Question

求sin^10°+cos^40°+sin10°cos40°

求sin^10°+cos^40°+sin10°cos40°

Answer 1

运用余弦定理可得
因为c^2=a^2+b^2-2abcosC
运用正弦定理可得
(2rsinC)^2=(2rsinA)^2+(2rsinB)^2-2(rsinA)(rsinB)cosC
所以(sinC)^2=(sinA)^2+(sinB)^2-2sinAsinBcosC

原式=sin^2(10°)+sin^2(50°)+sin10°sin50°
=sin^2(10°)+sin^2(50°)-2sin10°sin50°cos120°
=sin^2(120°)
=3/4

Answer 2

sin^10°+cos^40°+sin10°cos40°
=(1-cos20)/2+(1+cos80)/2+sin10*cos40
=1+(cos80-cos20)/2+(1/2)[sin(-30)+sin50]
=1-sin50sin30-1/4+(1/2)sin50
=3/4

利用公式:
cosA-cosB=-2sin[(A+B)/2]sin[(A-B)/2]
sinAcosB=(1/2)[sin(A+B)+sin(A-B)]