Tính giá trị biểu thức (không sử dụng máy tính cầm tay)
a) \(M=\dfrac{\sin48^{\circ}}{\cos42^{\circ}}-\cos60^{\circ}+\tan27^{\circ}.\tan63^{\circ}\)
b) \(N=\cot27^{\circ}.\cot60^{\circ}.\cot63^{\circ}+\sin^244+\sin^246\)
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HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
a) \(\cos \frac{{3\pi }}{7} = 0,22252\); \(\tan ( - {37^ \circ }25') = 0,765018\)
b) \(179^o23'30"\approx3,130975234\left(rad\right)\)
c) \(\frac{{7\pi }}{9} = {140^ \circ }\)
23 tháng 3 2022
a) Ta có: \(sin^2x+sin^2\left(90-x\right)=sin^2x+cos^2x=1.\)
áp dụng: A = 2
b)Ta có: \(cos\left(x\right)=-cos\left(180-x\right)\)
áp dụng: B = 0
c) Ta có: \(tan\left(x\right)\cdot tan\left(90-x\right)=\frac{sinx}{cosx}\cdot\frac{sin\left(90-x\right)}{cos\left(90-x\right)}=\frac{sinx}{cosx}\cdot\frac{cosx}{sinx}=1\)
áp dụng: C = 1
QT
Quoc Tran Anh Le
Giáo viên
21 tháng 9 2023
\(\begin{array}{l}\cos 75^\circ = \frac{{\sqrt 6 - \sqrt 2 }}{4}\\\tan \left( { - \frac{{19\pi }}{6}} \right) = - \frac{{\sqrt 3 }}{3}\end{array}\)
HQ
Hà Quang Minh
Giáo viên
12 tháng 1
a) Vì \(\Delta DEG \backsim \Delta MNP\) nên \(\widehat D = \widehat M,\,\,\widehat E = \widehat N,\,\,\widehat G = \widehat P\)
\( \Rightarrow \widehat D = \widehat M = 40^\circ \)
\( \to \) Chọn đáp án A.
b) Theo câu a) ta có \(\widehat E = \widehat N = 60^\circ \)
\( \to \) Chọn đáp án C.
c) Xét tam giác MNP có:
\(\begin{array}{l}\widehat M + \widehat N + \widehat P = 180^\circ \\ \Rightarrow 40^\circ + 60^\circ + \widehat P = 180^\circ \\ \Rightarrow \widehat P = 80^\circ \end{array}\)
\( \to \) Chọn đáp án D.
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
\(\begin{array}{l}\cos \left( {{{225}^ \circ }} \right) = \cos \left( {{{180}^ \circ } + {{45}^ \circ }} \right) = - \cos \left( {{{45}^ \circ }} \right) = - \frac{{\sqrt 2 }}{2}\\\sin \left( {{{225}^ \circ }} \right) = \sin \left( {{{180}^ \circ } + {{45}^ \circ }} \right) = - \sin \left( {{{45}^ \circ }} \right) = - \frac{{\sqrt 2 }}{2}\\\tan \left( {225^\circ } \right) = \frac{{\sin \left( {{{225}^ \circ }} \right)}}{{\cos \left( {{{225}^ \circ }} \right)}} = 1\\\cot \left( {225^\circ } \right) = \frac{1}{{\tan \left( {225^\circ } \right)}} = 1\end{array}\)
