Tính :
a) \(\cos225^0;\sin240^0;\cot\left(-15^0\right);\tan75^0\)
b) \(\sin\dfrac{7\pi}{15};\cos\left(-\dfrac{\pi}{12}\right);\tan\dfrac{13\pi}{12}\)
Những câu hỏi liên quan
bài 1: a) \(sin\left(2x+\dfrac{\pi}{6}\right)+sin\left(x-\dfrac{\pi}{3}\right)=0\)
b) \(sin\left(2x-\dfrac{\pi}{3}\right)-cos\left(x+\dfrac{\pi}{3}\right)=0\)
c) \(sin\left(2x+\dfrac{\pi}{3}\right)+cos\left(x-\dfrac{\pi}{6}\right)=0\)
a) \(sin\left(2x+\dfrac{\pi}{6}\right)+sin\left(x-\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{6}\right)=-sin\left(x-\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{3}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{6}=\dfrac{\pi}{3}-x+k\pi\\2x+\dfrac{\pi}{6}=\pi-\dfrac{\pi}{3}+x+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{\pi}{6}+k\pi\\x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{18}+\dfrac{k\pi}{3}\\x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)
b) \(sin\left(2x-\dfrac{\pi}{3}\right)-cos\left(x+\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{3}\right)=cos\left(x+\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{3}\right)=sin\left(\dfrac{\pi}{6}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{6}-x+k\pi\\2x-\dfrac{\pi}{3}=\pi-\dfrac{\pi}{6}+x+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{7\pi}{6}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=\dfrac{\pi}{6}+\left(k+1\right)\pi\end{matrix}\right.\)
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c: =>\(cos\left(x-\dfrac{pi}{6}\right)=-sin\left(2x+\dfrac{pi}{3}\right)\)
=>\(cos\left(x-\dfrac{pi}{6}\right)=sin\left(-2x-\dfrac{pi}{3}\right)\)
=>\(sin\left(-2x-\dfrac{pi}{3}\right)=sin\left(\dfrac{pi}{2}-x+\dfrac{pi}{6}\right)\)
=>\(sin\left(-2x-\dfrac{pi}{3}\right)=sin\left(-x+\dfrac{2}{3}pi\right)\)
=>\(\left[{}\begin{matrix}-2x-\dfrac{pi}{3}=-x+\dfrac{2}{3}pi+k2pi\\-2x-\dfrac{pi}{3}=pi+x-\dfrac{2}{3}pi+k2pi\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}-x=pi+k2pi\\-3x=\dfrac{2}{3}pi+k2pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-pi-k2pi\\x=-\dfrac{2}{9}pi-\dfrac{k2pi}{3}\end{matrix}\right.\)
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Chứng minh các hệ thức sau :
a) sinalpha+sinleft(alpha+dfrac{14}{3}piright)+sinleft(alpha-dfrac{8}{3}piright)0
b) dfrac{sin4a}{1+cos4a}.dfrac{cos2a}{1+cos2a}cotleft(dfrac{3}{2}pi-aright)
c) left(cos a-cos bright)^2-left(sin a-sin bright)^2-4sin^2dfrac{a-b}{2}cosleft(a+bright)
d) sin^2left(45^0+alpharight)-sin^2left(30^0-alpharight)-sin15^0cosleft(15^0+2alpharight)sin2alpha
Đọc tiếp
