Cho \(\sin a + \cos a = \dfrac{5}{4}\). Khi đó \(\sin a.\cos a\) có giá trị bằng
Ta có \(\sin a + \cos a = \dfrac{5}{4}\) \( \Leftrightarrow {\left( {\sin a + \cos a} \right)^2} = \dfrac{{25}}{{16}}\) \( \Leftrightarrow 1 + 2\sin a\cos a = \dfrac{{25}}{{16}} \) \(\Leftrightarrow \sin a\cos a = \dfrac{9}{{32}}\)
Biết \(\sin \alpha = \dfrac{{\sqrt 3 }}{2}\) và \(\dfrac{\pi }{2} < \alpha < \pi \). Tính giá trị của \(\cos \left( {2\alpha - \dfrac{\pi }{3}} \right)\).
Dễ thấy với \(\left\{ \begin{array}{l}\dfrac{\pi }{2} < \alpha < \pi \\\sin \alpha = \dfrac{{\sqrt 3 }}{2}\end{array} \right.\) \( \Rightarrow \alpha = \dfrac{{2\pi }}{3} \Rightarrow 2\alpha = \dfrac{{4\pi }}{3}\) \( \Rightarrow \cos \left( {2\alpha - \dfrac{\pi }{3}} \right) = \cos \pi = - 1\)
Cho \(\cos \alpha = \dfrac{1}{3}\). Tính giá trị của biểu thức \(P = \dfrac{{\sin 3\alpha - \sin \alpha }}{{\sin 2\alpha }}\).
Ta có \(P = \dfrac{{\sin 3\alpha - \sin \alpha }}{{\sin 2\alpha }} = \dfrac{{2.\cos 2\alpha .\sin \alpha }}{{2.\sin \alpha .\cos \alpha }} = \dfrac{{\cos 2\alpha }}{{\cos \alpha }} = \dfrac{{2.{{\cos }^2}\alpha - 1}}{{\cos \alpha }} = - \dfrac{7}{3}\).
Tính giá trị của biểu thức \(P = \dfrac{{\sin 2a.\sin a}}{{1 + \cos 2a}}\) biết \(\cos a = - \dfrac{2}{3}\).
Ta có \(P = \dfrac{{\sin 2a.\sin a}}{{1 + \cos 2a}} = \dfrac{{2\sin a\cos a.\sin a}}{{2{{\cos }^2}a}}\) \( = \dfrac{{2{{\sin }^2}a\cos a}}{{2{{\cos }^2}a}} = \dfrac{{2\cos a\left( {1 - {{\cos }^2}a} \right)}}{{2{{\cos }^2}a}} = - \dfrac{5}{6}\)
Giá trị của biểu thức \(T = \dfrac{{\cos \left( {a + b} \right)\cos \left( {a - b} \right) + 1}}{{{{\cos }^2}a + {{\cos }^2}b}}\) là
Thực nghiệm \(T = \dfrac{{\cos \left( {\pi + 0} \right)\cos \left( {\pi - 0} \right) + 1}}{{{{\cos }^2}\pi + {{\cos }^2}0}} = 1\)
Tự luận:
\(\begin{array}{l}T = \dfrac{{\cos \left( {a + b} \right)\cos \left( {a - b} \right) + 1}}{{{{\cos }^2}a + {{\cos }^2}b}}\\ = \dfrac{{\dfrac{1}{2}\left\{ {\cos \left[ {\left( {a + b} \right) + \left( {a - b} \right)} \right] + \cos \left[ {\left( {a + b} \right) - \left( {a - b} \right)} \right]} \right\} + 1}}{{{{\cos }^2}a + {{\cos }^2}b}}\\ = \dfrac{{\dfrac{1}{2}\left( {\cos 2a + \cos 2b} \right) + 1}}{{{{\cos }^2}a + {{\cos }^2}b}}\\ = \dfrac{1}{2}.\dfrac{{\left( {\cos 2a + \cos 2b} \right) + 2}}{{{{\cos }^2}a + {{\cos }^2}b}}\\ = \dfrac{1}{2}.\dfrac{{(\cos 2a + 1) + (\cos 2b + 1)}}{{{{\cos }^2}a + {{\cos }^2}b}}\\ = \dfrac{1}{2}.\dfrac{{2{{\cos }^2}a + 2{{\cos }^2}b}}{{{{\cos }^2}a + {{\cos }^2}b}} = 1\end{array}\)
Cho biểu thức: \(A = {\sin ^2}\left( {a + b} \right) - {\sin ^2}a - {\sin ^2}b\). Chọn đáp án đúng:
Ta có: \(A = {\left( {\sin a\cos b + \cos a\sin b} \right)^2} - {\sin ^2}a - {\sin ^2}b\)
\(\begin{array}{l} = {\sin ^2}a{\cos ^2}b + 2\sin a\cos a\sin b\cos b + {\cos ^2}a{\sin ^2}b - {\sin ^2}a - {\sin ^2}b\\ = {\sin ^2}a\left( {{{\cos }^2}b - 1} \right) + {\sin ^2}b\left( {{{\cos }^2}a - 1} \right) + 2\sin a\cos a\sin b\cos b\\ = 2\sin a\cos a\sin b\cos b - 2{\sin ^2}a{\sin ^2}b\\ = 2\sin a\sin b\left( {\cos a\cos b - \sin a\sin b} \right) = 2\sin a\sin b\cos \left( {a + b} \right)\end{array}\)
Tính giá trị biểu thức \(P = {\left( {\sin a + \sin b} \right)^2} + {\left( {\cos a + \cos b} \right)^2}\) biết \(a - b = \dfrac{\pi }{4}\).
