Question

Упростите выражение, срочно, завтра кр​

Answer 1

sin (2x) = 2sin(x)cos(x)

sin² (x) + cos² (x) = 1

log(a) b = c     a^c = b

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(sin(x) - cos(x) )² - 1 + 1/2*sin(2x) = sin² (x) - 2sin(x)cos(x) + cos²(x) - 1 + 1/2*sin(2x) = 1 - sin(2x)+1/2*sin(2x) - 1 = - 1/2*sin(2x)

(-1/2*sin(2x))² = 1/4*sin²(2x)

log(2) (1/4*sin²(2x) + 1/4*cos²(2x)) = log(2) 1/4*(sin²(2x) + cos²(2x)) = log(2) 1/4 = log(2) 2⁻² = -2

Answer 2

$log_{2}(((Sin\alpha-Cos\alpha)^{2}-1+\frac{1}{2}Sin2\alpha)^{2}+\frac{1}{4}Cos^{2}2\alpha)=log_{2} ((Sin^{2}\alpha-2Sin\alpha Cos\alpha+Cos^{2}\alpha-1+\frac{1}{2}Sin2\alpha)^{2}+\frac{1}{4}Cos^{2}2\alpha)=log_{2}((1-Sin2\alpha-1+\frac{1}{2}Sin2\alpha)^{2} +\frac{1}{4} Cos^{2}2\alpha)=log_{2} ((-\frac{1}{2}Sin2\alpha)^{2}+\frac{1}{4}Cos^{2}2\alpha)=log_{2}(\frac{1}{4}Sin^{2}2\alpha+\frac{1}{4}Cos^{2}2\alpha)=log_{2}\frac{1}{4}=-2$