Đáp án:
$D=2$
Giải thích các bước giải:
Ta có:
`\qquad sin(a+π/2)=sin a cos\ π/2 +cos a sin \ π/ 2`
`=sin a.0+cos a.1=cosa`
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`\qquad sin(a+π)=sin a.cos π +cos a.sin π`
`=sina.(-1)+cosa. 0=-sina`
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Áp dụng công thức trên ta có:
`\qquad sin^2(x+{2π}/4)=sin^2(x+π/2)=cos^2x`
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`\qquad sin^2(x+{3π}/4)=sin^2(x+π/4+π/2)=cos^2(x+π/4)`
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`\qquad sin^2(x+{4π}/4)=sin^2(x+π)=(-sinx)^2=sin^2x`
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`D=sin^2(x+π/4)+sin^2(x+{2π}/4)`
`\qquad +sin^2(x+{3π}/4)+sin^2(x+{4π}/4)`
`D=sin^2(x+π/4)+cos^2x`
`\qquad +cos^2(x+π/4)+sin^2x`
`D=[sin^2(x+π/4)+cos^2(x+π/4)]`
`\qquad +(sin^2x+cos^2x)`
`D=1+1=2`