sin²75°-sin²15°=?

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sin²75°-sin²15°=?
sin²15°-cos²15°=

利用二倍角公式cos2A=cos²A-sin²A∴sin²15°-cos²15°=-(cos²15°-sin²15°)=-cos30°=-√

cos²15°-sin²15°=

=cos(2×15°)=cos30°=√3/2

sin 15度乘sin 75度等于

sinα·sinβ=-(1/2)[cos(α+β)-cos(α-β)]sin15*sin75=-(1/2)(cos90-cos60)=1/4

sin(a+360°)=?

sin(a+360)=sina

△ABC中,sin²A+sin²B-sin²C=sinAsi

解题思路:第一问利用正弦定理求解,第二问先证明三角形是直角三角形,然后求出外接圆面积解题过程:

证明:(1) sin(360°-α)=-sinα

奇变偶不变,符号看象限.比如(1)中,360°等于180°乘2,2是偶数,所以函数名不变,后面先写上sinα,然后把α看成锐角,-α就是第四象限角,加360°后仍是第四象限角,第四象限角的sinα(看

sin(x+105°)cos(x-15°)-cos(x+105°)sin求化简

根据公式sin(a-b)=sinacosb-cosasinb所以原式=sin(x+105-x+15)=sin120°=√3/2

一道数学化简题sin 90°-α/sin α =

sin(90°-α)/sinα=cosα/sinα=cotα再问:是cosα还是cotα再答:cotα

sin²1°+sin²2°+sin²89°+sin²88°=

原式=sin²1+sin²2+cos²(90-89)+cos²(90-88)=(sin²1+cos²1)+(sin²2+cos&#

sin²1°+sin²2°+……sin²88°+sin²89°

sin²1°+sin²2°+……sin²88°+sin²89°=(sin²1°+sin²89°)+(sin²2°+sin²

化简cos(15°+a)sin(45°+a)-cos(75°-a)sin(45°-a)

由题意可得:cos(75°-a)=sin(15°+a)sin(45°-a)=cos(45°+a)所以cos(15°+a)sin(45°+a)-cos(75°-a)sin(45°-a)=cos(15°+

sin(15)×sin(30)×sin(75)等于多少

sin(15)×sin(30)×sin(75)=sin15°*cos15°*sin30°=1/2*(2sin15°*cos15°)*sin30°=1/2sin30°*sin30°=1/2*1/2*1/

计算:sin² 15°+tan15°·tan75°-cot15°·cot90°+sin² 75°的值

原式=(sin15)^2+tan15*tan(90-15)-cot15*0+[sin(90-15)]^2=(sin15)^2+tan15*cot15-cot15*0+(cos15)^2=[(sin15

计算:sin方10°+sin方20°+sin方30°+sin方40°+sin方50°+sin方60°+sin方70°+s

sin方10°+sin方20°+sin方30°+sin方40°+sin方50°+sin方60°+sin方70°+sin方80°+sin方90°=sin²10+sin²80+sin&

sin^2 (0°)+sin^2( 1°)+.+sin^2(90du) 求和

°省略原式=sin²0+sin²1+sin²2+……+sin²44+sin²45+cos²(90-46)+……+cos²(90-8

证明:sin(360°-α)=-sinα

sin(360°-a)=sin(-a)=-sina

计算:sin²1°+sin²2°+sin²3°...+sin²45°+sin&#

sin²1°+sin²2°+sin²3°...+sin²45°+sin²46°...+sin²89°=sin^2(90-89)+sin^2(

数列求和 sin²1°+sin²2°+sin²3°+.+sin²88°+sin&

sin(π/2-x)=cosx原式=sin^21°+……+sin^244°+1/2+cos^244°+……+cos^21°=44+1/2=89/2

证明sin(a+b)sin(a-b)=sin²a-sin² b,并用该式计算sin²20°

sin(a+b)sin(a-b)=(sinacosb+sinbcosa)(sinacosb-sinbcosa)=(sinacosb)^2+sinasinbcosacosb-sinasinbcosaco