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a) \(\cos \left( {a + b} \right) = \sin \left[ {\left( {\frac{\pi }{2} - a} \right) - b} \right] = \sin \left( {\frac{\pi }{2} - a} \right).\cos b - \cos \left( {\frac{\pi }{2} - a} \right).\sin b = \cos a.\cos b - \sin a.\sin b\)
b) \(\cos \left( {a - b} \right) = \cos \left[ {a + \left( { - b} \right)} \right] = \cos a.\cos \left( { - b} \right) - \sin a.\sin \left( { - b} \right) = \sin a.\sin b + \cos a.\cos b\)
\(a,y'=\left(tanx\right)'=\left(\dfrac{sinx}{cosx}\right)'\\ =\dfrac{\left(sinx\right)'cosx-sinx\left(cosx\right)'}{cos^2x}\\ =\dfrac{cos^2x+sin^2x}{cos^2x}\\ =\dfrac{1}{cos^2x}\\ b,\left(cotx\right)'=\left[tan\left(\dfrac{\pi}{2}-x\right)\right]'\\ =-\dfrac{1}{cos^2\left(\dfrac{\pi}{2}-x\right)}\\ =-\dfrac{1}{sin^2\left(x\right)}\)
\(\begin{array}{l}\cos \left( {a + b} \right) + \cos \left( {a - b} \right) = \cos a.\cos b - \sin a.\sin b + \sin a.\sin b + \cos a.\cos b = 2\cos a.\cos b\\\cos \left( {a + b} \right) - \cos \left( {a - b} \right) = \cos a.\cos b - \sin a.\sin b - \sin a.\sin b - \cos a.\cos b = - 2\sin a.\sin b\\\sin \left( {a + b} \right) + \sin \left( {a - b} \right) = \sin a.\cos b + \cos a.\sin b + \sin a.\cos b - \cos a.\sin b = 2\sin a.\cos b\end{array}\)
a) Ta có: \(\cos \left( {a + b} \right) + \cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b + \cos a\cos b - \sin a\sin b = 2\cos a\cos b\)
Suy ra: \(\cos a\cos b = \frac{1}{2}\left[ {\cos \left( {a - b} \right) + \cos \left( {a + b} \right)} \right]\;\)
b) Ta có: \(\sin \left( {a + b} \right) + \sin \left( {a - b} \right) = \sin a\cos b + \cos a\sin b + \sin a\cos b - \cos a\sin b = 2\sin a\cos b\)
Suy ra: \(\sin a\cos b = \frac{1}{2}\left[ {\sin \left( {a - b} \right) + \sin \left( {a + b} \right)} \right]\)
a) Ta có: VT = \(\cos \left( {\frac{\pi }{3} - \frac{\pi }{6}} \right) = \cos \frac{\pi }{{6}} = \frac{{\sqrt 3 }}{2}\)
\(VP = \cos \frac{\pi }{3}\cos \frac{\pi }{6} + \sin \frac{\pi }{3}\sin \frac{\pi }{6} = \frac{{1 }}{2}.\frac{{\sqrt 3 }}{2} + \frac{{\sqrt 3 }}{2}.\frac{1}{2} = \frac{{\sqrt 3 }}{2} = VT\)
Vậy \(\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b\)
b) Ta có: \(\cos \left( {a + b} \right) = \cos (a--b) = \cos a\cos \left( { - b} \right) + \sin a\sin \left( { - b} \right) = \cos a\cos b - \sin a\sin b\)
c) Ta có: \(\sin \left( {a - b} \right) = \cos \left[ {\frac{\pi }{2} - \left( {a - b} \right)} \right] = \cos \left[ {\left( {\frac{\pi }{2} - a} \right) + b} \right] = \cos \left( {\frac{\pi }{2} - a} \right)\cos b + \sin \left( {\frac{\pi }{2} - a} \right)\sin b\)
