\(P=sin^2x+c\text{os}\left(\frac{\pi}{3}-x\right)c\text{os}\left(\frac{\pi}{3}+x\right)\)không phụ thuộc vào x
Những câu hỏi liên quan
\(\frac{sin\left(2x+\frac{3\pi}{4}\right)}{c\text{os}\left(x+\frac{\pi}{4}\right)}+1=0\)
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\(\Leftrightarrow sin\left(2x+\frac{3\pi}{4}\right)+cos\left(x+\frac{\pi}{4}\right)=0\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{4}\right)=-sin\left(2x+\frac{3\pi}{4}\right)\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{4}\right)=cos\left(2x+\frac{5\pi}{4}\right)\)
\(\Rightarrow\left[{}\begin{matrix}2x+\frac{5\pi}{4}=x+\frac{\pi}{4}+k2\pi\\2x+\frac{5\pi}{4}=-x-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\pi+k2\pi\\x=-\frac{\pi}{2}+\frac{k2\pi}{3}\end{matrix}\right.\)
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\(\frac{c\text{os}\left(x+\frac{5\pi}{6}\right)}{c\text{os}\left(2x-\frac{\pi}{6}\right)}+tan\left(2x-\frac{\pi}{6}\right)=0\)
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\(\Leftrightarrow\frac{cos\left(x+\frac{5\pi}{6}\right)}{cos\left(2x-\frac{\pi}{6}\right)}+\frac{sin\left(2x-\frac{\pi}{6}\right)}{cos\left(2x-\frac{\pi}{6}\right)}=0\)
\(\Leftrightarrow cos\left(x+\frac{5\pi}{6}\right)+sin\left(2x-\frac{\pi}{6}\right)=0\)
\(\Leftrightarrow cos\left(x+\frac{5\pi}{6}\right)=-sin\left(2x-\frac{\pi}{6}\right)\)
\(\Leftrightarrow cos\left(x+\frac{5\pi}{6}\right)=cos\left(2x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{3}=x+\frac{5\pi}{6}+k2\pi\\2x+\frac{\pi}{3}=-x-\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=-\frac{7\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)
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\(c\text{os}\left(x-\frac{5\pi}{4}\right)=sin\left(x+\frac{3\pi}{4}\right)\)
\(\Leftrightarrow cos\left(x-\frac{5\pi}{4}\right)=cos\left(\frac{\pi}{2}-x-\frac{3\pi}{4}\right)=cos\left(x+\frac{\pi}{4}\right)\)
\(\Leftrightarrow x-\frac{5\pi}{4}=-x-\frac{\pi}{4}+k2\pi\)
\(\Leftrightarrow x=\frac{\pi}{2}+k\pi\)
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\(\int_0^{\frac{\Pi}{2}}c\text{os}^2x\left(1-sin^3x\right)dx\)
2) \(\int_0^{\frac{\Pi}{4}}\frac{sin\left(x-\frac{\Pi}{4}\right)}{sin2x+2\left(1+s\text{inx}+c\text{ox}\right)}dx\)
hộ mk vs nha
1)
\(I=\int\left(cos^2x-cos^2x\cdot sin^3x\right)dx\\ =\int cos^2x\cdot dx-\int cos^2x\cdot sin^3x\cdot dx\\ =\frac{1}{2}\int\left(cos2x+1\right)dx+\int cos^2x\left(1-cos^2x\right)d\left(cosx\right)\\ =\frac{1}{4}sin2x+\frac{1}{2}+\frac{cos^3x}{3}-\frac{cos^5x}{5}+C\)
....
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2) Xét riêng mẫu số:
\(sin2x+2\left(1+sinx+cosx\right)\\ =\left(sin2x+1\right)+2\left(sinx+cosx\right)+1\\ =\left(sinx+cosx\right)^2+2\left(sinx+cosx\right)+1\\ =\left(sinx+cosx+1\right)^2\\ =\left[\sqrt{2}cos\left(x-\frac{\pi}{4}\right)+1\right]^2\)
Khi đó:
\(I_2=\int\frac{sin\left(x-\frac{\pi}{4}\right)}{\left[\sqrt{2}cos\left(x-\frac{\pi}{4}\right)+1\right]^2}dx\\ =-\frac{1}{\sqrt{2}}\int\frac{d\left[\sqrt{2}cos\left(x-\frac{\pi}{4}\right)+1\right]}{\left[\sqrt{2}cos\left(x-\frac{\pi}{4}\right)+1\right]^2}\\ =\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}cos\left(x-\frac{\pi}{4}\right)+1}+C=\frac{1}{2cos\left(x-\frac{\pi}{4}\right)+1}\)
...
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Tìm tất cả các nghiệm của phương trình : \(\frac{2.sin\left(\frac{pi}{3}-2x\right)+2.sin2x+\sqrt{3}}{c\text{os}x}=4.c\text{os}4x\)
trên đoạn 50,55
\(c\text{os}3a=4cosa.c\text{os}\left(\frac{\pi}{3}-a\right).c\text{os}\left(\frac{\pi}{3}+a\right)\)
\(4cosa.cos\left(\frac{\pi}{3}-a\right).cos\left(\frac{\pi}{3}+a\right)=2cosa.\left(cos\frac{2\pi}{3}+cos2a\right)\)
\(=2cosa\left(cos2a-\frac{1}{2}\right)=2cosa.cos2a-cosa\)
\(=cos3a+cosa-cosa=cos3a\)
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\(2cos\left(\frac{\pi}{4}+x\right)cos\left(\frac{\pi}{4}-x\right)=c\text{os}2x\)
\(2cos\left(\frac{\pi}{4}+x\right)cos\left(\frac{\pi}{4}-x\right)=cos\left(\frac{\pi}{2}\right)+cos2x=0+cos2x=cos2x\)
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chứng minh rằng
a)
\(\frac{1-2\text{s}in^2x}{2cot\left(\frac{\pi}{4}+\alpha\right).c\text{os}^2\left(\frac{\pi}{4}-\alpha\right)}=1\)
b)
\(\frac{\frac{\sqrt{3}}{2}c\text{os}2\text{a}-\frac{1}{2}sin2\text{a}}{1-\frac{1}{2}c\text{os}2\text{a}-\frac{\sqrt{3}}{2}sin2\text{a}}=tan\left(a+\frac{\pi}{4}\right)\)
Giúp mình với ạ. Giải pt:
1) \(sin^2x\left(x+\frac{\pi}{4}\right)=\sqrt{2}s\text{inx}\)
2) \(3\sqrt{2}c\text{os}x-s\text{inx}=c\text{os}3x+3\sqrt{2}sinxsin2x\:\)