Chứng minh các hệ thức sau :
a) \(\sin\alpha+\sin\left(\alpha+\dfrac{14}{3}\pi\right)+\sin\left(\alpha-\dfrac{8}{3}\pi\right)=0\)
b) \(\dfrac{\sin4a}{1+\cos4a}.\dfrac{\cos2a}{1+\cos2a}=\cot\left(\dfrac{3}{2}\pi-a\right)\)
c) \(\left(\cos a-\cos b\right)^2-\left(\sin a-\sin b\right)^2=-4\sin^2\dfrac{a-b}{2}\cos\left(a+b\right)\)
d) \(\sin^2\left(45^0+\alpha\right)-\sin^2\left(30^0-\alpha\right)-\sin15^0\cos\left(15^0+2\alpha\right)=\sin2\alpha\)
Giải các phương trình :
a) cosleft(22^0-tright)cosleft(82^0-tright)+cosleft(112^0-tright)cosleft(172^0-tright)dfrac{1}{2}left(sin t+cos tright)
b) sin^2left(t+45^0right)-sin^2left(t-30^0right)-sin15^0cosleft(2t+15^0right)dfrac{1}{2}sin6t
c) sin^82x+cos^82xdfrac{41}{128}
d) sqrt{4cos^2+1}+sqrt{4sin^2x+3}4
e) tanleft(picot tright)cotleft(pisin tright)
Đọc tiếp
Giải các phương trình :
a) \(\cos\left(22^0-t\right)\cos\left(82^0-t\right)+\cos\left(112^0-t\right)\cos\left(172^0-t\right)=\dfrac{1}{2}\left(\sin t+\cos t\right)\)
b) \(\sin^2\left(t+45^0\right)-\sin^2\left(t-30^0\right)-\sin15^0\cos\left(2t+15^0\right)=\dfrac{1}{2}\sin6t\)
c) \(\sin^82x+\cos^82x=\dfrac{41}{128}\)
d) \(\sqrt{4\cos^2+1}+\sqrt{4\sin^2x+3}=4\)
e) \(\tan\left(\pi\cot t\right)=\cot\left(\pi\sin t\right)\)
chứng minh đẳng thức lượng giác
a) 2.\(cot\left(\dfrac{\pi}{2}-x\right)\)+ tan\(\left(\pi-x\right)\)= tan\(x\)
b) sin\(\left(\dfrac{5\pi}{2}-x\right)\)+ cos \(\left(13\pi+x\right)\) - sin\(\left(x-5\pi\right)\) = sin\(x\)
a: \(2\cdot cot\left(\dfrac{pi}{2}-x\right)+tan\left(pi-x\right)\)
\(=2\cdot tanx-tanx\)
=tan x
b: \(sin\left(\dfrac{5}{2}pi-x\right)+cos\left(13pi+x\right)-sin\left(x-5pi\right)\)
\(=sin\left(\dfrac{pi}{2}-x\right)+cos\left(pi+x\right)+sin\left(pi-x\right)\)
\(=cosx-cosx+sinx=sinx\)
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chứng minh đẳng thức lượng giác
a) 2.cot\(\left(\dfrac{\pi}{2}-x\right)\)+ tan\(\left(\pi-x\right)\) = tan\(x\)
b) \(sin\left(\dfrac{5\pi}{2}-x\right)\)+ cos\(\left(13\pi+x\right)\) - sin\(\left(x-5\pi\right)\) = sin\(x\)
\(a,VT=2.tanx+tan\left(-x\right)\\ =2tanx-tanx=tanx\)
\(b,VT=sin\left(2\pi+\dfrac{\pi}{2}-x\right)+cos\left(12\pi+\pi+x\right)-sin\left(x-4\pi-\pi\right)\\ =sin\left(\dfrac{\pi}{2}-x\right)+cos\left(\pi+x\right)+sin\left(\pi-x\right)\\ =cosx-cosx+sinx\\ =sinx=VP\)
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Giải các phương trình :
a) cos^2x+cos^22x-cos^23x-cos^24x0
b) cos4xcosleft(pi+2xright)-sin2xcosleft(dfrac{pi}{2}-4xright)dfrac{sqrt{2}}{2}sin4x
c) tanleft(120^0+3xright)-tanleft(140^0-xright)2sinleft(80^0+2xright)
