求证:cos(α+β)cos(α-β)=cos^2 α-sin^2 β sin(α+β)sin(α-β)=sin^2 α-sin^2 βcos(α+β)cos(α-β)=cos^2 α-sin^2 β和sin(α+β)sin(α-β)=sin^2 α-sin^2 β
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求证:cos(α+β)cos(α-β)=cos^2 α-sin^2 β sin(α+β)sin(α-β)=sin^2 α-sin^2 βcos(α+β)cos(α-β)=cos^2 α-sin^2 β和sin(α+β)sin(α-β)=sin^2 α-sin^2 β
求证:cos(α+β)cos(α-β)=cos^2 α-sin^2 β sin(α+β)sin(α-β)=sin^2 α-sin^2 β
cos(α+β)cos(α-β)=cos^2 α-sin^2 β和sin(α+β)sin(α-β)=sin^2 α-sin^2 β
求证:cos(α+β)cos(α-β)=cos^2 α-sin^2 β sin(α+β)sin(α-β)=sin^2 α-sin^2 βcos(α+β)cos(α-β)=cos^2 α-sin^2 β和sin(α+β)sin(α-β)=sin^2 α-sin^2 β
因为 cos(a+b) = cos(a)cos(b) - sin(a)sin(b)
所以 cos[(α+β)+(α-β)] = cos(α+β)cos(α-β) - sin(α+β)sin(α-β)
即 cos(2α) = cos(α+β)cos(α-β) - sin(α+β)sin(α-β) ---------- (1)
同理,因为 cos(a-b) = cos(a)cos(b) + sin(a)sin(b)
可得 cos[(α+β)-(α-β)] = cos(α+β)cos(α-β) + sin(α+β)sin(α-β)
即 cos(2β) = cos(α+β)cos(α-β) + sin(α+β)sin(α-β) ---------- (2)
(1)(2)两式相加,得
cos(2α) + cos(2β) = 2cos(α+β)cos(α-β) ---------- (3)
因为 cos(2α) = cos(α+α) = cos^2(a) - sin^2(a) = cos^2(a) - sin^2(a) + 1 - 1 = 2cos^2(a) - 1
同理 cos(2β) = cos^2(β) - sin^2(β) - 1 + 1 = 1 - 2sin^2(β)
所以(3)式左边变为 2cos^2(a) - 2sin^2(β)
即得 cos(α+β)cos(α-β) = cos^2(α) - sin^2(β)
原题得证.
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第二个等式的证明原理相同,都是先化出(3)式,然后得出最后结果.
(3)式的这个等式就叫做“和差化积”.
除了上面出现的(3)式外,还有
cos(2β) - cos(2α) = 2sin(α+β)sin(α-β) ---------- (4)
sin(2α) + sin(2β) = 2sin(α+β)cos(α-β)
sin(2α) - sin(2β) = 2cos(α+β)sin(α-β)
观察欲证明的第二个等式,可知,需要用到的“和差化积”等式为(4)式.