求1-cos^6x-sin^6x/1-cos^4x-sin^4x的值,
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求(1-sin^6x-cos^6x)/(1-sin^4x-cos^4x)的值
求1-cos^6x-sin^6x/1-cos^4x-sin^4x的值,
化简 1-sin^4x-cos^4x/1-sin^6x-cos^6x
化简(1-cos^4x-sin^4x)/(1-cos^6x-sin^6x)
求值:(1-sin^6 x-cos^6 x)/(1-sin^4 x-cos^4 x)
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cos x/sin x+sin x/(1+cos x) 化简.
证明一个sin&cos的等式证明 1+sin^2(x)+sin^4(x)+sin^6(x)=[1-sin^8(x)]/[cos^2(x)]
化简(1)√3sin x+cos x (2)√2(sin x-cos x) (3)√2cos x-√6sin x
已知sin(x+π/3)=1/4,求sin(2π/3-x)+cos²(π/6-x)
f'(sin x)=1-cos x 求f''(x)
1/cos(x) - tan(x) = 3 求sin(x)1/cos(x) - tan(x) = 3求sin(x)
第一题1/2cos x-√3/2sin x 第二题√3sin x+cos x 第三题√2[sin x-cos x] 第四题√2cos x-√6sin x
(1-cos^4X-sin^4x)∕(1-cos^6x-sin^6x)=化简(急)
求证:1-cos^4x-sin^4x/1-cos^6x-sin^6x=2/3
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sin(x+π/6)=1/2,求cos(π/3-x)
sin(x/2+派/6)=1/2,求cos(x+派/3)