d) tan^2dfrac{x}{2}+sin^2dfrac{x}{2}tandfrac{x}{2}+cos^2dfrac{x}{2}+cot^2dfrac{x}{2}+sin x4
e) dfrac{sin2t+2cos^2t-1}{cot t-cot3t+sin3t-sin t}cos t
Đọc tiếp
Giải các phương trình :
a) \(\cos^2x+\cos^22x-\cos^23x-\cos^24x=0\)
b) \(\cos4x\cos\left(\pi+2x\right)-\sin2x\cos\left(\dfrac{\pi}{2}-4x\right)=\dfrac{\sqrt{2}}{2}\sin4x\)
c) \(\tan\left(120^0+3x\right)-\tan\left(140^0-x\right)=2\sin\left(80^0+2x\right)\)
d) \(\tan^2\dfrac{x}{2}+\sin^2\dfrac{x}{2}\tan\dfrac{x}{2}+\cos^2\dfrac{x}{2}+\cot^2\dfrac{x}{2}+\sin x=4\)
e) \(\dfrac{\sin2t+2\cos^2t-1}{\cot t-\cot3t+\sin3t-\sin t}=\cos t\)
Tính :
a) \(4\left(\cos24^0+\cos48^0-\cos84^0-\cos12^0\right)\)
b) \(96\sqrt{3}\sin\dfrac{\pi}{48}\cos\dfrac{\pi}{48}\cos\dfrac{\pi}{24}\cos\dfrac{\pi}{12}\cos\dfrac{\pi}{6}\)
c) \(\tan9^0-\tan63^0+\tan81^0-\tan27^0\)
Tính giá trị của biểu thức sau : B= \(\dfrac{tan\left(\dfrac{21\pi}{2}-x\right).cos\left(38\pi-x\right).sin\left(x-7\pi\right)}{sin\left(\dfrac{13\pi}{2}-x\right).cos\left(x-2023\pi\right)}\)
bài 1:
a) \(cos\left(2x-\dfrac{\pi}{6}\right)+cos\left(x+\dfrac{\pi}{3}\right)=0\)
b) \(cos\left(2x+30^o\right)+sin\left(x-30^o\right)=0\)
a: \(cos\left(2x-\dfrac{\Omega}{6}\right)+cos\left(x+\dfrac{\Omega}{3}\right)=0\)
=>\(cos\left(2x-\dfrac{\Omega}{6}\right)+sin\left(\dfrac{\Omega}{6}-x\right)=0\)
=>\(cos\left(2x-\dfrac{\Omega}{6}\right)=-sin\left(\dfrac{\Omega}{6}-x\right)=sin\left(x-\dfrac{\Omega}{6}\right)\)
=>\(cos\left(2x-\dfrac{\Omega}{6}\right)=cos\left(\dfrac{\Omega}{2}-x+\dfrac{\Omega}{6}\right)\)
=>\(cos\left(2x-\dfrac{\Omega}{6}\right)=cos\left(-x+\dfrac{2}{3}\Omega\right)\)
=>\(\left[{}\begin{matrix}2x-\dfrac{\Omega}{6}=-x+\dfrac{2\Omega}{3}+k2\Omega\\2x-\dfrac{\Omega}{6}=x-\dfrac{2}{3}\Omega+k2\Omega\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}3x=\dfrac{5}{6}\Omega+k2\Omega\\x=-\dfrac{1}{2}\Omega+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{18}\Omega+\dfrac{k2\Omega}{3}\\x=-\dfrac{1}{2}\Omega+k2\Omega\end{matrix}\right.\)
b: \(cos\left(2x+30^0\right)+sin\left(x-30^0\right)=0\)
=>\(cos\left(2x+30^0\right)=-sin\left(x-30^0\right)\)
=>\(cos\left(2x+30^0\right)=sin\left(-x+30^0\right)\)
=>\(cos\left(2x+30^0\right)=cos\left(60^0+x\right)\)
=>\(\left[{}\begin{matrix}2x+30^0=x+60^0+k\cdot360^0\\2x+30^0=-x-60^0+k\cdot360^0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=30^0+k\cdot360^0\\3x=-90^0+k\cdot360^0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=30^0+k\cdot360^0\\x=-30^0+k\cdot120^0\end{matrix}\right.\)
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