Đề bài
Không dùng máy tính cầm tay, tính các giá trị lượng giác của các góc:
a, \(\frac{{5\pi }}{{12}}\)
b, \(-{\rm{ }}{555^0}\)
Phương pháp giải - Xem chi tiết
Áp dụng công thức
\(\begin{array}{l}\sin \left( {a + b} \right) = \sin a\cos b + \cos asinb\\sin\left( {a - b} \right) = \sin a\cos b - \cos asinb\\\cos \left( {a + b} \right) = \cos a\cos b - \sin asinb\\\cos \left( {a - b} \right) = \cos a\cos b + \sin asinb\\\tan \left( {a + b} \right) = \frac{{\tan a + \tan b}}{{1 - \tan a\tan b}}\\\tan \left( {a - b} \right) = \frac{{\tan a - \tan b}}{{1 + \tan a\tan b}}\end{array}\)
Lời giải chi tiết
a, Ta có:
\(\begin{array}{l}\cos \frac{{5\pi }}{{12}} = \cos \left( {\frac{\pi }{4} + \frac{\pi }{6}} \right) = \cos \frac{\pi }{4}\cos \frac{\pi }{6} - \sin \frac{\pi }{4}sin\frac{\pi }{6}\\ = \frac{{\sqrt 2 }}{2}.\frac{{\sqrt 3 }}{2} - \frac{{\sqrt 2 }}{2}.\frac{1}{2} = \frac{{\sqrt 6 - \sqrt 2 }}{4}\end{array}\)
\(\begin{array}{l}\sin \frac{{5\pi }}{{12}} = \sin \left( {\frac{\pi }{4} + \frac{\pi }{6}} \right) = \sin \frac{\pi }{4}\cos \frac{\pi }{6} + \cos \frac{\pi }{4}sin\frac{\pi }{6}\\ = \frac{{\sqrt 2 }}{2}.\frac{{\sqrt 3 }}{2} + \frac{{\sqrt 2 }}{2}.\frac{1}{2} = \frac{{\sqrt 6 + \sqrt 2 }}{4}\end{array}\)
\(\tan \frac{{5\pi }}{{12}} = \frac{{sin\frac{{5\pi }}{{12}}}}{{cos\frac{{5\pi }}{{12}}}} = 2 + \sqrt 3 \)
\(\cot \frac{{5\pi }}{{12}} = \frac{1}{{\tan \frac{{5\pi }}{{12}}}} = \frac{1}{{2 + \sqrt 3 }}\)
b, Ta có:
\(\begin{array}{l}\cos \left( { - {{555}^0}} \right) = \cos \left( {3\pi + \frac{\pi }{{12}}} \right) = - \cos \frac{\pi }{{12}} = - \cos \left( {\frac{\pi }{3} - \frac{\pi }{4}} \right)\\ = - \left( {\cos \frac{\pi }{3}\cos \frac{\pi }{4} + \sin \frac{\pi }{3}sin\frac{\pi }{4}} \right) = - \frac{{\sqrt 6 + \sqrt 2 }}{4}\end{array}\)
\(\begin{array}{l}\sin \left( { - {{555}^0}} \right) = \sin \left( {3\pi + \frac{\pi }{{12}}} \right) = sin\frac{\pi }{{12}} = sin\left( {\frac{\pi }{3} - \frac{\pi }{4}} \right)\\ = \sin \frac{\pi }{3}\cos \frac{\pi }{4} - \cos \frac{\pi }{3}sin\frac{\pi }{4}\\ = \frac{{\sqrt 3 }}{2}.\frac{{\sqrt 2 }}{2} - \frac{1}{2}.\frac{{\sqrt 2 }}{2} = \frac{{\sqrt 6 - \sqrt 2 }}{4}\end{array}\)
\(\tan \left( { - {{555}^0}} \right) = \frac{{\sin \left( { - {{555}^0}} \right)}}{{\cos \left( { - {{555}^0}} \right)}} = - 2 + \sqrt 3 \)
\(\cot \left( { - {{555}^0}} \right) = \frac{1}{{ - 2 + \sqrt 3 }} = - 2 - \sqrt 3 \)
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