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  • sinx+sin^2x=1求cos^2x+cosx的6次方+cosx的8次方的值!

    sinx+sin^2x=1求cos^2x+cosx的6次方+cosx的8次方的值!
    sinx+sin^2x=1
    求1) cos^2x+cosx的6次方+cosx的8次方的值
    2) 3cos^2x+cosx的8次方的值
    數(shù)學(xué)人氣:618 ℃時(shí)間:2019-10-29 11:54:20
    優(yōu)質(zhì)解答
    (1)因?yàn)?sinx+sin^2x=1
    又因?yàn)?
    sin^2x+cos^2x=1
    兩式相減得:
    sinx-cos^2x=0
    sinx=cos^2x
    則:cos^2x+cos^6x+cos^8x
    =cos^2x+(cos^2x)^3+(cos^2x)^4
    =sinx+(sinx)^3+(sinx)^4
    =sinx+[(sinx)^3+(sinx)^4]
    =sinx+sin^2x[sinx+sin^2x]
    =sinx+sin^2x[cos^2x+sin^2x]
    =sinx+sin^2x
    =cos^2x+sin^2x
    =1
    (2)3cos^2x+cos^8x
    =3cos^2x+(cos^2x)^4
    =3sinx+(sinx)^4
    =3sinx+(sin^2x)^2
    =3sinx+(1-sinx)^2
    =3sinx+(1+sin^2x-2sinx)
    =sin^2x+sinx+1
    =sin^2x+cos^2x+1
    =2
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