化简1-cos4α-sin4α/1-cos5α-sin6α(数字是次数,如图所示)
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化简1-cos4α-sin4α/1-cos5α-sin6α(数字是次数,如图所示)
化简(1-cos⁴α-sin⁴α)/(1-cos⁵α-sin⁶α)
原式=[1-(cos⁴α+sin⁴α)]/[1-cos⁵α-(1-cos²α)³]
=[1-(cos²α+sin²α)²+2sin²αcos²α]/[1-cos⁵α-(1-3cos²α+3cos⁴α-cos⁶α)]
=(2sin²αcos²α)/(cos⁶α-cos⁵α-3cos⁴α+3cos²α)
=(2sin²α)/(cos⁴α-cos³α-3cos²α+3)
=(2sin²α)/[(cosα-1)(cos³α-3cosα-3)]
=[2(1-cos²α)]/[(cosα-1)(cos³α-3cosα-3)]
=[-2(cosα+1)(cosα-1)]/[(cosα-1)(cos³α-3cosα-3)]
=-2(cosα+1)/(cos³α-3cosα-3)
原式=[1-(cos⁴α+sin⁴α)]/[1-cos⁵α-(1-cos²α)³]
=[1-(cos²α+sin²α)²+2sin²αcos²α]/[1-cos⁵α-(1-3cos²α+3cos⁴α-cos⁶α)]
=(2sin²αcos²α)/(cos⁶α-cos⁵α-3cos⁴α+3cos²α)
=(2sin²α)/(cos⁴α-cos³α-3cos²α+3)
=(2sin²α)/[(cosα-1)(cos³α-3cosα-3)]
=[2(1-cos²α)]/[(cosα-1)(cos³α-3cosα-3)]
=[-2(cosα+1)(cosα-1)]/[(cosα-1)(cos³α-3cosα-3)]
=-2(cosα+1)/(cos³α-3cosα-3)
化简:1-sin6α(六次方)-cos6(六次方)α/1-sin4(四次方)α-cos4(四次方)α
化简1+sin4α+cos4α/1+sin4α -cos4α
化简(1-cos4α+sin4α)/(1+cos4α+sin4α)
怎样化简(1+sin4α+cos4α)÷(1+sin4-cos4α)
化简cos4α/2-sin4α/2
求(1-sin6次幂a-cos6次幂a)/(1-sin4次幂a-cos4次幂a)
[sin4α/(1+cos4α)]×[cos2α/(1+cos2α)]×[cosα/(1+cosα)]=?
证明:(1-cos4α)/sin4α*cos2α/(1+cos2α)=tanα
sin2α+cos2α=1,sin4α+cos4α为什么等于1
求证:sin4α+cos4α=1-2sin2αcos2α
1sinα+cosα=根号2求sin4α-cos4α
sin4α-cos4α=sin2α-cos2α求证