已知函数f(x)=sin(2x+φ)(0<φ<π)的图象经过点(π/12,1)求φ的值在三角形ABC中,若a∧2+b∧2-c∧2=ab,且f(A/2+π/12)=√2/2,求sinB
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![已知函数f(x)=sin(2x+φ)(0<φ<π)的图象经过点(π/12,1)求φ的值在三角形ABC中,若a∧2+b∧2-c∧2=ab,且f(A/2+π/12)=√2/2,求sinB](/uploads/image/z/5892049-1-9.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dsin%282x%2B%CF%86%29%280%EF%BC%9C%CF%86%EF%BC%9C%CF%80%29%E7%9A%84%E5%9B%BE%E8%B1%A1%E7%BB%8F%E8%BF%87%E7%82%B9%28%CF%80%2F12%2C1%29%E6%B1%82%CF%86%E7%9A%84%E5%80%BC%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2C%E8%8B%A5a%E2%88%A72%2Bb%E2%88%A72-c%E2%88%A72%3Dab%2C%E4%B8%94f%28A%2F2%2B%CF%80%2F12%29%3D%E2%88%9A2%2F2%2C%E6%B1%82sinB)
已知函数f(x)=sin(2x+φ)(0<φ<π)的图象经过点(π/12,1)求φ的值在三角形ABC中,若a∧2+b∧2-c∧2=ab,且f(A/2+π/12)=√2/2,求sinB
已知函数f(x)=sin(2x+φ)(0<φ<π)的图象经过点(π/12,1)
求φ的值
在三角形ABC中,若a∧2+b∧2-c∧2=ab,且f(A/2+π/12)=√2/2,求sinB
已知函数f(x)=sin(2x+φ)(0<φ<π)的图象经过点(π/12,1)求φ的值在三角形ABC中,若a∧2+b∧2-c∧2=ab,且f(A/2+π/12)=√2/2,求sinB
fx=sin(2x+φ)
经过点(π/12,1)
sin(π/6+φ)=1
∴π/6+φ=π/2+2kπ,k∈Z
∴φ=π/3+2kπ,k∈Z
∵0<φ<π
∴k=0
∴φ=π/3
f(A/2+π/12)=√2/2=sin(3π/4) 则A+π/6+π/3=3π/4 则A=π/4 =45°
cosC=(a^2+b^2-c^2)/2ab cosC=1/2,C=60°
则B=180-45°-60°=75°
sin75°=sin(45°+30°)=sin45°cos30°+cos45°sin30°=(√6+√2)/4
把(π/12,1)代入函数解析式得:
sin(2*π/12+φ)=1
sin(π/6+φ)=1
π/6+φ=2kπ+π/2
φ=2kπ+π/3
∵0<φ<π
∴φ=π/3
f(x)=sin(2x+φ)(0<φ<π)的图象经过点(π/12,1)
点代入得
1=sin(π/6+φ)
因此π/6+φ=π/2
φ=π/3
a^2+b^2-c^2=ab
cosC=(a^2+b^2-c^2)/(2ab)=1/2
C=π/3
f(A/2+π/12)=√2/2
sin[2(A/2+π/12)+φ]=√2/2
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f(x)=sin(2x+φ)(0<φ<π)的图象经过点(π/12,1)
点代入得
1=sin(π/6+φ)
因此π/6+φ=π/2
φ=π/3
a^2+b^2-c^2=ab
cosC=(a^2+b^2-c^2)/(2ab)=1/2
C=π/3
f(A/2+π/12)=√2/2
sin[2(A/2+π/12)+φ]=√2/2
sin(A+π/6+π/3)=√2/2
sin(A+π/2)=√2/2
A=π/4
sinB=sin(A+C)
=sinAcosC+cosAsinC
=√2/2(√3/2+1/2)
=(√6+√2)/4
收起
π/6+φ=π/2 , φ=π/3
f(A/2+π/12)=√2/2=sin[2(A/2+π/12)+π/3]=sin(A+π/2)=cosA , A=π/4
a^2+b^2-c^2=ab, c^2=a^2+b^2-ab
c^2=a^2+b^2-2abcosC=a^2+b^2-ab , cosC=1/2 , C=π/3
sinB=sin[π-(π/4+π/3)]=sin(π/4+π/3)=√2/2(1/2+√3/2)=(√2+√6)/4