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\(F=cos\left(\frac{\pi}{4}+a\right)\cdot cos\left(\frac{\pi}{4}-a\right)\)
\(=\frac12\cdot\left\lbrack cos\left(\frac{\pi}{4}+a-\frac{\pi}{4}+a\right)+cos\left(\frac{\pi}{4}+a+\frac{\pi}{4}+a\right)\right\rbrack\)
\(=\frac12\cdot\left\lbrack cos\left(2a\right)+cos\left(\frac{\pi}{2}\right)\right\rbrack=\frac12\cdot cos2a\)
\(G=\sin\left(\frac{\pi}{3}+a\right)\cdot cos\left(\frac{\pi}{3}-a\right)\)
\(=\frac12\cdot\left\lbrack\sin\left(\frac{\pi}{3}+a+\frac{\pi}{3}-a\right)+\sin\left(\frac{\pi}{3}+a-\frac{\pi}{3}+a\right)\right\rbrack\)
\(=\frac12\cdot\left\lbrack\sin\left(\frac23\pi\right)+\sin2a\right\rbrack=\frac12\cdot\left\lbrack\frac12+\sin2a\right\rbrack\)
\(H=cos\left(\frac{\pi}{2}-a\right)\cdot\sin\left(\frac{\pi}{2}+a\right)\)
\(=\frac12\cdot\left\lbrack\sin\left(\frac{\pi}{2}+a+\frac{\pi}{2}-a\right)+\sin\left(\frac{\pi}{2}+a-\frac{\pi}{2}+a\right)\right\rbrack\)
\(=\frac12\cdot\left\lbrack\sin\left(\pi\right)+\sin2a\right\rbrack=\frac12\left\lbrack2\cdot\sin a\cdot cosa\right\rbrack=\sin a\cdot cosa\)
\(I=\sin\left(\frac{\pi}{4}+a\right)-cos\left(\frac{\pi}{4}-a\right)\)
\(=\sin\left(\frac{\pi}{4}+a\right)-\sin\left(\frac{\pi}{2}-\frac{\pi}{4}+a\right)=\sin\left(\frac{\pi}{4}+a\right)-\sin\left(\frac{\pi}{4}+a\right)\)
=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
1. Đặt \(\left\{{}\begin{matrix}u=x\\dv=\dfrac{dx}{sin^2x}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=dx\\v=-cotx\end{matrix}\right.\)
Do đó I= \(-x.cotx+\int cotxdx\)= \(-xcotx+ln\left|sinx\right|\)
2. Đặt \(\left\{{}\begin{matrix}u=x+1\\dv=\dfrac{dx}{e^x}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=dx\\v=-e^{-x}\end{matrix}\right.\)
Do đó I= \(-\left(x+1\right)e^{-x}+\int e^{-x}dx\)=\(-\left(x+1\right)e^{-x}-e^{-x}\)
=\(-\left(x+2\right)e^{-x}\)
\(\cos a=\dfrac{-12}{13}\)
\(\sin b=\dfrac{4}{5}\)
\(\sin\left(a+b\right)=\sin a\cos b+\sin b\cos a\)
\(=\dfrac{5}{13}\cdot\dfrac{3}{5}+\dfrac{4}{5}\cdot\dfrac{-12}{13}=\dfrac{-45}{65}=\dfrac{-9}{13}\)
I 3 = 2 3 , I 5 = 8 15