求证cos α/(1+sin α) -sinα/(1+cos α)=2(cos α-sin α)/(1+sin α+cos α)
2个回答
展开全部
证明:
令α=2t;
等式左边=
cos α/(1+sin α) -sinα/(1+cos α)
=cos2t/(1+sin2t)-sin2t/(1+cos2t)
=(cos²t-sin²t)/(cos²t+sin²t+2sint*cost)-2sint*cost/2cos²t -----降倍升幂
=(cost-sint)/(cost+sint)-sint/cost -----化简、通分
=(cos²t-sin²t-2sintcost)/cost(cost+sint) -----升倍降幂
=(cos2t-sin2t)/[1/2(cos2t+1)+1/2sin2t]
=2(cosα-sinα)/(1+sinα+cosα)
=等式右边
证毕.
令α=2t;
等式左边=
cos α/(1+sin α) -sinα/(1+cos α)
=cos2t/(1+sin2t)-sin2t/(1+cos2t)
=(cos²t-sin²t)/(cos²t+sin²t+2sint*cost)-2sint*cost/2cos²t -----降倍升幂
=(cost-sint)/(cost+sint)-sint/cost -----化简、通分
=(cos²t-sin²t-2sintcost)/cost(cost+sint) -----升倍降幂
=(cos2t-sin2t)/[1/2(cos2t+1)+1/2sin2t]
=2(cosα-sinα)/(1+sinα+cosα)
=等式右边
证毕.
展开全部
令α=2t;
等式左边=
cos α/(1+sin α) -sinα/(1+cos α)
=cos2t/(1+sin2t)-sin2t/(1+cos2t)
=(cos²t-sin²t)/(cos²t+sin²t+2sint*cost)-2sint*cost/2cos²t
=(cost-sint)/(cost+sint)-sint/cost
=(cos²t-sin²t-2sintcost)/cost(cost+sint)
=(cos2t-sin2t)/[1/2(cos2t+1)+1/2sin2t]
=2(cosα-sinα)/(1+sinα+cosα)
=等式右边
证毕.
等式左边=
cos α/(1+sin α) -sinα/(1+cos α)
=cos2t/(1+sin2t)-sin2t/(1+cos2t)
=(cos²t-sin²t)/(cos²t+sin²t+2sint*cost)-2sint*cost/2cos²t
=(cost-sint)/(cost+sint)-sint/cost
=(cos²t-sin²t-2sintcost)/cost(cost+sint)
=(cos2t-sin2t)/[1/2(cos2t+1)+1/2sin2t]
=2(cosα-sinα)/(1+sinα+cosα)
=等式右边
证毕.
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