Why does this expression simplify to 1-sin(x)?

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ok, i have this and know what it should simplify to, but why?

(1-sin(x))(sin(x))-(cos(x))(0-cos(x))


i know it simplifies to

1-sin(x)

but my book doesn't explain what exactly happens. could someone please shed some light and explain what exactly happens
 
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I think you mistyped something. It doesn't simplify to that. Try for example x =pi/4.
 
Defennder said:
I think you mistyped something. It doesn't simplify to that. Try for example x =pi/4.

I'm parsing his message as:

(1-sinx)(sin2x+cos2x)

Does that make sense?

Edit: No, I misread!
 
thats what i kept seeing.

i see it simplifies to one.

the answer in the book shows the answer i previously gave over (1-sinx)^2

then the final answer being

1 over (1-sinx)
 
[(1-sinx)(-sinx)-cosx(0-cosx)]/(1-sinx)^2

then
(1-sinx)/(1-sinx)^2

then the final being
(1)/(1-sinx)


PS sorry for the double posting
 
Jbright1406 said:
[(1-sinx)(-sinx)-cosx(0-cosx)]/(1-sinx)^2
That minus sign, which was absent in your original post, makes the whole world of difference. To simplify, just expand the numerator out and apply a trigo identity.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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