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\(\begin{array}{l}1.\,\,\,\,\cos a.\cos b = \frac{1}{2}\left[ {\cos \left( {a + b} \right) + \cos \left( {a - b} \right)} \right] \Leftrightarrow 2\cos a.\cos b = \cos \left( {a + b} \right) + \cos \left( {a - b} \right)\\ \Leftrightarrow 2\cos \frac{{u + v}}{2}.\cos \frac{{u - v}}{2} = \cos u + \cos v\\2.\,\,\,\,\sin a.\sin b = - \frac{1}{2}.\left[ {\cos \left( {a + b} \right) - \cos \left( {a - b} \right)} \right] \Leftrightarrow - 2.\sin a.\sin b = \cos \left( {a + b} \right) - \cos \left( {a - b} \right)\\ \Leftrightarrow - 2.\sin \frac{{u + v}}{2}.\sin \frac{{u - v}}{2} = \cos u - \cos v\\3.\,\,\,\,\sin a.\cos b = \frac{1}{2}\left[ {\sin \left( {a + b} \right) + \sin \left( {a - b} \right)} \right] \Leftrightarrow 2\sin a.\cos b = \sin \left( {a + b} \right) + \sin \left( {a - b} \right)\\ \Leftrightarrow 2\sin \frac{{u + v}}{2}.\cos \frac{{u - v}}{2} = \sin u + \sin v\\4.\,\,\,\,\sin \left( {a + b} \right) - \sin \left( {a - b} \right) = \sin a.\cos b + \cos a.\sin b - \sin a.\cos b + \cos a.\sin b = 2\cos a.\sin b\\ \Leftrightarrow \sin u - \sin v = 2.\cos \frac{{u + v}}{2}.\sin \frac{{u - v}}{2}\end{array}\)
a) Các hằng đẳng thức lượng giác cơ bản:
sin2α + cos2α = 1
1 + tan2α = 1/(cos2α); α ≠ π/2 + kπ, k ∈ Z
1 + cot2α = 1/(sin2α); α ≠ kπ, k ∈ Z
tanα.cotα = 1; α ≠ kπ/2, k ∈ Z
b) Công thức cộng:
cos(a - b) = cosa cosb + sina sinb
cos(a + b) = cosa cosb - sina sinb
sin(a - b) = sina cosb - cosa sinb
sin(a + b) = sina.cosb + cosa.sinb
c) Công thức nhân đôi:
sin2α = 2 sinα cosα
cos2α = cos2α - sin2α = 2cos2α - 1 = 1 - 2sin2α
d) Công thức biến đổi tích thành tổng:
cos a cosb = 1/2 [cos(a - b) + cos(a + b) ]
sina sinb = 1/2 [cos(a - b) - cos(a + b) ]
sina cosb = 1/2 [sin(a - b) + sin(a + b) ]
Công thức biến đổi tổng thành tích:
a) \(PQ = n.\cos a,PQ = m.\cos b\)
b) \(MQ = n.\sin a,PN = m.\sin b \Rightarrow MN = n.\sin a + m.\sin b\)
\(\begin{array}{l}{S_{MPQ}} = \frac{1}{2}m.\cos b.n.\sin a = \frac{1}{2}m.n.\cos b.\sin a\\{S_{NPQ}} = \frac{1}{2}n.\cos a.m.\sin b = \frac{1}{2}m.n.\cos a.\sin b\\{S_{MNP}} = \frac{1}{2}m.n.\sin \left( {a + b} \right)\end{array}\)
c) \({S_{MNP}} = {S_{MPQ}} + {S_{NPQ}} \Rightarrow \frac{1}{2}m.n.\cos b.\sin a + \frac{1}{2}m.n.\cos a.\sin b = \frac{1}{2}m.n.\sin \left( {a + b} \right)\)
\( \Rightarrow \sin \left( {a + b} \right) = \sin a.\cos b + \cos a.\sin b\)
d) \(\sin \left( {a - b} \right) = \sin \left[ {a + \left( { - b} \right)} \right] = \sin a.\cos \left( { - b} \right) + \cos a.\sin \left( { - b} \right) = \sin a.\cos b - \cos a.\sin b\)
Ta có:
\(\begin{array}{l}\cos \alpha \cos \beta = \cos \frac{{\alpha + \beta }}{2}\cos \frac{{\alpha - \beta }}{2}\\ = \frac{1}{2}\left[ {\cos \left( {\frac{{\alpha + \beta }}{2} + \frac{{\alpha - \beta }}{2}} \right) + \cos \left( {\frac{{\alpha + \beta }}{2} - \frac{{\alpha - \beta }}{2}} \right)} \right]\\ = \frac{1}{2}\left( {\cos \alpha + \cos \beta } \right)\end{array}\)
\(\begin{array}{l}\sin \alpha \sin \beta = \sin \frac{{\alpha + \beta }}{2}\sin \frac{{\alpha - \beta }}{2}\\ = \frac{1}{2}\left[ {\cos \left( {\frac{{\alpha + \beta }}{2} - \frac{{\alpha - \beta }}{2}} \right) - \cos \left( {\frac{{\alpha + \beta }}{2} + \frac{{\alpha - \beta }}{2}} \right)} \right]\\ = \frac{1}{2}\left( {\cos \beta - \cos \alpha } \right)\end{array}\)
\(\begin{array}{l}\sin \alpha \cos \beta = \sin \frac{{\alpha + \beta }}{2}\cos \frac{{\alpha - \beta }}{2}\\ = \frac{1}{2}\left[ {\sin \left( {\frac{{\alpha + \beta }}{2} + \frac{{\alpha - \beta }}{2}} \right) + \sin \left( {\frac{{\alpha + \beta }}{2} - \frac{{\alpha - \beta }}{2}} \right)} \right]\\ = \frac{1}{2}\left( {\sin \alpha + \sin \beta } \right)\end{array}\)
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)
\(\tan \left( {a - b} \right) = \tan \left( {a + \left( { - b} \right)} \right) = \frac{{\tan a + \tan \left( { - b} \right)}}{{1 - \tan a.\tan \left( { - b} \right)}} = \frac{{\tan a - \tan b}}{{1 + \tan a.\tan b}}\)
Ta có: \(u = a - b;v = a + b\).
Suy ra \(u + v = 2a \to a = \frac{{u + v}}{2}\)
\(u - v = 2b \to b = \frac{{u - v}}{2}\)
Ta có: \(\cos u + \cos v = 2\cos \frac{{u + v}}{2}\cos \frac{{u - v}}{2}\)
\(\cos u - \cos v = - 2\sin \frac{{u + v}}{2}\sin \frac{{u - v}}{2}\)
\(\sin u + \sin v = 2\sin \frac{{u + v}}{2}\cos \frac{{u - v}}{2}\)
\(\sin u - \sin v = 2\cos \frac{{u + v}}{2}\sin \frac{{u - v}}{2}\)