Không dùng máy tinh hãy tính:
\(sin\left(\alpha-\frac{\pi}{3}\right)cos\left(\frac{\pi}{4}-\alpha\right)+cos\left(\alpha-\frac{\pi}{3}\right)sin\left(\frac{\pi}{4}-\alpha\right)\)
Tính \(\sin \left( {\alpha + \frac{\pi }{6}} \right),\cos \left( {\frac{\pi }{4} - \alpha } \right)\) biết \(\sin \alpha = - \frac{5}{{13}},\pi < \alpha < \frac{{3\pi }}{2}\)
\(\cos \alpha = - \sqrt {1 - {{\left( { - \frac{5}{{13}}} \right)}^2}} = - \frac{{12}}{{13}}\) (vì \(\pi < \alpha < \frac{{3\pi }}{2}\))
\(\sin \left( {\alpha + \frac{\pi }{6}} \right) = \sin \alpha \cos \frac{\pi }{6} + \cos \alpha sin\frac{\pi }{6} = \frac{{ - 12 + 5\sqrt 3 }}{{26}}\)
\(\cos \left( {\frac{\pi }{4} - \alpha } \right) = \cos \frac{\pi }{4}\cos \alpha + \sin \frac{\pi }{4}sin\alpha = \frac{{ - 17\sqrt 2 }}{{26}}\)
Cho \(\sin\alpha+\cos\alpha=\frac{\sqrt{6}}{2},a\in\left(0;\frac{\pi}{4}\right)\)
Tính giá trị biểu thức: \(P=\cos\left(\alpha+\frac{\pi}{4}\right)+\sqrt{2\left(1-\sin\alpha\cos\alpha+\sin\alpha-\cos\alpha\right)}\)
Cho \(\alpha\in\left(\frac{\Pi}{2};\Pi\right)\) và \(sin\alpha=\frac{3}{5}\). Tính \(A=\frac{sin\left(\frac{7\Pi}{2}-\alpha\right)}{sin\left(\frac{\Pi}{4}+\alpha\right)-cos\alpha}\)
\(a\in\left(\frac{\pi}{2};\pi\right)\Rightarrow cosa< 0\Rightarrow cosa=-\sqrt{1-sin^2a}=-\frac{4}{5}\)
\(A=\frac{sin\left(4\pi-\frac{\pi}{2}-a\right)}{sin\left(a+\frac{\pi}{4}\right)-cosa}=\frac{-sin\left(a+\frac{\pi}{2}\right)}{sin\left(a+\frac{\pi}{4}\right)-cosa}=\frac{-cosa}{sina.cos\frac{\pi}{4}+cosa.sin\frac{\pi}{4}-cosa}\)
\(=\frac{-\frac{4}{5}}{\frac{3}{5}.\frac{\sqrt{2}}{2}-\frac{4}{5}.\frac{\sqrt{2}}{2}-\frac{4}{5}}=...\)
Cho \(\sin \alpha = \frac{{12}}{{13}}\) và \(\cos \alpha = - \frac{5}{{13}}\). Tính \(\sin \left( { - \frac{{15\pi }}{2} - \alpha } \right) - \cos \left( {13\pi + \alpha } \right)\)
Ta có:
\(\begin{array}{l}\sin \left( { - \frac{{15\pi }}{2} - \alpha } \right) - \cos \left( {13\pi + \alpha } \right) = \sin \left( { -\frac{{16\pi }}{2} +\frac{{\pi }}{2} + \alpha } \right) - \cos \left( {12\pi + \pi + \alpha } \right) = \sin \left( {-8\pi + \frac{\pi }{2} - \alpha } \right) - \cos \left( { \pi + \alpha } \right) \\ = \sin \left( {\frac{\pi }{2} - \alpha } \right) + \cos \left( \alpha \right) = \cos \left( \alpha \right) + \cos \left( \alpha \right) = 2\cos \left( \alpha \right) = 2.\left( { - \frac{5}{{13}}} \right) = \frac{{ - 10}}{{13}}\end{array}\)
Cho góc \(\alpha \) thỏa mãn \(\frac{\pi }{2} < \alpha < \pi ,\cos \alpha = - \frac{1}{{\sqrt 3 }}\). Tính giá trị của các biểu thức sau:
a) \(\sin \left( {\alpha + \frac{\pi }{6}} \right)\);
b) \(\cos \left( {\alpha + \frac{\pi }{6}} \right);\)
c) \(\sin \left( {\alpha - \frac{\pi }{3}} \right)\);
d) \(\cos \left( {\alpha - \frac{\pi }{6}} \right)\).
