已知向量a=(2cos(x/2),tan(x/2+π/4)),b=(根号2sin(x/2+π/4),tan(x/2-π/
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/07/26 23:56:34
已知向量a=(2cos(x/2),tan(x/2+π/4)),b=(根号2sin(x/2+π/4),tan(x/2-π/4)),令f(x)=a×b,是否存在实数x∈[0,π]使f(x)+f’(x)=0,其中f’(x)是f(x)的导函数?若存在,则求出x的值,若不存在,则证明之
![已知向量a=(2cos(x/2),tan(x/2+π/4)),b=(根号2sin(x/2+π/4),tan(x/2-π/](/uploads/image/z/6468500-20-0.jpg?t=%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8Fa%3D%282cos%28x%2F2%29%2Ctan%28x%2F2%2B%CF%80%2F4%29%29%2Cb%3D%28%E6%A0%B9%E5%8F%B72sin%28x%2F2%2B%CF%80%2F4%29%2Ctan%28x%2F2-%CF%80%2F)
f(x)=a●b= 2cos(x/2)* √2*sin(x/2+π/4)+ tan(x/2+π/4)* tan(x/2-π/4)
=2√2 sin(x/2+π/4)* cos(x/2)+ tan(x/2+π/4)* tan(x/2-π/4)
=√2(sin(x+π/4)+sin(π/4))+[cos(π/2)-cos(x+π/2)]/[ cos(x+π/2)+ cos(π/2)]
=√2*(sin(x)*cos(π/4)+cos(x)*sin(π/4)+ 1/2*√2)+[-cos(x+π/2)]/[ cos(x+π/2)]
=√2*(sin(x)* 1/2*√2+cos(x)* 1/2*√2+ 1/2*√2)-1
= sin(x)+ cos(x)
f ’(x)=cos(x)-sin(x)
则f(x)+f’(x)=2*cos(x)=0,x=π/2
即存在实数x∈[0,π]使f(x)+f’(x)=0,且x=π/2
=2√2 sin(x/2+π/4)* cos(x/2)+ tan(x/2+π/4)* tan(x/2-π/4)
=√2(sin(x+π/4)+sin(π/4))+[cos(π/2)-cos(x+π/2)]/[ cos(x+π/2)+ cos(π/2)]
=√2*(sin(x)*cos(π/4)+cos(x)*sin(π/4)+ 1/2*√2)+[-cos(x+π/2)]/[ cos(x+π/2)]
=√2*(sin(x)* 1/2*√2+cos(x)* 1/2*√2+ 1/2*√2)-1
= sin(x)+ cos(x)
f ’(x)=cos(x)-sin(x)
则f(x)+f’(x)=2*cos(x)=0,x=π/2
即存在实数x∈[0,π]使f(x)+f’(x)=0,且x=π/2
已知向量a=(2cos(x/2),tan(x/2+π/4)),b=(根号2sin(x/2+π/4),tan(x/2-π/
已知向量a=(2cosx/2,1+tan^2x),b=(根号2sin(π/4+x/2),cos^2x),令f(x)=a*
已知向量a=(2cosx/2,1+tan²x),b=(根号2sin(x/2+π/4),cos²x);
已知向量a=(2cosX/2,tan(X/2+π/4)),b=(根号二sin(X/2+π/4)),tan=(X/2—π/
已知向量a=(2cosx/2,tan(x/2+π/4)),b=(√2sin(x/2+π/4),tan(x/2-π/4))
已知向量→a=(2cosx/2,tan(x/2+π/4)),→b=(√2sin(x/2+π/4),tan(x/2-π/4
已知向量a=(sin x,-2)与b=(1,cos x)垂直,求tan 2x的值?
已知tan=2,求(cos x+sin x)/(cos x-sin x)+sin^2x
f(x)=1/cos²x-tan²x+根号2*sin(2x-π/4)
已知:tan a,tan b是方程X^2+4X+3=0的两个根,求3cos^2(a+b)+sin(a+b)*cos(a+
已知(x)=[sin(π-x)cos(2π-x)tan(-x+π)]/[cos(-π/2+x)]
已知sin(π/2-x)+sin(π-x)/cos(-x)+sin(2π-x)=2009,则tan(x+5π/4)等于?