Toán Lớp 10: Cho tam giác ABC chứng minh rằng :
a)sin (a + b )= sinc
b)cos ( a+b ) = -cosc
c)tan (a+b) = -tanc
d)cot ( a+b) = -cotc
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Toán Lớp 10: Cho tam giác ABC chứng minh rằng :
a)sin (a + b )= sinc
b)cos ( a+b ) = -cosc
c)tan (a+b) = -tanc
d)cot ( a+b) = -cotc
Comments ( 1 )
\Delta ABC\\
\Leftrightarrow A + B + C = {180^0}\\
a)\\
\sin \left( {A + B} \right)\\
= \sin \left( {{{180}^0} – C} \right)\\
= \sin {180^0}.\cos C – \cos {180^0}.\sin C\\
= \sin C\\
b)\cos \left( {A + B} \right)\\
= \cos \left( {{{180}^0} – C} \right)\\
= \cos {180^0}.\cos C + \sin {180^0}.\sin C\\
= – \cos C\\
c)\tan \left( {A + B} \right)\\
= \tan \left( {{{180}^0} – C} \right)\\
= \dfrac{{\sin \left( {{{180}^0} – C} \right)}}{{\cos \left( {{{180}^0} – C} \right)}}\\
= \dfrac{{\sin C}}{{ – \cos C}}\\
= – \tan C\\
d)\cot \left( {A + B} \right)\\
= \cot \left( {{{180}^0} – C} \right)\\
= \dfrac{{\cos \left( {{{180}^0} – C} \right)}}{{\sin \left( {{{180}^0} – C} \right)}}\\
= \dfrac{{ – \cos C}}{{\sin C}}\\
= – \cot C
\end{array}$