.
    He used LaTeX though so his looks way cooler.
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    (Original post by ConstatSententia)
    My method of starting from the RHS is far simpler as it doesn't require seeing the "trick" of adding 1 etc.
    If this is the entirety of the identity I agree, I was however assuming that this is one step in a more complicated identity, in which case manipulating the first expression into the second may be a priority.
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    (Original post by joostan)
    If this is the entirety of the identity I agree, I was however assuming that this is one step in a more complicated identity, in which case manipulating the first expression into the second may be a priority.
    noli sine ratione rem postulare.
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    Thanks for your help everyone
 
 
 
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