Với \(sina=\dfrac{1}{\sqrt{3}}\) với \(0< a< \dfrac{\pi}{2}\)
\(sin^2a+cos^2a=1\)
\(\Leftrightarrow cosa=\sqrt{1-sin^2a}=\sqrt{1-\dfrac{1}{3}}=\sqrt{\dfrac{2}{3}}\)
\(cos\left(a+\dfrac{\pi}{3}\right)=cosa.cos\dfrac{\pi}{3}-sina.sin\dfrac{\pi}{3}=\sqrt{\dfrac{2}{3}}.\dfrac{1}{2}-\dfrac{1}{\sqrt{3}}.\dfrac{\sqrt{3}}{2}=-0.09\)
Chứng minh rằng:
a) \(sin\left(a+b\right).sin\left(a-b\right)=sin^2a-sin^2b=cos^2b-cos^2a\)
b) \(4sin\left(x+\dfrac{\Pi}{3}\right).sin\left(x-\dfrac{\Pi}{3}\right)=4sin^2x-3\)
c) \(sin\left(x+\dfrac{\Pi}{4}\right)-sin\left(x-\dfrac{\Pi}{4}\right)=\sqrt{2}cosx\)
d) \(\dfrac{1}{sin10^0}-\dfrac{\sqrt{3}}{cos10^0}=4\)
Cho \(sina=\dfrac{3}{5},cosb=-\dfrac{5}{13}\)và \(\dfrac{\pi}{2}< a,b< \pi\)
Tính \(cos\dfrac{a}{2};sin\dfrac{b}{2};tan\left(a+b\right);sin\left(a-b\right)\)
GIÚP VỚI MÌNH ĐANG CẦN GẤP
Rút gọn:
C= \(sin^2\dfrac{\pi}{3}+sin^2\dfrac{5\pi}{6}+sin^2\dfrac{\pi}{9}+sin^2\dfrac{11\pi}{18}+sin^2\dfrac{13\pi}{18}+sin^2\dfrac{2\pi}{9}\)
D=\(cos\left(x-\dfrac{\pi}{3}\right).cos\left(x+\dfrac{\pi}{4}\right)+cos\left(x+\dfrac{\pi}{6}\right).cos\left(x+\dfrac{3\pi}{4}\right)\)
Cho \(\alpha\) , \(\beta\in\left(0;\dfrac{\pi}{2}\right)\) và sin \(\alpha\) = \(\dfrac{1}{\sqrt{5}}\) ; Cos \(\alpha\) = \(\dfrac{1}{\sqrt{10}}\) . Tính Cos \(\left(\alpha+\beta\right)\)
Đơn giản các biểu thức sau:
G = \(cos\left(\alpha-5\pi\right)+sin\left(-\dfrac{3\pi}{2}+\alpha\right)-tan\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\dfrac{3\pi}{2}-\alpha\right)\)
H = \(cot\left(\alpha-2\pi\right).cos\left(\alpha-\dfrac{3\pi}{2}\right)+cos\left(\alpha-6\pi\right)-2sin\left(\alpha-\pi\right)\)
Tính \(\cos\left(\dfrac{28\pi}{3}\right)+\sin\left(\dfrac{37\pi}{6}\right)+\tan\left(-\dfrac{13\pi}{4}\right)\)
Đơn giản biểu thức sau:
\(G=Cos\left(\alpha-5\pi\right)+sin\left(-\dfrac{3\pi}{2}+\alpha\right)-tan\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\dfrac{3\pi}{2}-\alpha\right)\)
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Tính giá trị của biểu thức:
A= \(\dfrac{cos\dfrac{9\pi}{5}-cos\dfrac{6\pi}{5}+cos\dfrac{11\pi}{5}}{cos\dfrac{3\pi}{10}-sin\dfrac{6\pi}{5}}.tan\dfrac{16\pi}{5}\).
Nếu cosa+sina=\(\sqrt{2}\) \(\left(0< a< \dfrac{\pi}{2}\right)\) thì a bằng
