Tính gtri bthuc
A= sin 23\(^o\)\(-\) cos 67\(^o\)
B= \(\dfrac{tan70^0.tan45^0.tan20^0}{cot70^0.cot45^0.cot20^0}\)
tính
A=sin10+sin40-cos50-cos80
B=cos15+cos35-sin65-sin75
C=\(\dfrac{tan27.tan63}{cot63.cot27}\)
D= \(\dfrac{cot20.cot45.cot70}{tan20.tan45.tan70}\)
Giải:
\(A=\sin10+\sin40-\cos50-\cos80\)
\(\Leftrightarrow A=\cos80+\cos50-\cos50-\cos80\)
\(\Leftrightarrow A=0\)
Vậy ...
\(B=\cos15+\cos25-\sin65-\sin75\)
\(\Leftrightarrow B=\sin75+\sin65-\sin65-\sin75\)
\(\Leftrightarrow B=0\)
Vậy ...
\(C=\dfrac{\tan27.\tan63}{\cot63.\cot27}\)
\(\Leftrightarrow C=\dfrac{\tan27.\tan63}{\tan27.\tan63}\)
\(\Leftrightarrow C=1\)
Vậy ...
\(D=\dfrac{\cot20.\cot45.\cot70}{\tan20.\tan45.\tan70}\)
\(\Leftrightarrow D=\dfrac{\cot20.\cot45.\cot70}{\cot70.\cot45.\cot20}\)
\(\Leftrightarrow D=1\)
Vậy ...
bài 1 tính giá trị biểu thức sau
A = sin 10 + sin 40 - cos 55 - cos 80
B = cos 15 + cos 35 - sin 55 - sin 75
C = \(\frac{tan27^0.tan63^0}{cot63^0.cot27^0}\)
D = \(\frac{cot20^0cot45^0cot70^0}{tan20^0tan45^0tan70^0}\)
\(D=tan10^0.tan70^0+tan70^0.tan130^0+tan130^0.tan190^0\text{ }\)
\(E=\dfrac{cot225^0-cot79^0.cot71^0}{cot259^0+cot151}\)
\(F=cos^275^0-sin^275^0\)
\(F=cos^275^0-sin^275^0\)
\(=1-sin^275^0-sin^275^0\)
\(=1-2\cdot sin^275^0=cos150=-\dfrac{\sqrt{3}}{2}\)
cho góc nhọn x bt tgx=3 tính 0
B=cos2(36o)+cos2(54o)- cos2(61o)-cos2(29o)/sin2(24o)+sin2(66o)+sin2(25o)+sin2(650) + 3cotg(67o) / 2tg(23o)
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: \(cos\left(2x+30^0\right)+sin\left(x-30^0\right)=0\)
=>\(cos\left(2x+30^0\right)=-sin\left(x-30^0\right)\)
=>\(cos\left(2x+30^0\right)=sin\left(-x+30^0\right)\)
=>\(cos\left(2x+30^0\right)=cos\left(60^0+x\right)\)
=>\(\left[{}\begin{matrix}2x+30^0=x+60^0+k\cdot360^0\\2x+30^0=-x-60^0+k\cdot360^0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=30^0+k\cdot360^0\\3x=-90^0+k\cdot360^0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=30^0+k\cdot360^0\\x=-30^0+k\cdot120^0\end{matrix}\right.\)
Rút gọn biểu thức :
a) \(4a^2\cos^260^0+2ab.\cos^2180^0+\dfrac{4}{3}\cos^230^0\)
b) \(\left(a\sin90^0+b\tan45^0\right)\left(a\cos0^0+b\cos180^0\right)\)
a)
\(4a^2cos^260^o+2ab.cos^2180^o+\dfrac{4}{3}cos^230^o\)
\(=4a^2.\left(\dfrac{1}{2}\right)^2+2ab.\left(-1\right)^2+\dfrac{4}{3}.\left(\dfrac{\sqrt{3}}{2}\right)^2\)
\(=4a^2.\dfrac{1}{4}+2ab+\dfrac{4}{3}.\dfrac{3}{4}\)
\(=a^2+2ab+1\).
b)
\(\left(asin90^o+btan45^o\right)\left(acos0^o+bcos180^o\right)\)
\(=\left(a+b\right)\left(a-b\right)=a^2-b^2\).
