三角函数证明题(1)3+cos4a-4cos2a=8sin~4次方 a(2)tanatanb/tan2a-tana +
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三角函数证明题
(1)3+cos4a-4cos2a=8sin~4次方 a
(2)tanatanb/tan2a-tana + 根号3(sin~2 a-cos~2 a)=2sin(2a-π/3)
(1)3+cos4a-4cos2a=8sin~4次方 a
(2)tanatanb/tan2a-tana + 根号3(sin~2 a-cos~2 a)=2sin(2a-π/3)
1.倍角公式:
cos2α=cos^2(α)-sin^2(α)=2cos^2(α)-1=1-2sin^2(α)
3+cos4a-4cos2a
=3+(2cos^2(2a)-1)-4(1-2sin^2(a))
=3+2cos^2(2a)-1-4+8sin^2(a)
=-2+2(1-2sin^2(a))^2 +8sin^2(a)
=-2+2(1-4sin^2(a)+4sin^4(a))+8sin^2(a)
=-2+2-8sin^2(a)+8sin^4(a)+8sin^2(a)
=8sin^4(a)
2.题目有误:应为tanatan2a/tan2a-tana + 根号3(sin~2 a-cos~2 a)=2sin(2a-π/3)
tan2a=2tana/(1-tan²a)
tana/tan2a=(1-tan²a)/2
(tanatan2a)/(tan2a-tana)
上下除以tan2a
=tana/(1-tana/tan2a)
tana/tan2a=(1-tan²a)/2
所以整理得 =2tana/(1+tan²a)
=2(sina/cosa)/(1+sin²a/cos²a)
上下乘cos²a
=2sinacosa/(sin²a+cos²a)
=sin2a
√3(sin²a-cos²a)=-√2(cos²a-sin²a)
所以左边=sin2a-√3cos2a
=√[(1²+(√3)²]sin(2a-z)
=2sin(2a-z)
其中tanz=√3/1=√3
z=π/3
所以左边=2sin(2a-π/3)=右边
cos2α=cos^2(α)-sin^2(α)=2cos^2(α)-1=1-2sin^2(α)
3+cos4a-4cos2a
=3+(2cos^2(2a)-1)-4(1-2sin^2(a))
=3+2cos^2(2a)-1-4+8sin^2(a)
=-2+2(1-2sin^2(a))^2 +8sin^2(a)
=-2+2(1-4sin^2(a)+4sin^4(a))+8sin^2(a)
=-2+2-8sin^2(a)+8sin^4(a)+8sin^2(a)
=8sin^4(a)
2.题目有误:应为tanatan2a/tan2a-tana + 根号3(sin~2 a-cos~2 a)=2sin(2a-π/3)
tan2a=2tana/(1-tan²a)
tana/tan2a=(1-tan²a)/2
(tanatan2a)/(tan2a-tana)
上下除以tan2a
=tana/(1-tana/tan2a)
tana/tan2a=(1-tan²a)/2
所以整理得 =2tana/(1+tan²a)
=2(sina/cosa)/(1+sin²a/cos²a)
上下乘cos²a
=2sinacosa/(sin²a+cos²a)
=sin2a
√3(sin²a-cos²a)=-√2(cos²a-sin²a)
所以左边=sin2a-√3cos2a
=√[(1²+(√3)²]sin(2a-z)
=2sin(2a-z)
其中tanz=√3/1=√3
z=π/3
所以左边=2sin(2a-π/3)=右边
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