Cho y = trên và F(x) là một nguyên hàm của hàm số xf ‘(x) thỏa mãn F(0) = 0. Biết thỏa mãn tan a = 3. Tính F(a) – 10a2 + 3a
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cosx >0
=>\(-\frac{\pi}{2}+k2\pi
mà \(x\in\left\lbrack-\pi;\pi\right\rbrack\)
nên \(-\frac{\pi}{2}
cosx <0
=>\(\frac{\pi}{2}+k2\pi
mà \(x\in\left\lbrack-\pi;\pi\right\rbrack\)
nên \(-\pi
\(F=cos\left(\frac{\pi}{4}+a\right)\cdot cos\left(\frac{\pi}{4}-a\right)\)
\(=\frac12\cdot\left\lbrack cos\left(\frac{\pi}{4}+a-\frac{\pi}{4}+a\right)+cos\left(\frac{\pi}{4}+a+\frac{\pi}{4}+a\right)\right\rbrack\)
\(=\frac12\cdot\left\lbrack cos\left(2a\right)+cos\left(\frac{\pi}{2}\right)\right\rbrack=\frac12\cdot cos2a\)
\(G=\sin\left(\frac{\pi}{3}+a\right)\cdot cos\left(\frac{\pi}{3}-a\right)\)
\(=\frac12\cdot\left\lbrack\sin\left(\frac{\pi}{3}+a+\frac{\pi}{3}-a\right)+\sin\left(\frac{\pi}{3}+a-\frac{\pi}{3}+a\right)\right\rbrack\)
\(=\frac12\cdot\left\lbrack\sin\left(\frac23\pi\right)+\sin2a\right\rbrack=\frac12\cdot\left\lbrack\frac12+\sin2a\right\rbrack\)
\(H=cos\left(\frac{\pi}{2}-a\right)\cdot\sin\left(\frac{\pi}{2}+a\right)\)
\(=\frac12\cdot\left\lbrack\sin\left(\frac{\pi}{2}+a+\frac{\pi}{2}-a\right)+\sin\left(\frac{\pi}{2}+a-\frac{\pi}{2}+a\right)\right\rbrack\)
\(=\frac12\cdot\left\lbrack\sin\left(\pi\right)+\sin2a\right\rbrack=\frac12\left\lbrack2\cdot\sin a\cdot cosa\right\rbrack=\sin a\cdot cosa\)
\(I=\sin\left(\frac{\pi}{4}+a\right)-cos\left(\frac{\pi}{4}-a\right)\)
\(=\sin\left(\frac{\pi}{4}+a\right)-\sin\left(\frac{\pi}{2}-\frac{\pi}{4}+a\right)=\sin\left(\frac{\pi}{4}+a\right)-\sin\left(\frac{\pi}{4}+a\right)\)
=0
\(K=cos\left(\frac{\pi}{6}-x\right)-\sin\left(\frac{\pi}{3}+x\right)\)
\(=\sin\left(\frac{\pi}{2}-\frac{\pi}{6}+x\right)-\sin\left(\frac{\pi}{3}+x\right)=\sin\left(\frac{\pi}{3}+x\right)-\sin\left(\frac{\pi}{3}+x\right)\)
=0
\(sin\left(x-\dfrac{\pi}{2}\right)+cos\left(x-\pi\right)+tan\left(\dfrac{5\pi}{2}-x\right)+tan\left(x-\dfrac{\pi}{2}\right)\)
\(=-sin\left(\dfrac{\pi}{2}-x\right)+cos\left(\pi-x\right)+tan\left(2\pi+\dfrac{\pi}{2}-x\right)-tan\left(\dfrac{\pi}{2}-x\right)\)
\(=-cosx-cosx+tan\left(\dfrac{\pi}{2}-x\right)-cotx\)
\(=-2cosx+cotx-cotx=-2cosx\)
pi<x<3/2pi
=>cosx<0
pi<x<3/2pi
=>pi/2<1/2x<3/4pi
=>cos(x/2)<0
1+tan^2x=1/cos^2x
=>1/cos^2x=1+8=9
=>cosx=-1/3
\(cosx=2\cdot cos^2\left(\dfrac{x}{2}\right)-1\)
=>\(2\cdot cos^2\left(\dfrac{x}{2}\right)=\dfrac{2}{3}\)
=>\(cos^2\left(\dfrac{x}{2}\right)=\dfrac{1}{3}\)
=>cos(x/2)=1/căn 3
a) √2 cos(x - π/4)
= √2.(cosx.cos π/4 + sinx.sin π/4)
= √2.(√2/2.cosx + √2/2.sinx)
= √2.√2/2.cosx + √2.√2/2.sinx
= cosx + sinx (đpcm)
b) √2.sin(x - π/4)
= √2.(sinx.cos π/4 - sin π/4.cosx )
= √2.(√2/2.sinx - √2/2.cosx )
= √2.√2/2.sinx - √2.√2/2.cosx
= sinx – cosx (đpcm).
\(A=\frac{1}{2}+\frac{1}{2}cos2x+\frac{1}{2}+\frac{1}{2}cos\left(2x+\frac{4\pi}{3}\right)+\frac{1}{2}+\frac{1}{2}cos\left(2x-\frac{4\pi}{3}\right)\)
\(=\frac{3}{2}+\frac{1}{2}cos2x+cos2x.cos\frac{4\pi}{3}\)
\(=\frac{3}{2}+\frac{1}{2}cos2x-\frac{1}{2}cos2x=\frac{3}{2}\)
\(B=\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos\left(2x+\frac{4\pi}{3}\right)+\frac{1}{2}-\frac{1}{2}cos\left(2x-\frac{4\pi}{3}\right)\)
\(=\frac{3}{2}-\frac{1}{2}cos2x-cos2x.cos\frac{4\pi}{3}\)
\(=\frac{3}{2}-\frac{1}{2}cos2x+\frac{1}{2}cos2x=\frac{3}{2}\)



Đáp án A
Phương pháp: Sử dụng phương pháp tích phân từng phần tính F(x)
Cách giải:
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