Question

Докажите, что
а) 2sin^2(2t) (синус в квадрате двух t) = 1+sin ((3pi)/2-4t)
б)sin^2((3pi/4)+2t)= (1-sin4t)/2
в)1-sin(t)=2sin(t)^2(45- t/2)

Answer 1

Ответ:

Объяснение:

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Answer 2

$Formyla:\\\\sin^2x=\frac{\big{1-cos2x}}{\big2}\\\\a)\:\:\:2sin^2(2t)=1+sin\Big(\frac{\:\big{3\pi}}{\big2}-4t\Big)=1-cos4t\\\\2sin^2(2t)=2*\frac{\big{1-cos4t}}{\big2}=1-cos4t\\\\b)\:\:\:sin^2\Big(\:\frac{\big{3\pi}}{\big4}+2t\Big)=\frac{\big{1-sin4t}}{\big2}$

$sin^2\Big(\:\frac{\big{3\pi}}{\big4}+2t\Big)=\frac{\big{1-cos\Big(2*\Big(\:\frac{\big{3\pi}}{\big4}+2t\Big)\Big)}}{\big2}=\frac{\big{1-cos\Big(\:\frac{\big{3\pi}}{\big2}+4t\Big)}}{\big2}=\\\\=\:\frac{\big{1-sin4t}}{\big2}$

$c)\:\:\:1-sint=2sin^2\Big(\:\frac{\big\pi}{\big4}+\frac{\big{t}}{\big2}\:\Big)\\\\2sin^2\Big(\:\frac{\big\pi}{\big4}+\frac{\big{t}}{\big2}\:\Big)=2*\frac{\big{1-cos\Big(2*\Big(\:\frac{\big\pi}{\big4}+\frac{\big{t}}{\big2}\:\Big)\Big)}}{\big2}=\big{1-cos\Big(\:\frac{\big\pi}{\big2}+t\Big)}=1-sint$