设θ为锐角,若cos(θ-4分之3π)=5分之3,则sin(θ 4分之π)=

（sinθ+cosθ）²=sin²θ+cos²θ+2sinθcosθ=1+2*（1/4）=3/2所以sinθ+cosθ=根号（3/2）=（根号6）/2

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（

要求y的范围,只要求y^2的范围y^2=(cosθ)^4*(1-cosθ^2)设（cosθ）^2=t(

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

α+β=θ+φ=π/2所以有cosβ=sin(π/2-β)=sinα,cosθ=sin(π/2-θ)=sinφa*b=cosα*cosθ+cosβ*cosφ=cosα*sinφ+sinα*cosφ=s

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

证明由题设可知tanθ=sinθ/cosθ=(sina-cosa)/(sina+cosa).结合sina+cosa＞0,tanθ＞0可得sina-cosa＞0∴可设:sina-cosa=psinθsi

(cos-sin)/(cos+sin)=3+2√2两边平方（1-2sinθcosθ)/(1+2sinθcosθ)=17+12√2（1-2sinθcosθ)=(1+2sinθcosθ)(17+12√2)

a是锐角π/2

a为锐角,cos(a+π/6)=4/5所以,sin(a+π/6)=3/5sin(2a+π/3)=2sin(a+π/6)cos(a+π/6)=2×(3/5)×(4/5)=24/25cos(2a+π/3)

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

cos(2a+π/3)=2cos²(a+π/6)-1=7/25a为锐角,则：2a+π/3∈(π/3,π/2)∴sin(2a+π/3)=24/25sin(2a+π/12)=sin[(2a+π/

50分之17倍根号2

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

证明:因为θ是锐角,即0

sinθ-cosθ=1/2两边同时平方：sinθ^2-2sinθcosθ+cosθ^2=1/4-2sinθcosθ=-3/42sinθcosθ=3/4则sinθ＋cosθ=√（sinθ＋cosθ）^2

sinθ+cosθ=√2(sinθ*√2/2+cosθ*√2/2)又因为有sin(α+β)=sinαcosβ+cosαsinβ,且sin45=cos45=√2/2所以sinθ*√2/2+cosθ*√2