\( = \left( {\cos \frac{\pi }{2}\cos a + \sin \frac{\pi }{2}\sin a} \right)\cos b + \sin \left( {\frac{\pi }{2} - a} \right)\sin b = \sin a\cos b + \cos a\sin b\)
Ta có:
\(\begin{array}{l}2a = \left( {a + b} \right) + \left( {a - b} \right) \Rightarrow \tan 2a = \tan \left[ {\left( {a + b} \right) + \left( {a - b} \right)} \right]\\2b = \left( {a + b} \right) - \left( {a - b} \right) \Rightarrow \tan 2b = \tan \left[ {\left( {a + b} \right) - \left( {a - b} \right)} \right]\end{array}\)
\(\begin{array}{l}\tan \left[ {\left( {a + b} \right) + \left( {a - b} \right)} \right] = \frac{{\tan \left( {a + b} \right) + \tan \left( {a - b} \right)}}{{1 - \tan \left( {a + b} \right).\tan \left( {a - b} \right)}} = \frac{{3 + 2}}{{1 - 3.2}} = - 1\\\tan \left[ {\left( {a + b} \right) - \left( {a - b} \right)} \right] = \frac{{\tan \left( {a + b} \right) - \tan \left( {a - b} \right)}}{{1 + \tan \left( {a + b} \right).\tan \left( {a - b} \right)}} = \frac{{3 - 2}}{{1 + 3.2}} = \frac{1}{7}\end{array}\)
Vậy \(\tan 2a = - 1,\,\,\,\tan 2b = \frac{1}{7}\)
a: A=(sinx+cosx)^2-1=m^2-1
b: B=căn (sinx+cosx)^2-4sinxcosx=căn m^2-4(m^2-1)=căn -3m^2+4
c: C=(sin^2x+cos^2x)^2-2(sinx*cosx)^2=1-2m^2
Cách 1:
Ta có: \(tan\alpha=\sqrt{2}\Rightarrow\left\{{}\begin{matrix}\dfrac{sin\alpha}{cos\alpha}=\sqrt{2}\\1+\left(\sqrt{2}\right)^2=\dfrac{1}{cos^2\alpha}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}sin\alpha=\sqrt{2}\cdot cos\alpha\\cos^2\alpha=\dfrac{1}{3}\end{matrix}\right.\)
\(P=\dfrac{sin\alpha-cos\alpha}{sin^3\alpha+3cos^3\alpha+2sin\alpha}\)
\(=\dfrac{\sqrt{2}\cdot cos\alpha-cos\alpha}{\left(\sqrt{2}\cdot cos\alpha\right)^3+3cos^3\alpha+2\cdot\sqrt{2}\cdot cos\alpha}\)
\(=\dfrac{cos\alpha\left(\sqrt{2}-1\right)}{2\sqrt{2}\cdot cos^3\alpha+3cos^3\alpha+2\sqrt{2}\cdot cos\alpha}\)
\(=\dfrac{cos\alpha\left(\sqrt{2}-1\right)}{cos\alpha\left(2\sqrt{2}\cdot cos^2\alpha+3cos^2\alpha+2\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{2}-1}{2\sqrt{2}\cdot cos^2\alpha+3cos^2\alpha+2\sqrt{2}}\)
Thay \(cos^2\alpha=\dfrac{1}{3}\) vào \(P\) ta có:
\(P=\dfrac{\sqrt{2}-1}{2\sqrt{2}\cdot\dfrac{1}{3}+3\cdot\dfrac{1}{3}+2\sqrt{2}}=\dfrac{\sqrt{2}-1}{1+\dfrac{8}{3}\sqrt{2}}\)
\(=\dfrac{3\left(\sqrt{2}-1\right)}{3\left(1+\dfrac{8}{3}\sqrt{2}\right)}=\dfrac{3\left(\sqrt{2}-1\right)}{3+8\sqrt{2}}\)
\(=\dfrac{3\left(\sqrt{2}-1\right)}{3+2^3\sqrt{2}}=\dfrac{a\left(\sqrt{b}-1\right)}{a+b^3\sqrt{b}}\)
\(\Rightarrow\left\{{}\begin{matrix}a=3\\b=2\end{matrix}\right.\Rightarrow a+b=5\)
Chọn đáp án A.