Tính:
a) \(sin42^0-cos48^0\)
b) \(sin^261^0+sin^229^0\)
c) \(tan40^0.tan45^0.tan50^0\)
a) Ta có: \(sin\alpha=cos\left(90-\alpha\right)\Rightarrow sin42=cos48\)
\(\Rightarrow sin42-cos48=0\)
b) Ta có: \(sin\alpha=cos\left(90-\alpha\right)\Rightarrow sin61=cos29\Rightarrow sin^261=cos^229\)
\(\Rightarrow sin^261+sin^229=sin^229+cos^229=1\)
c) Ta có: \(tan\alpha=\dfrac{1}{tan\left(90-\alpha\right)}\Rightarrow tan40=\dfrac{1}{tan50}\)
\(\Rightarrow tan40.tan50=1\) mà \(tan45=1\Rightarrow tan40.tan45.tan50=1\)
\(sin42^0-cos48^0=sin42^0-sin\left(90^0-48^0\right)=sin42^0-sin42^0=0\)
\(sin^261^0+sin^229^0=sin^261^0+cos^2\left(90^0-29^0\right)=sin^261^0+cos^261^0=1\)
\(tan40^0.tan50^0.tan45^0=tan40^0.cot\left(90^0-50^0\right).1=tan40^0.cot40^0=1\)
Sử dụng các công thức:
\(cosa=sin\left(90^0-a\right)\) ; \(sina=cos\left(90^0-a\right)\) ; \(tana=cot\left(90^0-a\right)\) ; \(tana.cota=1\)
Bài 2: ( chỉ ghi kết quả )
a) Tìm x biết : \(\dfrac{1}{2-\dfrac{3}{4+\dfrac{5}{6-\dfrac{7}{8+\dfrac{9}{10}}}}}=\dfrac{1}{x+\dfrac{1}{3+\dfrac{1}{5}}}+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{2}}}\)
b) Tính: ( kết quả lấy 4 chữ số thập phân)
\(P=\dfrac{\sin^390^0-\cot^330^0-\cos^245^0+\tan20^0}{2\sqrt{7}+\sin108^0\cos32^0\tan64^0}\)
\(\dfrac{1}{2-\dfrac{3}{4+\dfrac{5}{6-\dfrac{7}{8+\dfrac{9}{10}}}}}=\dfrac{1}{x+\dfrac{1}{3+\dfrac{1}{5}}}+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{2}}}\\ \Leftrightarrow\dfrac{767}{1070}=\dfrac{1}{x+\dfrac{5}{16}}+\dfrac{3}{5}\\ \Leftrightarrow\dfrac{25}{214}=\dfrac{1}{x+\dfrac{5}{16}}\\ \Rightarrow x+\dfrac{5}{16}=\dfrac{214}{25}\Rightarrow x=\dfrac{3299}{400}\)
Không dùng máy tính hãy tính:
A= Sin2100+Sin2200+Sin2450+Sin2700+Sin2800
B=Sin2420+Sin2430+Sin2440+Sin2450+Sin2460+Sin2470+Sin2490
C= tan350*tan400*tan450*tan500*tan550( dấu "*" là nhân nha)
D= cos2150-cos2250+cos2350-cos2450+cos2550-cos2650+cos2750
Giúp mình nha!
a: \(=\left(sin^210^0+sin^280^0\right)+\left(sin^220^0+sin^270^0\right)+sin^245^0\)
\(=1+1+\dfrac{1}{2}=\dfrac{5}{2}\)
b: \(=\left(sin^242^0+sin^248^0\right)+\left(sin^243^0+sin^247^0\right)+...+sin^245^0\)
=1+1+1+1/2
=3,5
c: \(=tan35^0\cdot tan55^0\cdot tan40^0\cdot tan50^0\cdot tan45^0=1\)
d: \(=\left(cos^215^0+cos^275^0\right)-\left(cos^225^0+cos^265^0\right)+\left(cos^235^0+cos^255^0\right)-\dfrac{1}{2}\)
=1-1+1-1/2
=1/2