作业帮已知sin(α+β)=3/5,sinβ=1/2,α+β,β均为锐角,求sinα的值.已知sin(α+β)=3/5,cos（α-β）=1/3,α+β,α-β均为锐角,求sin2α、cos2β.求[（sin5π/12）+（cos5π/12）]·[（sin5π/12）-（cos5π/12）]的值.求1+2cos²θ-cos2
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/17 16:28:44
![已知sin(α+β)=3/5,sinβ=1/2,α+β,β均为锐角,求sinα的值.已知sin(α+β)=3/5,cos（α-β）=1/3,α+β,α-β均为锐角,求sin2α、cos2β.求[（sin5π/12）+（cos5π/12）]·[（sin5π/12）-（cos5π/12）]的值.求1+2cos²θ-cos2](/uploads/image/z/5171170-58-0.jpg?t=%E5%B7%B2%E7%9F%A5sin%28%CE%B1%2B%CE%B2%29%3D3%2F5%2Csin%CE%B2%3D1%2F2%2C%CE%B1%2B%CE%B2%2C%CE%B2%E5%9D%87%E4%B8%BA%E9%94%90%E8%A7%92%2C%E6%B1%82sin%CE%B1%E7%9A%84%E5%80%BC.%E5%B7%B2%E7%9F%A5sin%28%CE%B1%2B%CE%B2%29%3D3%2F5%2Ccos%EF%BC%88%CE%B1-%CE%B2%EF%BC%89%3D1%2F3%2C%CE%B1%2B%CE%B2%2C%CE%B1-%CE%B2%E5%9D%87%E4%B8%BA%E9%94%90%E8%A7%92%2C%E6%B1%82sin2%CE%B1%E3%80%81cos2%CE%B2.%E6%B1%82%5B%EF%BC%88sin5%CF%80%2F12%EF%BC%89%2B%EF%BC%88cos5%CF%80%2F12%EF%BC%89%5D%C2%B7%5B%EF%BC%88sin5%CF%80%2F12%EF%BC%89-%EF%BC%88cos5%CF%80%2F12%EF%BC%89%5D%E7%9A%84%E5%80%BC.%E6%B1%821%2B2cos%26%23178%3B%CE%B8-cos2)
xNG_%Jxcܴy(fgvܘR/k%5PZH~@H&8|CWy cJr^X=~XWI~Xg'k;õPz,Z%=:L*ѳR|xw{Q0SL vFjEZ rsRHς H ޞow>d|H07W?B^UGZeF8?7Nu߯] nԪHq\J6TH䍍z};/t>~yp wMf`|j*x67>95O:kH;WY'/ǯЏ9BWG!/=XbG93ZS" h ,S\b_Qv$VI$ӟ %g-eQVZQ@v5R: (}Չ Y\lB!Q0FQõ`[H Lg#sQ
g^^rBcI@aDIBb#B>Yz[اO?CBL$cJmTPtC/ƾ8C(jCBs~:m% g3\ @%i(y,%uڐR [ NP!z0q?>#%-aL8n%=,圅B3W>n
!dmtc{4?ߍw7N
cOmJntRx܍zA\* 19x l5KӍ9z(=nE+ϓ'\[R|V
xP?mAx~X[q O_Ж͟SF\h+D !T0WP? oŃ-hAx9ͭLw^ k- ]O5ʯߗhq/&+ըZ:dY^㭵˽ O67;_,ş_=3љ#ti_ %Y9
已知sin(α+β)=3/5,sinβ=1/2,α+β,β均为锐角,求sinα的值.已知sin(α+β)=3/5,cos（α-β）=1/3,α+β,α-β均为锐角,求sin2α、cos2β.求[（sin5π/12）+（cos5π/12）]·[（sin5π/12）-（cos5π/12）]的值.求1+2cos²θ-cos2
已知sin(α+β)=3/5,sinβ=1/2,α+β,β均为锐角,求sinα的值.
已知sin(α+β)=3/5,cos（α-β）=1/3,α+β,α-β均为锐角,求sin2α、cos2β.
求[（sin5π/12）+（cos5π/12）]·[（sin5π/12）-（cos5π/12）]的值.
求1+2cos²θ-cos2θ的值.（写过程）
已知sin(α+β)=3/5,sinβ=1/2,α+β,β均为锐角,求sinα的值.已知sin(α+β)=3/5,cos（α-β）=1/3,α+β,α-β均为锐角,求sin2α、cos2β.求[（sin5π/12）+（cos5π/12）]·[（sin5π/12）-（cos5π/12）]的值.求1+2cos²θ-cos2
~请首先关注【我的采纳率】
~如果不懂,请继续追问!
~如果你认可我的回答,请及时点击【采纳为最佳回答】按钮~
~如还有新的问题,在您采纳后还可以继续求助我二次!
~您的采纳是我前进的动力~~
O(∩_∩)O,记得好评和采纳,互相帮助
祝学习进步!
第二题第二问那个是cos2β=cos[(α+β)-(α-β)]