Ta có \(P = {\left( {\sin a + \sin b} \right)^2} + {\left( {\cos a + \cos b} \right)^2}\)
\(\begin{array}{l} = {\sin ^2}a + 2\sin a\sin b + {\sin ^2}b + {\cos ^2}a + 2\cos a\cos b + {\cos ^2}b\\ = 2 + 2\left( {\sin a.\sin b + \cos a.\cos b} \right)\\ = 2 + 2.\cos \left( {a - b} \right) = 2 + 2\cos \dfrac{\pi }{4} = 2 + \sqrt 2 \end{array}\)
Mệnh đề nào dưới đây là đúng?
Ta có \(\sin \left( {a + b} \right)\sin \left( {a - b} \right) = \dfrac{1}{2}\left( {\cos 2b - \cos 2a} \right)\)
\( = \dfrac{1}{2}\left[ {\left( {2{{\cos }^2}b - 1} \right) - \left( {2{{\cos }^2}a - 1} \right)} \right] = {\cos ^2}b - {\cos ^2}a\).
Cho góc \(\alpha \) thỏa mãn \(\tan \alpha = 2\). Tính giá trị biểu thức \(P = \dfrac{{1 + \cos \alpha + \cos 2\alpha }}{{\sin \alpha + \sin 2\alpha }}\).
Ta có \(P = \dfrac{{1 + \cos \alpha + \cos 2\alpha }}{{\sin \alpha + \sin 2\alpha }}\) \( = \dfrac{{2{{\cos }^2}\alpha + \cos \alpha }}{{\sin \alpha + 2\sin \alpha .\cos \alpha }} \) \(= \dfrac{{\cos \alpha \left( {1 + 2\cos \alpha } \right)}}{{\sin \alpha \left( {1 + 2\cos \alpha } \right)}} \) \(= \cot \alpha = \dfrac{1}{2}\)
Ta có \(\cos \alpha = - \dfrac{2}{{\sqrt 5 }} \Rightarrow {\cos ^2}\alpha = \dfrac{4}{5}\)\( \Rightarrow {\sin ^2}\alpha = 1 - \dfrac{4}{5} = \dfrac{1}{5}\)
Do \(\pi < \alpha < \dfrac{{3\pi }}{2} \Rightarrow \sin \alpha < 0\)\( \Rightarrow \sin \alpha = - \dfrac{1}{{\sqrt 5 }} \Rightarrow \tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }} = \dfrac{1}{2}\)
Tính \(\sin \dfrac{{2\pi }}{7} + \sin \dfrac{{4\pi }}{7} + \sin \dfrac{{6\pi }}{7}\)
Ta có:
\(\begin{array}{l}\sin \dfrac{\pi }{7}\left( {\sin \dfrac{{2\pi }}{7} + \sin \dfrac{{4\pi }}{7} + \sin \dfrac{{6\pi }}{7}} \right) = \sin \dfrac{\pi }{7}\sin \dfrac{{2\pi }}{7} + \sin \dfrac{\pi }{7}\sin \dfrac{{4\pi }}{7} + \sin \dfrac{\pi }{7}\sin \dfrac{{6\pi }}{7}\\ = \dfrac{1}{2}\left( {\cos \dfrac{\pi }{7} - \cos \dfrac{{3\pi }}{7}} \right) + \dfrac{1}{2}\left( {\cos \dfrac{{3\pi }}{7} - \cos \dfrac{{5\pi }}{7}} \right) + \dfrac{1}{2}\left( {\cos \dfrac{{5\pi }}{7} - \cos \dfrac{{7\pi }}{7}} \right)\\ = \dfrac{1}{2}\cos \dfrac{\pi }{7} + \dfrac{1}{2} = {\cos ^2}\dfrac{\pi }{{14}}\\\sin \dfrac{\pi }{7} = 2\sin \dfrac{\pi }{{14}}\cos \dfrac{\pi }{{14}} \Rightarrow \left( {\sin \dfrac{{2\pi }}{7} + \sin \dfrac{{4\pi }}{7} + \sin \dfrac{{6\pi }}{7}} \right) = \dfrac{1}{2}\cot \dfrac{\pi }{{14}}\end{array}\)
