cos2x cos^2sin^2的不定积分是多少?

（1）∵函数f(x)=12sin2xsinφ+cos2xcosφ-12sin(π2+φ)(0<φ<π)，∴f（x）=12sin2xsin∅+1+cos2x2•cos∅-12cos∅=12s

楼上都错了,图像没问题这个表达式实际是个常数,你可以运行TrigReduce[Sin[x]Sin[x+2]-Sin[x+1]^2]看看,结果为1/2(-1+Cos[2])只不过Plot的自动选择坐标系

(2sina+cosa)=-5(sina-3cosa)7sina=14cosasina=2cosa

∵y＝cos2xcosπ5−2sinxcosxsin6π5y＝cos2xcosπ5−sin2xsin6π5=cos2xcosπ5−sin2xsinπ5=cos（2x+π5）∴2x+π5∈[2kπ-π，

左边=(sinacosb+cosasinb)(sinacosb-cosasinb)=sin²acos²b-cos²asin²b=sin²a(1-sin

2sin2α+2sinαcosα1+tanα=2sinα(sinα+cosα)1+sinαcosα=2sinαcosα(sinα+cosα)sinα+cosα=2sinαcosα=k．当0＜α＜π4时

证明：∵sin（2α+β）-2cos（α+β）sinα=sin[（α+β）+α]-2cos（α+β）sinα=sin（α+β）cosα+cos（α+β）sinα-2cos（α+β）sinα=sin（α

=sin^2(x)*[cos^2(x)-1]=-sin^4(x)再答：别忘了负号再问：嗯谢谢

2(sin120)^2+(sin30)^2=2(sin60)^2+(sin30)^2=2(cos30)^2+(sin30)^2=2cos²30+sin²30=cos²30

（Ⅰ）f（x）=12sin2xsinφ+cos2xcosφ-sin（π2+φ）=12sin2xsinφ+12（1+cos2x）cosφ-12cosφ=12sin2xsinφ+12cos2xcosφ=1

sin^2/(sin-cos)-(sin+cos)/(tan^2-1)=sin^2/(sin-cos)-(sin+cos)/[(sin^2/cos^2)-1]=sin^2/(sin-cos)-(sin

(tan^2-sin^2)cot^2=(sin^2/cos^2-sin^2)cot^2=sin^2(1/cos^2-1)cot^2=sin^2(1-cos^2)/cos^2*cot^2=[sin^2*

/>利用积化和差公式,达到裂项的效果.2sinka*sin(a/2)=-cos[(k+1/2)a]+[cos(k-1/2)a]∴2sin(a/2)*(sina+sin2a+sin3a+...+sinn

Sin[a+b]Sin[a-b]积化和差公式得=1/2(-Cos[2a]+Cos[2b])余弦二倍角公式得=Sin[a]^2-Sin[b]^2Sin[80°]Sin[40°]=Sin[60°+20°]

应该是sinA+sinB=2sin[(A+B)/2]cos[(A-B)/2]A=(A+B)/2+(A-B)/2.B=(A+B)/2-(A-B)/2所以sin(A+B)/2cos(A-B)/2+cos(

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

答：sin^2a+sin^2(a+60)+sin^2（a+120）=3/2.证明：左边=sin^2a+sin^2(a+60)+sin^2（a+120）=sin^2a+(sinacos60+cosasi

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

∵sin(π4-x)=513(0＜x＜π4)，∴cos（π4-x）=1213，∴cos2xcos(π4+x)=sin(π2-2x)sin(π4-x)=2sin(π4-x)cos(π4-x)