Linear transformation rotation

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Homework Statement


T: R2-->R2 first reflects points through -3pi/4 radian (clockwise) and then reflects points through the horizontal x1-axis. [Hint T(e1)= (-1/sqrt2, 1/sqrt2)



The Attempt at a Solution


I just don't understand why the points would be (-1/sqrt2, 1/sqrt2). If it's -3pi/4, why wouldn't it be (-(sqrt2)/2, -(sqrt2)/2)?
 
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Did you miss the part about the reflection or are you rotating the wrong way? Rotating (1, 0) -3pi/4 radians (clockwise as it says) gives you (-1/sqrt(2), -1/sqrt(2)) but then the reflection through the horizontal axis changes that to (-1/sqrt(2),+1/sqrt(2)).
 
Doesn't -3pi/4 radians correlate to (-(sqrt2)/2, -(sqrt2)/2) though (on a unit circle)? Where is the sqrt2 on the bottom coming from?
 
Are you aware that 2= (sqrt(2))*(sqrt(2))? 1/sqrt(2) is exactly the same as sqrt(2)/2.
 
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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