已知向量a=(sin(α+π/6),1),b=(4,4cosα-根号3),若a⊥b,则sin(α+4π/3)等于a⊥b,则a*b=0向量;sin(α+π/6)+cosα=√3/4sin(α+π/3)=1/4,所以sin(α+4π/3)=-1/4请问sin(α+π/6)+cosα=√3/4→sin(α+π/3)=1/4是怎么得到的
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![已知向量a=(sin(α+π/6),1),b=(4,4cosα-根号3),若a⊥b,则sin(α+4π/3)等于a⊥b,则a*b=0向量;sin(α+π/6)+cosα=√3/4sin(α+π/3)=1/4,所以sin(α+4π/3)=-1/4请问sin(α+π/6)+cosα=√3/4→sin(α+π/3)=1/4是怎么得到的](/uploads/image/z/535629-21-9.jpg?t=%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8Fa%3D%28sin%28%CE%B1%2B%CF%80%2F6%29%2C1%29%2Cb%3D%284%2C4cos%CE%B1-%E6%A0%B9%E5%8F%B73%29%2C%E8%8B%A5a%E2%8A%A5b%2C%E5%88%99sin%28%CE%B1%2B4%CF%80%2F3%29%E7%AD%89%E4%BA%8Ea%E2%8A%A5b%2C%E5%88%99a%2Ab%3D0%E5%90%91%E9%87%8F%EF%BC%9Bsin%28%CE%B1%2B%CF%80%2F6%29%2Bcos%CE%B1%3D%E2%88%9A3%2F4sin%28%CE%B1%2B%CF%80%2F3%29%3D1%2F4%2C%E6%89%80%E4%BB%A5sin%28%CE%B1%2B4%CF%80%2F3%29%3D-1%2F4%E8%AF%B7%E9%97%AEsin%28%CE%B1%2B%CF%80%2F6%29%2Bcos%CE%B1%3D%E2%88%9A3%2F4%E2%86%92sin%28%CE%B1%2B%CF%80%2F3%29%3D1%2F4%E6%98%AF%E6%80%8E%E4%B9%88%E5%BE%97%E5%88%B0%E7%9A%84)
已知向量a=(sin(α+π/6),1),b=(4,4cosα-根号3),若a⊥b,则sin(α+4π/3)等于a⊥b,则a*b=0向量;sin(α+π/6)+cosα=√3/4sin(α+π/3)=1/4,所以sin(α+4π/3)=-1/4请问sin(α+π/6)+cosα=√3/4→sin(α+π/3)=1/4是怎么得到的
已知向量a=(sin(α+π/6),1),b=(4,4cosα-根号3),若a⊥b,则sin(α+4π/3)等于
a⊥b,则a*b=0向量;sin(α+π/6)+cosα=√3/4
sin(α+π/3)=1/4,所以sin(α+4π/3)=-1/4
请问sin(α+π/6)+cosα=√3/4→sin(α+π/3)=1/4是怎么得到的
已知向量a=(sin(α+π/6),1),b=(4,4cosα-根号3),若a⊥b,则sin(α+4π/3)等于a⊥b,则a*b=0向量;sin(α+π/6)+cosα=√3/4sin(α+π/3)=1/4,所以sin(α+4π/3)=-1/4请问sin(α+π/6)+cosα=√3/4→sin(α+π/3)=1/4是怎么得到的
sin(α+π/6)=sinα·cos(π/6)+cosα·sin(π/6)=(√3/2)sinα+(1/2)cosα;
所以sin(α+π/6)+cosα=(√3/2)sinα+(3/2)cosα;
即(√3/2)sinα+(3/2)cosα=√3/4
sinα+(√3)cosα=1/2;
(1/2)sinα+(√3/2)cosα=1/4
即sinα·cos(π/3)+cosα·sin(π/3)=1/4;
即sin(α+π/3)=1/4
dfwrw
对sin(α+π/6)+cosα=√3/4进行配方得
sin[(α+π/3)-π/6]+cos[(α+π/3)-π/3]=√3/4
展开可以求出sin(a+π/3)=1/4