(sinx+cosx)2 -sin(2x)=1
(sinx+cosx)2−sin2x=1sin2x+cos2x+2sinx∗cosx−sin2x=1sin2x+cos2x+2sinx∗cosx−(2sinx∗cosx)=11+2sinx∗cosx−(2sinx∗cosx)=11=1(\sin{x}+\cos{x})^2-\sin{2x}=1\\ \sin^2{x}+\cos^2{x}+2\sin{x}*\cos{x}-\sin{2x}=1\\ \sin^2{x}+\cos^2{x}+2\sin{x}*\cos{x}-(2\sin{x}*\cos{x})=1\\ 1+2\sin{x}*\cos{x}-(2\sin{x}*\cos{x})=1\\ 1=1(sinx+cosx)2−sin2x=1sin2x+cos2x+2sinx∗cosx−sin2x=1sin2x+cos2x+2sinx∗cosx−(2sinx∗cosx)=11+2sinx∗cosx−(2sinx∗cosx)=11=1
Answer:x∈(−∞;∞)x \in (-\infty;\infty)x∈(−∞;∞)
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