\(M = \dfrac{{\sin a}}{{{{\sin }^3}a + 2{{\cos }^3}a}} \Rightarrow \dfrac{1}{M} = \dfrac{{{{\sin }^3}a + 2{{\cos }^3}a}}{{\sin a}} = {\sin ^2}a + 2\dfrac{{{{\cos }^3}a}}{{\sin a}}\)
Do \(\tan a = \dfrac{{\sin a}}{{\cos a}} = 2 \Rightarrow \dfrac{{\cos a}}{{\sin a}} = \dfrac{1}{2}\) \( \Rightarrow \dfrac{1}{M} = {\sin ^2}a + 2.\dfrac{1}{2}.{\cos ^2}a = {\sin ^2}a + {\cos ^2}a = 1\)
\( \Rightarrow M = 1\)
Với mọi a, biểu thức : $A = \cos \alpha {\rm{ + }}\cos \left( {\alpha + \dfrac{\pi }{5}} \right) + ... + \cos \left( {\alpha + \dfrac{{9\pi }}{5}} \right)$ nhận giá trị bằng :
$A = \cos \alpha {\rm{ + }}\cos \left( {\alpha + \dfrac{\pi }{5}} \right) + ... + \cos \left( {\alpha + \dfrac{{9\pi }}{5}} \right)$$A = \left[ {\cos \alpha + \cos \left( {\alpha + \dfrac{{9\pi }}{5}} \right)} \right] + ... + \left[ {\cos \left( {\alpha + \dfrac{{4\pi }}{5}} \right) + \cos \left( {\alpha + \dfrac{{5\pi }}{5}} \right)} \right]$$A = 2\cos \left( {\alpha + \dfrac{{9\pi }}{{10}}} \right)\cos \dfrac{{9\pi }}{{10}} + 2\cos \left( {\alpha + \dfrac{{9\pi }}{{10}}} \right)\cos \dfrac{{7\pi }}{{10}} + ... + 2\cos \left( {\alpha + \dfrac{{9\pi }}{{10}}} \right)\cos \dfrac{\pi }{{10}}$$A = 2\cos \left( {\alpha + \dfrac{{9\pi }}{{10}}} \right)\left( {\cos \dfrac{{9\pi }}{{10}} + \cos \dfrac{{7\pi }}{{10}} + \cos \dfrac{{5\pi }}{{10}} + \cos \dfrac{{3\pi }}{{10}} + \cos \dfrac{\pi }{{10}}} \right)$$A = 2\cos \left( {\alpha + \dfrac{{9\pi }}{{10}}} \right)\left( {2\cos \dfrac{\pi }{2}\cos \dfrac{{2\pi }}{5} + 2\cos \dfrac{\pi }{2}\cos \dfrac{\pi }{5} + \cos \dfrac{\pi }{2}} \right)$$ \Leftrightarrow A = 2\cos \left( {\alpha + \dfrac{{9\pi }}{{10}}} \right).0 = 0.$
Cho \(H = \dfrac{{\sin {{15}^0} + \sin {{45}^0} + \sin {{75}^0}}}{{\cos {{15}^0} + \cos {\rm{4}}{{\rm{5}}^0} + \cos {\rm{7}}{{\rm{5}}^0}}}\). Khi đó:
\(A = \dfrac{{\sin {{15}^o} + \sin {{45}^o} + \sin {{75}^o}}}{{\cos {{15}^o} + \cos {{45}^o} + \cos {{75}^o}}}\)\( = \dfrac{{\left( {\sin {{15}^o} + \sin {{75}^o}} \right) + \sin {{45}^o}}}{{\left( {\cos {{15}^o} + \cos {{75}^o}} \right) + \cos {{45}^o}}}\)\( = \dfrac{{2\sin {{45}^o}.\cos {{30}^o} + \sin {{45}^o}}}{{2\cos {{45}^o}.\cos {{30}^o} + \cos {{45}^o}}}\)
\(\,\,\,\, = \dfrac{{\sin {{45}^o}\left( {2\cos {{30}^o} + 1} \right)}}{{\cos {{45}^o}\left( {2\cos {{30}^o} + 1} \right)}} = \dfrac{{\sin {{45}^o}}}{{\cos {{45}^o}}}\)\( = \tan {45^o} = 1\)
Giá trị của biểu thức \({\cos ^3}x\cos 3x - {\sin ^3}x\sin 3x - \dfrac{3}{4}\cos 4x\).
