已知函数f(x)=2sinxcosx+cos2x(x属于R) 若θ为锐角,且f(θ+π/8)=3分之根号2,求tan2θ的值f(x)=sin2x+cos2x=√2sin(2x+π/4)f(θ+π/8)=√2sin(2θ+π/4+π/4) ←为什么要化为这个?=√2cos2θ=√2/3不是应该是f(θ+π/8)=√2sin
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![已知函数f(x)=2sinxcosx+cos2x(x属于R) 若θ为锐角,且f(θ+π/8)=3分之根号2,求tan2θ的值f(x)=sin2x+cos2x=√2sin(2x+π/4)f(θ+π/8)=√2sin(2θ+π/4+π/4) ←为什么要化为这个?=√2cos2θ=√2/3不是应该是f(θ+π/8)=√2sin](/uploads/image/z/10752479-71-9.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D2sinxcosx%2Bcos2x%28x%E5%B1%9E%E4%BA%8ER%29+%E8%8B%A5%CE%B8%E4%B8%BA%E9%94%90%E8%A7%92%2C%E4%B8%94f%28%CE%B8%2B%CF%80%2F8%29%3D3%E5%88%86%E4%B9%8B%E6%A0%B9%E5%8F%B72%2C%E6%B1%82tan2%CE%B8%E7%9A%84%E5%80%BCf%28x%29%3Dsin2x%2Bcos2x%3D%E2%88%9A2sin%282x%2B%CF%80%2F4%29f%28%CE%B8%2B%CF%80%2F8%29%3D%E2%88%9A2sin%282%CE%B8%2B%CF%80%2F4%2B%CF%80%2F4%29+%E2%86%90%E4%B8%BA%E4%BB%80%E4%B9%88%E8%A6%81%E5%8C%96%E4%B8%BA%E8%BF%99%E4%B8%AA%3F%3D%E2%88%9A2cos2%CE%B8%3D%E2%88%9A2%2F3%E4%B8%8D%E6%98%AF%E5%BA%94%E8%AF%A5%E6%98%AFf%28%CE%B8%2B%CF%80%2F8%29%3D%E2%88%9A2sin)
已知函数f(x)=2sinxcosx+cos2x(x属于R) 若θ为锐角,且f(θ+π/8)=3分之根号2,求tan2θ的值f(x)=sin2x+cos2x=√2sin(2x+π/4)f(θ+π/8)=√2sin(2θ+π/4+π/4) ←为什么要化为这个?=√2cos2θ=√2/3不是应该是f(θ+π/8)=√2sin
已知函数f(x)=2sinxcosx+cos2x(x属于R) 若θ为锐角,且f(θ+π/8)=3分之根号2,求tan2θ的值
f(x)=sin2x+cos2x=√2sin(2x+π/4)
f(θ+π/8)=√2sin(2θ+π/4+π/4) ←为什么要化为这个?
=√2cos2θ
=√2/3
不是应该是f(θ+π/8)=√2sin(2θ+π/4+π/8)=√2/3
已知函数f(x)=2sinxcosx+cos2x(x属于R) 若θ为锐角,且f(θ+π/8)=3分之根号2,求tan2θ的值f(x)=sin2x+cos2x=√2sin(2x+π/4)f(θ+π/8)=√2sin(2θ+π/4+π/4) ←为什么要化为这个?=√2cos2θ=√2/3不是应该是f(θ+π/8)=√2sin
f(x)=2sinxcosx+cos2x
=sin2x+cos2x
=√2sin(2x+π/4)
∵f(θ+π/8)=√2/3
∴f(θ+π/8)=√2sin[2^(θ+π/8)+π/4]
=√2sin(2θ+π/2)
=√2cos2θ=√2/3
∴cos2θ=1/3
又 θ为锐角
所以 0
f(x) = sin2x + cos2x = sqrt(2)sin(2x+pi/4)
f(θ+π/8) = sqrt(2) sin(2θ + pi/4+pi/4) = sqrt(2)cos(2θ) = sqrt(2)/3
cos(2θ) = 1/3
sin(2θ ) = sqrt(8)/3
tan(2θ) = sqrt(8)
f(x)=sin2x+cos2x=√2sin(2x+π/4)
∴f(θ+π/8)=√2sin[2(θ+π/8)+π/4]
=√2sin(2θ+π/4+π/4)
=√2sin(2θ+π/2)
然后可使用诱导公式
sin(θ+π/2)=cosθ
∴f(θ+π/8)=√2sin(2θ+π/2)
=√2cos2θ=√2/3
解得tan2θ=2√2