Cách 2:
\(P=\dfrac{sin\alpha-cos\alpha}{sin^3\alpha+3cos^3\alpha+2sin\alpha}=\dfrac{\left(sin\alpha-cos\alpha\right)\div cos^3\alpha}{\left(sin^3\alpha+3cos^3\alpha+2sin\alpha\right)\div cos^3\alpha}\)
\(=\dfrac{\dfrac{sin\alpha}{cos^3\alpha}-\dfrac{1}{cos^2\alpha}}{\dfrac{sin^3\alpha}{cos^3\alpha}+3+2\cdot\dfrac{sin\alpha}{cos^3\alpha}}=\dfrac{\dfrac{sin\alpha}{cos\alpha}\cdot\dfrac{1}{cos^2\alpha}-\dfrac{1}{cos^2\alpha}}{tan^3\alpha+3+2\cdot\dfrac{sin\alpha}{cos\alpha}\cdot\dfrac{1}{cos^2\alpha}}\)
\(=\dfrac{tan\alpha\cdot\left(1+tan^2\alpha\right)-\left(1+tan^2\alpha\right)}{tan^3\alpha+3+2tan\alpha\cdot\left(1+tan^2\alpha\right)}\)
Thay \(tan\alpha=\sqrt{2}\) vào ta có:
\(P=\dfrac{\sqrt{2}\cdot\left[1+\left(\sqrt{2}\right)^2\right]-\left[1+\left(\sqrt{2}\right)^2\right]}{\left(\sqrt{2}\right)^3+3+2\sqrt{2}\cdot\left[1+\left(\sqrt{2}\right)^2\right]}=\dfrac{3\sqrt{2}-3}{2\sqrt{2}+3+6\sqrt{2}}\)
\(=\dfrac{3\left(\sqrt{2}-1\right)}{3+8\sqrt{2}}=\dfrac{3\left(\sqrt{2}-1\right)}{3+2^3\sqrt{2}}=\dfrac{a\left(\sqrt{b}-1\right)}{a+b^3\sqrt{b}}\)
\(\Rightarrow\left\{{}\begin{matrix}a=3\\b=2\end{matrix}\right.\Rightarrow a+b=3+2=5\)
Chọn đáp án A
a) \(\tan \left( {a + b} \right) = \frac{{\sin \left( {a + b} \right)}}{{\cos \left( {a + b} \right)}} = \frac{{\sin a.\cos b + \cos a.\sin b}}{{\cos a.\cos b - \sin a.\sin b}}\)
\(\begin{array}{l} = \frac{{\sin a.\cos b + \cos a.\cos b}}{{\cos a.\cos b - \sin a.\sin b}} = \frac{{\sin a.\cos b}}{{\cos a.\cos b - \sin a.\sin b}} + \frac{{\cos a.\sin b}}{{\cos a.\cos b - \sin a.\sin b}}\\ = \frac{{\frac{{\sin a.\cos b}}{{\cos a.\cos b}}}}{{\frac{{\cos a.\cos b - \sin a.\sin b}}{{\cos a.\cos b}}}} + \frac{{\frac{{\cos a.\sin b}}{{\cos a.\cos b}}}}{{\frac{{\cos a.\cos b - \sin a.\sin b}}{{\cos a.\cos b}}}} = \frac{{\tan a}}{{1 - \tan a.\tan b}} + \frac{{\tan b}}{{1 - \tan a.\tan b}}\\ = \frac{{\tan a + \tan b}}{{1 - \tan a.\tan b}}\end{array}\)
\( \Rightarrow \tan \left( {a + b} \right) = \frac{{\tan a + \tan b}}{{1 - \tan a.\tan b}}\)
b)
\(\tan \left( {a - b} \right) = \tan \left( {a + \left( { - b} \right)} \right) = \frac{{\tan a + \tan \left( { - b} \right)}}{{1 - \tan a.\tan \left( { - b} \right)}} = \frac{{\tan a - \tan b}}{{1 + \tan a.\tan b}}\)