已知sin(派 6-x)=3 5,则cos(x 3)=

根据诱导公式,sin(x-3π/2)cos(π-x)tan(π+x)=(cosx)(-cosx)(-tanx)=sinxcosx,∵sinx=3/5,∴cosx=±4/5,原式=±12/25.

再答：亲，满意请采纳再问：谢谢亲再问：再问：可以帮我解决第二问吗再答：恩，我看下再问：麻烦了。O(∩_∩)O再答：再答：再问：谢谢亲再答：😊😊再问：美丽的姑凉谢谢你再答

y=sin(2x-π/3)x∈[π/6,π/2]所以2x-π/3∈[0,2π/3]当2x-π/3=0即x=π/6时函数y有最小值为0当2x-π/3=π/2即5π/12时函数y有最大值为1所以值域为【0

稍等再答：sin(π/3-x)=sin[π/2-(x+π/6)]=cos(x+π/6)所以原式=1再答：快采纳

f(x)=(1+cos(2x-π/3))/2-(1-cos2x)/2=(cos(2x-π/3)+cos2x)/2f(π/12)=(cos(-π/6)+cosπ/6)/2=cosπ/6=√3/2

sinx=-5/13x∈(П,3П/2)cosx=√1-(-5/13)^2=-12/13sin(x-П/4)=sinxcosП/4-cosxsinпП/4=√2/2(-5/13+12/13)=7√2/

当x=11π/4时,1/3x-π/6=3π/4所以,f（11π/4）=sin3π/4=二分之根号二希望我的答案能让您满意,如有不明白的地方,请继续发问

∵-π/6≤x≤13π/12∴0≤2x+π/3≤5π/22sin(2x+π/3)-1=asin(2x+π/3)=(a+1)/2令t=2x+π/3∈[0,5π/2]sint=(a+1)/2（画出sint

Sin(2x+派/6)=3/5,2x+派/6=2Kπ+37°或2Kπ+143°,所以x1=Kπ+3点5°或Kπ+56点5°（其中派/6=30°,一般情况下两直角边为3和4的直角三角形,其两锐角约为37

f(x)=2sin(2x+π/6)=2(sin2x根号3/2+cos2x乘以1/2)=根号3乘以sin2x+cos2x所以有根号3乘以sin2x1+cos2x1=6/5；sin2x1平方+cos2x1

根号3乘以sin2x1+cos2x1=2/5,tan2x1=-根号3/3,cos2x1的平方=3/4,cos2x1=-根号3/2

f(x)=2sin(派-x)cosx=2sinxcosx=sin2x最小正周期=2pi/2=pi(pi就是“派”）f(-pi/6)=sin(-pi/3)=-(根号3)/2f(pi/2)=sin(pi)

sina=5/13且a是第二象限角,则可知cosa=-12/13,f(a/2)=2sin(a+π/6)=2sina*cosπ/6+2sinπ/6*cosa=2*5/13*√3/2+2*1/2*(-12

cos[(3/4派)+x]=cos[π-(π/4-x)]=-cos(π/4-x)=-cos[(π/2-(π/4+x)]=-sin(π/4+x)=1/5

sin(x+π/3)sin(x+11π/6)=1/2则有：sin(x+π/3)sin[(x-π/6)+2π]=1/2sin(x+π/3)sin(x-π/6)=1/2-----(1)由于：sin(x+π

y=sin(x+π/6)sin(x-π/6)+acosx=(3/4)(sinx)^2-(1/4)(cosx)^2+acosx=-(cosx)^2+acosx+3/4=-(cosx-a/2)^2+a^2

f(x)=2sin(x-π/6)cosx+2cos²x=(2sinxcosπ/6-2cosxsinπ/6)cosx+2cos²x=√3sinxcosx-cos²x+2co

f(x)=sinx-cosx=√2sin(x-4/π)(1).T=2π(2).f(x)max=√2f(x)min=-√2(3).sina+cosa=√2cos(a-π/4)cos(a-π/4)=√[1

f(x)=2sin2(π/6x)+sin(π/3x+π/6)-1=sin(π/3x+π/6)-[1-2sin2(π/6x)]=sin(π/3x+π/6)-cos(π/3x)=sin(π/3x)*cos

已知f(x)=2sin(πx/8＋π/4)＋1g(x)=f(x)＋f(-x)=2sin(πx/8＋π/4)＋1+2sin(-πx/8＋π/4)＋1=2[sin(πx/8＋π/4)+sin(-πx/8＋