解答
证明 cos(x)cos(3x)=1−4sin2(x)
解答
真
求解步骤
cos(x)cos(3x)=1−4sin2(x)
调整左侧cos(x)cos(3x)
使用三角恒等式改写
cos(x)cos(3x)
利用以下特性:cos(3x)=4cos3(x)−3cos(x)
cos(3x)
使用三角恒等式改写
cos(3x)
改写为=cos(2x+x)
使用角和恒等式: cos(s+t)=cos(s)cos(t)−sin(s)sin(t)=cos(2x)cos(x)−sin(2x)sin(x)
使用倍角公式: sin(2x)=2sin(x)cos(x)=cos(2x)cos(x)−2sin(x)cos(x)sin(x)
化简 cos(2x)cos(x)−2sin(x)cos(x)sin(x):cos(x)cos(2x)−2sin2(x)cos(x)
cos(2x)cos(x)−2sin(x)cos(x)sin(x)
2sin(x)cos(x)sin(x)=2sin2(x)cos(x)
2sin(x)cos(x)sin(x)
使用指数法则: ab⋅ac=ab+csin(x)sin(x)=sin1+1(x)=2cos(x)sin1+1(x)
数字相加:1+1=2=2cos(x)sin2(x)
=cos(x)cos(2x)−2sin2(x)cos(x)
=cos(x)cos(2x)−2sin2(x)cos(x)
=cos(x)cos(2x)−2sin2(x)cos(x)
使用倍角公式: cos(2x)=2cos2(x)−1=(2cos2(x)−1)cos(x)−2sin2(x)cos(x)
使用毕达哥拉斯恒等式: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=(2cos2(x)−1)cos(x)−2(1−cos2(x))cos(x)
乘开 (2cos2(x)−1)cos(x)−2(1−cos2(x))cos(x):4cos3(x)−3cos(x)
(2cos2(x)−1)cos(x)−2(1−cos2(x))cos(x)
=cos(x)(2cos2(x)−1)−2cos(x)(1−cos2(x))
乘开 cos(x)(2cos2(x)−1):2cos3(x)−cos(x)
cos(x)(2cos2(x)−1)
使用分配律: a(b−c)=ab−aca=cos(x),b=2cos2(x),c=1=cos(x)2cos2(x)−cos(x)1
=2cos2(x)cos(x)−1cos(x)
化简 2cos2(x)cos(x)−1⋅cos(x):2cos3(x)−cos(x)
2cos2(x)cos(x)−1cos(x)
2cos2(x)cos(x)=2cos3(x)
2cos2(x)cos(x)
使用指数法则: ab⋅ac=ab+ccos2(x)cos(x)=cos2+1(x)=2cos2+1(x)
数字相加:2+1=3=2cos3(x)
1⋅cos(x)=cos(x)
1cos(x)
乘以:1⋅cos(x)=cos(x)=cos(x)
=2cos3(x)−cos(x)
=2cos3(x)−cos(x)
=2cos3(x)−cos(x)−2(1−cos2(x))cos(x)
乘开 −2cos(x)(1−cos2(x)):−2cos(x)+2cos3(x)
−2cos(x)(1−cos2(x))
使用分配律: a(b−c)=ab−aca=−2cos(x),b=1,c=cos2(x)=−2cos(x)1−(−2cos(x))cos2(x)
使用加减运算法则−(−a)=a=−2⋅1cos(x)+2cos2(x)cos(x)
化简 −2⋅1⋅cos(x)+2cos2(x)cos(x):−2cos(x)+2cos3(x)
−2⋅1cos(x)+2cos2(x)cos(x)
2⋅1⋅cos(x)=2cos(x)
2⋅1cos(x)
数字相乘:2⋅1=2=2cos(x)
2cos2(x)cos(x)=2cos3(x)
2cos2(x)cos(x)
使用指数法则: ab⋅ac=ab+ccos2(x)cos(x)=cos2+1(x)=2cos2+1(x)
数字相加:2+1=3=2cos3(x)
=−2cos(x)+2cos3(x)
=−2cos(x)+2cos3(x)
=2cos3(x)−cos(x)−2cos(x)+2cos3(x)
化简 2cos3(x)−cos(x)−2cos(x)+2cos3(x):4cos3(x)−3cos(x)
2cos3(x)−cos(x)−2cos(x)+2cos3(x)
对同类项分组=2cos3(x)+2cos3(x)−cos(x)−2cos(x)
同类项相加:2cos3(x)+2cos3(x)=4cos3(x)=4cos3(x)−cos(x)−2cos(x)
同类项相加:−cos(x)−2cos(x)=−3cos(x)=4cos3(x)−3cos(x)
=4cos3(x)−3cos(x)
=4cos3(x)−3cos(x)
=cos(x)4cos3(x)−3cos(x)
化简 cos(x)4cos3(x)−3cos(x):4cos2(x)−3
cos(x)4cos3(x)−3cos(x)
分解 4cos3(x)−3cos(x):cos(x)(4cos2(x)−3)
4cos3(x)−3cos(x)
使用指数法则: ab+c=abaccos3(x)=cos(x)cos2(x)=4cos(x)cos2(x)−3cos(x)
因式分解出通项 cos(x)=cos(x)(4cos2(x)−3)
=cos(x)cos(x)(4cos2(x)−3)
约分:cos(x)=4cos2(x)−3
=4cos2(x)−3
使用毕达哥拉斯恒等式: cos2(x)+sin2(x)=1cos2(x)=1−sin2(x)=4(1−sin2(x))−3
化简 4(1−sin2(x))−3:−4sin2(x)+1
4(1−sin2(x))−3
乘开 4(1−sin2(x)):4−4sin2(x)
4(1−sin2(x))
使用分配律: a(b−c)=ab−aca=4,b=1,c=sin2(x)=4⋅1−4sin2(x)
数字相乘:4⋅1=4=4−4sin2(x)
=4−4sin2(x)−3
化简 4−4sin2(x)−3:−4sin2(x)+1
4−4sin2(x)−3
对同类项分组=−4sin2(x)+4−3
数字相加/相减:4−3=1=−4sin2(x)+1
=−4sin2(x)+1
=−4sin2(x)+1
=−4sin2(x)+1
=1−4sin2(x)
我们已展示,在两侧可以有相同的形式⇒真