Ta có:
a) \(\sin \left( {\alpha + \frac{\pi }{6}} \right) = \sin \alpha \cos \frac{\pi }{6} + \cos \alpha \sin \frac{\pi }{6} = \frac{{\sqrt 6 }}{3}.\frac{{\sqrt 3 }}{2} + \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{1}{2} = \frac{{ - \sqrt 3 + 3\sqrt 2 }}{6}\)
b) \(\cos \left( {\alpha + \frac{\pi }{6}} \right) = \cos \alpha .\cos \frac{\pi }{6} - \sin \alpha \sin \frac{\pi }{6} = \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} - \frac{{\sqrt 6 }}{3}.\frac{1}{2} = - \frac{{3 + \sqrt 6 }}{6}\)
c) \(\sin \left( {\alpha - \frac{\pi }{3}} \right) = \sin \alpha \cos \frac{\pi }{3} - \cos \alpha \sin \frac{\pi }{3} = \frac{{\sqrt 6 }}{3}.\frac{1}{2} - \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} = \frac{{3 + \sqrt 6 }}{6}\)
d) \(\cos \left( {\alpha - \frac{\pi }{6}} \right) = \cos \alpha \cos \frac{\pi }{6} + \sin \alpha \sin \frac{\pi }{6} = \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} + \frac{{\sqrt 6 }}{3}.\frac{1}{2} = \frac{{ - 3 + \sqrt 6 }}{6}\)
Rút gọn các biểu thức sau:
a) \(\frac{1}{{\tan \alpha + 1}} + \frac{1}{{\cot \alpha + 1}}\)
b) \(\cos \left( {\frac{\pi }{2} - \alpha } \right) - \sin \left( {\pi + \alpha } \right)\)
c) \(\sin \left( {\alpha - \frac{\pi }{2}} \right) + \cos \left( { - \alpha + 6\pi } \right) - \tan \left( {\alpha + \pi } \right)\cot \left( {3\pi - \alpha } \right)\)
\(a,\dfrac{1}{tan\alpha+1}+\dfrac{1}{cot\alpha+1}\\ =\dfrac{cot\alpha+1+tan\alpha+1}{\left(tan\alpha+1\right)\left(cot\alpha+1\right)}\\ =\dfrac{tan\alpha+cot\alpha+2}{tan\alpha\cdot cot\alpha+tan\alpha+cot\alpha+1}\\ =\dfrac{tan\alpha+cot\alpha+2}{tan\alpha+cot\alpha+2}\\ =1\)
\(b,cos\left(\dfrac{\pi}{2}-\alpha\right)-sin\left(\pi+\alpha\right)\\ =sin\alpha+sin\alpha\\ =2sin\alpha\)
\(c,sin\left(\alpha-\dfrac{\pi}{2}\right)+cos\left(-\alpha+6\pi\right)-tan\left(\alpha+\pi\right)cot\left(3\pi-\alpha\right)\\ =-sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\alpha\right)-tan\left(\alpha\right)cot\left(\pi-\alpha\right)\\ =-cos\left(\alpha\right)+cos\left(\alpha\right)+tan\left(\alpha\right)\cdot cot\left(\alpha\right)\\ =1\)
Biết \(sin\alpha=\frac{3}{5}\) và \(\frac{\pi}{2}< \alpha< \pi\). Tính \(P=1-2sin^2\left(\frac{\pi}{4}-\alpha\right)+sin2\alpha+cos\left(\pi-2\alpha\right)-6tan\left(\frac{\pi}{2}-\alpha\right)\)
\(\frac{\pi}{2}< a< \pi\Rightarrow cosa< 0\)
\(\Rightarrow cosa=-\sqrt{1-sin^2a}=-\frac{4}{5}\)
\(P=1-\left[1-cos\left(\frac{\pi}{2}-2a\right)\right]+sin2a-cos2a-6cota\)
\(=sin2a+sin2a-cos2a-6cota\)
\(=2sin2a-cos2a-6cota\)
\(=4sina.cosa-\left(cos^2a-sin^2a\right)-\frac{6cosa}{sina}\) (thay số và bấm máy)
Câu 1 : Chọn đẳng thức đúng và chứng minh :
\(A.cos^2\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)=\frac{1-sin\alpha}{2}\) \(B.cos^2\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)=\frac{1+sin\alpha}{2}\)
\(C.cos^2\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)=\frac{1-cos\alpha}{2}\) \(D.cos^2\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)=\frac{1+cos\alpha}{2}\)
Theo mk là A đúng
ta có : cos2x = \(\frac{1+cos2x}{2}\)
=> cos2(\(\frac{\pi}{4}\)+\(\frac{\alpha}{2}\))= \(\frac{1+cos\left(\frac{\pi}{2}+\alpha\right)}{2}\) = \(\frac{1-sinx}{2}\)
Cho cos \(\alpha\)=\(-\frac{4}{5}\) và \(-\pi< \alpha< \frac{-3}{2}\pi\). Tính \(\sin2\alpha;\)\(\cos2a;\sin\left(\frac{5\pi}{2}-\alpha\right);\tan\left(\alpha+\frac{\pi}{4}\right);\cos\frac{\alpha}{2}\)
--.-- \(-\pi>-\frac{3}{2}\pi\) mà
Chắc nhầm đề rồi, phải là \(-\pi>a>-\frac{3}{2}\pi\)mới đúng chứ
\(-\pi>a>-\frac{3}{2}\pi\Leftrightarrow\pi>a>\frac{1}{2}\pi\)
\(\cos a=-\frac{4}{5}\Rightarrow\sin a=\frac{3}{5}\)
\(\sin2a=2\sin a.\cos a=2.\frac{3}{5}.\frac{-4}{5}=-\frac{24}{25}\)
\(\cos2a=2\cos^2a-1=\frac{7}{25}\)
\(\sin\left(\frac{5\pi}{2}-a\right)=\sin\left(\frac{\pi}{2}-a\right)=\cos a=-\frac{4}{5}\)
\(\sin\left(a+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}.\frac{3}{5}-\frac{4}{5}.\frac{\sqrt{2}}{2}=-\frac{\sqrt{2}}{10}\)
\(\cos\left(a+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}.\frac{-4}{5}-\frac{\sqrt{2}}{2}.\frac{3}{5}=-\frac{7\sqrt{2}}{10}\)
\(\Rightarrow\tan\left(a+\frac{\pi}{4}\right)=\frac{1}{7}\)
\(\cos^2\left(\frac{a}{2}\right)=\frac{1+\cos a}{2}=\frac{1}{10}\Leftrightarrow\left|\cos\frac{a}{2}\right|=\frac{\sqrt{10}}{10}\)
Mà \(\frac{\pi}{2}>\frac{a}{2}>\frac{\pi}{4}\)
\(\Rightarrow\cos\frac{a}{2}=\frac{\sqrt{10}}{10}\)