Thực nghiệm \({\cos ^3}\pi \cos 3\pi - {\sin ^3}\pi \sin 3\pi - \dfrac{3}{4}\cos 4\pi = \dfrac{1}{4}\)
Ta có: \({\left( {\sin \alpha + \cos \alpha \;} \right)^2} = {\sin ^2}\alpha + {\cos ^2}\alpha + 2\sin \alpha \cos \alpha = 1 + \sin 2\alpha = 1 + a\)
Vì \({0^o} < \alpha < {90^o} \Rightarrow {0^o} < 2\alpha < {180^o} \Rightarrow a > 0 \Rightarrow 1 + a > 0\)
Mặt khác \({0^o} < \alpha < {90^o} \Rightarrow \sin \alpha + \cos \alpha > 0 \Rightarrow \sin \alpha + \cos \alpha = \sqrt {a + 1} \)
Giá trị của biểu thức $A = {\sin ^2}\dfrac{\pi }{8} + {\sin ^2}\dfrac{{3\pi }}{8} + {\sin ^2}\dfrac{{5\pi }}{8} + {\sin ^2}\dfrac{{7\pi }}{8}$ bằng
\(A = \dfrac{{1 - \cos \dfrac{\pi }{4}}}{2} + \dfrac{{1 - \cos \dfrac{{3\pi }}{4}}}{2} + \dfrac{{1 - \cos \dfrac{{5\pi }}{4}}}{2} + \dfrac{{1 - \cos \dfrac{{7\pi }}{4}}}{2}\)\( = 2 - \dfrac{1}{2}\left( {\cos \dfrac{\pi }{4} + \cos \dfrac{{3\pi }}{4} + \cos \dfrac{{5\pi }}{4} + \cos \dfrac{{7\pi }}{4}} \right)\)\( = 2 - \dfrac{1}{2}\left( {\cos \dfrac{\pi }{4} + \cos \dfrac{{3\pi }}{4} - \cos \dfrac{{3\pi }}{4} - \cos \dfrac{\pi }{4}} \right) = 2.\)
Cho biểu thức \(A = {\cos ^2}\left( {x - a} \right) + {\cos ^2}x - 2\cos a\cos x\cos \left( {a - x} \right)\). Rút gọn biểu thức A ta được
Ta có: \(A = \cos \left( {x - a} \right)\left[ {\cos \left( {x - a} \right) - 2\cos a\cos x} \right] + {\cos ^2}x\)
\(\begin{array}{l} = \cos \left( {x - a} \right)\left( { - \cos x\cos a + \sin x\sin a} \right) + {\cos ^2}x = - \cos \left( {x - a} \right).\cos \left( {x + a} \right) + {\cos ^2}x\\ = - \dfrac{1}{2}\left( {\cos 2x + \cos 2a} \right) + \dfrac{{1 + \cos 2x}}{2} = \dfrac{{1 - \cos 2a}}{2} = {\sin ^2}a\end{array}\)
Ta có: \(\cos 2\alpha = 1 - 2{\sin ^2}\alpha = 1 - {2.0,6^2} = 0,28\)
Tính \(C = \cos \dfrac{{2\pi }}{{11}} + \cos \dfrac{{4\pi }}{{11}} + \cos \dfrac{{6\pi }}{{11}} + \cos \dfrac{{8\pi }}{{11}} + \cos \dfrac{{10\pi }}{{11}}\)
Với $k = 1,2,3,4,5$ ta có
\(\cos \dfrac{{\left( {2k} \right)\pi }}{{11}}\sin \dfrac{\pi }{{11}} = \dfrac{1}{2}\left[ {\sin \dfrac{{\left( {2k + 1} \right)\pi }}{{11}} - \sin \dfrac{{\left( {2k - 1} \right)\pi }}{{11}}} \right]\)
\(\begin{array}{l} \Rightarrow C.\sin \dfrac{\pi }{{11}} = \dfrac{1}{2}\left[ {\left( {\sin \dfrac{{3\pi }}{{11}} - \sin \dfrac{\pi }{{11}}} \right) + \left( {\sin \dfrac{{5\pi }}{{11}} - \sin \dfrac{{3\pi }}{{11}}} \right) + ... + \left( {\sin \dfrac{{11\pi }}{{11}} - \sin \dfrac{{9\pi }}{{11}}} \right)} \right]\\ = - \dfrac{1}{2}\sin \dfrac{\pi }{{11}}\\ \Rightarrow C = - \dfrac{1}{2}\end{array}\)
