√2sin(x-π/4)+√6cos(x-π/4)化简

390466322022-10-04 11:39:543条回答

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lxxf 共回答了12个问题 | 采纳率83.3%
√2sin(x-π/4)+√6cos(x-π/4)
=2√2[(1/2)sin(x-π/4)+(√3/2)cos(x-π/4)]【提√(a²+b²)】
=2√2[sin(x-π/4)cos(π/3)+cos(x-π/4)sin(π/3)]
=2√2sin(x-π/4+π/3)
=2√2sin(x+π/12)
【如果熟悉辅助角公式更简单:asinα+bcosα=√(a²+b²)sin(α+φ),其中tanφ=b/a】
1年前
ccf_jcc 共回答了1576个问题 | 采纳率
答:
√2sin(x-π/4)+√6cos(x-π/4)
=2√2*[(1/2)sin(x-π/4)+(√3/2)cos(x-π/4)]
=2√2*sin(x-π/4+π/3)
=2√2*sin(x+π/12)
1年前
yxjht9098 共回答了231个问题 | 采纳率
√2sin(x-π/4)+√6cos(x-π/4)
=2√2[(1/2)sin(x-π/4)+(√3/2)cos(x-π/4)]
=2√2[sin(x-π/4)cos(π/3)+cos(x-π/4)sin(π/3)]
=2√2sin(x-π/4+π/3)
=2√2sin(x+π/12)
1年前

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