求高中三角函数推导
sin(a+b)+cos(a-b)==2cos(a+π/4)cos(b+π/4)
是怎么被推导出来的?
如果是
sin(a-b)+cos(a+b)
则可以推导成上述形式吗? 展开
利用三角函坦棚数化简而来,不建议直接推广(唯信搭本来三角函数的公指拿式就比较多,不要给自己的大脑增加太多负担!)。
sin(A+B)+cos(A-B)
=(sinAcosB+cosAsinB)+(cosAcosB+sinAsinB)
=sinAcosB+cosAsinB+cosAcosB+sinAsinB
=sinA(sinB+cosB)+cosA(sinB+cosB)
=(sinB+cosB)(sinA+cosA)
={√2·[(√2/2)sinB+(√2/2)cosB]}·{√2·[(√2/2)sinA+(√2/2)cosA]}
={√2·[sin(π/4}sinB+cos(π/4)cosB]}·{√2·[sin(π/4)sinA+cos(π/4)cosA]}
=√2cos[B-(π/4)]·√2cos[A-(π/4)]
=2cos[B-(π/4)]·cos[A-(π/4)]
【或者=2sin[A+(π/4)]cos[B+(π/4)]】
sin(a+b)+cos(a-b)=
=sin a·cos b+sin b·cos a+cos a·cos b+sin a·sin b
=(sin a+cos a)(sin b+cos b)
=2[(√2/2)sin a+(√2/2)cos a][(√缓茄2/2)sin b+(√2/2)cos b]
=2[sin(π/4)sin a+cos(π/4)cos a][sin(π/4)sin b+cos(π/4)cos b]
=2cos(a-π/4)cos(b-π/4)
(弯悉你好像记错了……)
sin(a-b)+cos(a+b)=
=sin a·cos b-sin b·cos a+cos a·cos b-sin a·sin b
=(sin a+cos a)(cos b-sin b)
=2[(√2/2)sin a+(√2/2)cos a][(√2/2)sin b-(√扰闹察2/2)cos b]
=2[sin(π/4)sin a+cos(π/4)cos a][sin(π/4)sin b-cos(π/4)cos b]
=-2cos(a-π/4)cos(b+π/4) 或 2sin(a+π/4)sin(b-π/4)
2022-06-09