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# C2 Trig: proving identities watch

1. 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
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.
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:
.
4. (Original post by joostan)
It's a simple manipulation, they've simply added one, then subtracted it again. So:
.
My method of starting from the RHS is far simpler as it doesn't require seeing the "trick" of adding 1 etc.
5. (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.
6. (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.
7. (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.
8. Thanks for your help everyone

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