设x为锐角,若cos(x π 6)=3 5
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解决方案,cosxcos30°-sinxsin30°+的sinx=0/5√3/2cosx1/2sinx=0/5COS(X-30°)=8/10COS(X-60°)=COS2(X-30°)=2cos^2(
f(x)=cos²x+sinxcosx=1/2*(sin2x+cos2x)+1/2=√2/2*sin(2x+π/4)+1/2f(A)=√2/2*sin(2A+π/4)+1/2=1sin(2A
cos(X-cosX)=sin(π/2-X+cosX)带入原式sin(x+sinx)=sin(π/2-x+cosx)x+sinx=π/2-x+cosxsinx-cosx=π/2-2x对于原式左侧有si
(1)f(x)=sin²x-sin(2x-π/6)=(1-cos2x)/2-(sin2xcosπ/6-cos2xsinπ/6)=1/2-(cos2x)/2-sin2xcosπ/6+(cos2
1)f(x)=1+cos(2x+π/3)-(1+cos2x)/2=1/2-sin2x根号3/2最小值1/2-根号3/2最小正周期π2)c带入得sinC=根号3/2C=π/3A=π-B-C=2π/3-a
α∈(0,π/2)α+π/6∈(π/6,2π/3)cos(α+π/6)=4/5>0∴α+π/6∈(π/6,π/2)∴2α+π/3∈(π/3,π)cos(2α+π/3)=2cos²(α+π/6
sin(2α+π/12)=sin(α+π/6+α-π/12)=sin(α+π/6)cos(α-π/12)+cos(α+π/6)sin(α-π/12)cos(α+π/6)=4/5,α为锐角,则sin(α
设b=a+π/6,sinb=3/5,sin2b=2sinbcosb=24/25,cos2b=7/25sin(2a+π/12)=sin(2a+π/3-π/4)=sin(2b-π/4)=sin2bcosπ
50分之17倍根号2
a为锐角,cos(a+π/6)=3/5则:sin(a+π/6)=4/5sin(a-π/12)=sin[(a+π/6)-π/4]=sin(a+π/6)cos(π/4)-cos(a+π/6)sin(π/4
sin(2a+π/12)=sin【2(a+π/6)—π/4】=sin2(a+π/6)cos(π/4)—cos2(a+π/6)sin(π/4)因为cos(a+π/6)=4/5,a为锐角,所以0
设A+π/6=α,则2A+π/12=2α-π/4cosα=4/5,α为锐角,sinα=3/5,所以sin2α=2sinαcosα=24/25,cos2α=2(cosα)^2-1=7/25sin(2A+
设α为锐角,若cos(α+π/6)=4/5,求sin(2α+π/12),cos(α+π/6)=4/5sin(α+π/6)=3/5,sin(2α+π/3)=24/25cos(2α+π/3)=7/25si
设x为锐角,cos(x+π除以6)=五分之四=4/5,∵4/5>√2/2,cos(π除以4)=√2/2∴0
f(x)=cos(√3x+A)f'(x)=-√3sin(√3x+A)f'(x)+f(x)=cos(√3x+A)-√3sin(√3x+A)-(f'(-x)+f(-x))=-cos(-√3x+A)+√3s
cos(x-π/4)=sin2x+acosxcos(π/4)+sinxsin(π/4)=2sinxcosx+a(根号2/2)*cosx+(根号2/2)*sinx=2sinxcosx+a(根号2/2)(
∵sin(πcosx)=cos(πsinx)=sin(π/2-πsinx)∴πcosx=π/2-πsinx或πcosx+(π/2-πsinx)=π即cosx+sinx=1/2或cosx-sinx=1/
f(a)=6cos²a-√3sin2a=3(1+cos2a)-√3sin2a=3cos2a-√3sin2a+3=2√3(√3/2cos2a-1/2sin2)+3=2√3cos(2a+π/6)