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sin11.5°

因为一个2π是一个周期 sin(11π/5)=sin(10π/5+1π/5)=sin(2π+1π/5)=sin(π/5)

sin112.5=sin(225/2),其中sin225=sin(180+45)=-sin45=-(根号2)/2,由sin225=2sin112.5cos112.5=2sin112.5*根号[1-(sin112.5)^2]得sin112.5 补充: 最后sin112.50) 补充: sin22.5=2sin11.25cos11.25=2sin11.25根号[1-(sin11.25)^2]解得sin11.25...

sin11.5=0.19936793441719718108604 sin3.5=0.06104853953485687203667013038475

此类方程设sin(θ/5)=x,利用多倍角公式可以将 11sin(11θ/5)-3sin(3θ/5) = 0 化成关于x的高次方程(只要是奇次方就可以),然后求解,但估计该题难解。

解: cos25/12π·cos11/6π-sin11/12π·sin5/6π =cos(π/12)cos(π/6)-sin(π/12)sin(π/6) =cos(π/12-π/6) =cos(π/12) =cos(π/3-π/4) =cos(π/3)cos(π/4)-sin(π/3)sin(π/4) =1/2*√2/2-√3/2*√2/2 =(√2-√6)/4

1:sin(-11π/6)+cos12π/5 *tan4π =sin(π/6)+cos(2π/5)tan0 =1/2; 2:sin11ts40*cos(-690)+tan1845 =sin60°cos30°+tan45º =√3/2*(√3/2)+1 =3/4+1 =7/4;

(1) tan 32°=0.625 (2) cos 24.53°=0.910 (3) sin62°11′=0.884 (4) tan39° 39′ 39〃=0.829 (5)sin256°+cos2°=0.029

sina-sinb=2cos[(a+b)/2]sin[(a-b)/2] 所以4sin(3ωt+45º)-4 sin(3ωt+15º)=8cos(3wt+30º)sin15=2cos(3wt+30)*(根号6-根号2) (根据正弦函数性质) 有效值=振幅/2倍根号2 3+2(根号6-根号2)/根号2=....找不到答案不知道哪里错了 ...

作∠B的平分线交AC于D,则∠DBA=1/2∠B=∠C,∠A=∠A, ∴△ABC∽△ADB ∴AB/AC=AD/AB 又∵AB=5 ∴5/(AD+DC)=AD/5 ① 由AD平分∠B,角平分线定理可得:AB/AD=BC/CD即5/AD=11/CD ② 联立①②解得AD=(5√5)/4,DC=(11√5)/4 ∴AC=AD+DC=4√5 ∴cos∠B=(AB²+BC²-AC...

sin(5π/4)×cos(4π/5)×tan(11π/6) =sin(π+π/4)×cos(π-π/5)×tan(π+5π/6) =-sin(π/4)×[-cos(π/5)]×tan(5π/6) =√2/2×cos36°×tan150° =√2/2×(-√3/3)cos36° =-√6/6×cos36° 36°是锐角,所以cos36°是正,所以这个式子是负号 cos150°×tan330°×sin...

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