若sin(x 50°) cos(x 20°)=根3-搜答案-大师作文网

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（

两边平方得sin²x/2-2sinx/2cosx/2+cos²x/2=1/91-sinx=1/9sinx=8/9

虽然它们都小于1举个例子,sinx=0.7,cos=0.5.0.7^2>0.5^2,即使小于0,还是符合

根据公式sin（a-b）=sinacosb-cosasinb所以原式=sin(x+105-x+15）=sin120°=√3/2

sin^2x+cos^2x=1

原式=cos[(x-25)-(65-x)]=cos(2x-90)=sin2x

这个方程无解取区间[-π,π]-1

开方可以消掉

cos(180°+x)*sin(x+360°)/[sin(-x-180°)*cos(-180°-x)]=-cosx×sinx/[-sin(180°+x)][cos(180°+x)]=-cosx×sin

f(x)=sin(x+α)-2cos(x-α)是奇函数f(-x)=-f(x)sin(-x+a)-2cos(-x-a)=-(sin(x+a)-2cos(x-a))-sin(x-a)-2cos(x+a)=

∫cos2xdx/(sin^2xcos^2x)=4∫cos2xdx/(2sinxcosx)^2=4∫cos2xdx/(sin2x)^2=2∫cos2xd(2x)/(sin2x)^2=2∫d(sin2x

套公式啊,sin(10°+x)cos(50°-x)+sin(50°-x)cos(10°+x)=sin(10°+x+50°-x)=sin60°=√3/2.再问：诶……那cos呢……再答：sin(A+B)

2cos x (sin x -cos x)+1

2cosx(sinx-cosx)+1=2sinxcosx-2cosx^2+1=sin2x+1-2cosx^2=sin2x-cos2x=√2sin(2x-π/4)

sin(1/x)-cos(1/x)/x

该函数在x=0处的左右极限都没有比如x=1/(2npi+pi/2)时,f(x)=1x=1/(2npi-pi/2)时,f(x)=-1取n->无穷大所以在x=0处没有右极限,左极限同理

sinx=2cosx,sin^2x=4cos^2xsin^2x=4-4sin^2x,sin^2x=4/5(cosx+sinx)/(cosx-sinx)+sin^2x=(1+tanx)/(1-tanx)

算错了,答案应该是： ±3分之根号5

cos(33°-x)sin(63°-x)-cos(x+57°)sin(27°+x)=cos[90°-(x+57°)]sin[90°-(x+27°]-cos(x+57°)sin(x+27°)=sin(x

f(sin15)=f(cos75)=cos150=-cos30=-√3/2

(x*sinx*cosx)'=(1/2xsin2x)'=1/2(sin2x+xcos2x*2)=1/2sin2x+xcos2x

利用积化和差公式sinAcosB=(1/2)*[sin(A+B)+sin(A-B)]以及sin(-A)=-sinA的性质可得sin(x)(cos(x)+cos(3x)+cos(5x)+cos(7x))