\(\begin{array}{l}\cos \left( { - {{225}^ \circ }} \right) = \cos \left( {{{225}^ \circ }} \right) = \cos \left( {{{180}^ \circ } + {{45}^ \circ }} \right) = - \cos \left( {{{45}^ \circ }} \right) = - \frac{{\sqrt 2 }}{2}\\\sin \left( { - {{225}^ \circ }} \right) = - \sin \left( {{{225}^ \circ }} \right) = - \sin \left( {{{180}^ \circ } + {{45}^ \circ }} \right) = \sin \left( {{{45}^ \circ }} \right) = \frac{{\sqrt 2 }}{2}\\\tan \left( { - 225^\circ } \right) = \frac{{\sin \left( {{{225}^ \circ }} \right)}}{{\cos \left( {{{225}^ \circ }} \right)}} = - 1\\\cot \left( { - 225^\circ } \right) = \frac{1}{{\tan \left( {225^\circ } \right)}} = - 1\end{array}\)
\(\begin{array}{l}\cos \left( { - {{1035}^ \circ }} \right) = \cos \left( {{{1035}^ \circ }} \right) = \cos \left( {{{6.360}^ \circ } - {{45}^ \circ }} \right) = \cos \left( { - {{45}^ \circ }} \right) = \cos \left( {{{45}^ \circ }} \right) = \frac{{\sqrt 2 }}{2}\\\sin \left( { - {{1035}^ \circ }} \right) = - \sin \left( {{{1035}^ \circ }} \right) = - \sin \left( {{{6.360}^ \circ } - {{45}^ \circ }} \right) = - \sin \left( { - {{45}^ \circ }} \right) = \sin \left( {{{45}^ \circ }} \right) = \frac{{\sqrt 2 }}{2}\\\tan \left( { - 1035^\circ } \right) = \frac{{\sin \left( { - {{1035}^ \circ }} \right)}}{{\cos \left( { - {{1035}^ \circ }} \right)}} = 1\\\cot \left( { - 1035^\circ } \right) = \frac{1}{{\tan \left( { - 1035^\circ } \right)}} = - 1\end{array}\)
\(\begin{array}{l}\cos \left( {\frac{{5\pi }}{3}} \right) = \cos \left( {\pi + \frac{{2\pi }}{3}} \right) = - \cos \left( {\frac{{2\pi }}{3}} \right) = \frac{1}{2}\\\sin \left( {\frac{{5\pi }}{3}} \right) = \sin \left( {\pi + \frac{{2\pi }}{3}} \right) = - \sin \left( {\frac{{2\pi }}{3}} \right) = - \frac{{\sqrt 3 }}{2}\\\tan \left( {\frac{{5\pi }}{3}} \right) = \frac{{\sin \left( {\frac{{5\pi }}{3}} \right)}}{{\cos \left( {\frac{{5\pi }}{3}} \right)}} = - \sqrt 3 \\\cot \left( {\frac{{5\pi }}{3}} \right) = \frac{1}{{\tan \left( {\frac{{5\pi }}{3}} \right)}} = - \frac{{\sqrt 3 }}{3}\end{array}\)
\(\begin{array}{l}\cos \left( {\frac{{19\pi }}{2}} \right) = \cos \left( {8\pi + \frac{{3\pi }}{2}} \right) = \cos \left( {\frac{{3\pi }}{2}} \right) = \cos \left( {\pi + \frac{\pi }{2}} \right) = - \cos \left( {\frac{\pi }{2}} \right) = 0\\\sin \left( {\frac{{19\pi }}{2}} \right) = \sin \left( {8\pi + \frac{{3\pi }}{2}} \right) = \sin \left( {\frac{{3\pi }}{2}} \right) = \sin \left( {\pi + \frac{\pi }{2}} \right) = - \sin \left( {\frac{\pi }{2}} \right) = - 1\\\tan \left( {\frac{{19\pi }}{2}} \right)\\\cot \left( {\frac{{19\pi }}{2}} \right) = \frac{{\cos \left( {\frac{{19\pi }}{2}} \right)}}{{\sin \left( {\frac{{19\pi }}{2}} \right)}} = 0\end{array}\)
\(\begin{array}{l}\cos \left( { - \frac{{159\pi }}{4}} \right) = \cos \left( {\frac{{159\pi }}{4}} \right) = \cos \left( {40.\pi - \frac{\pi }{4}} \right) = \cos \left( { - \frac{\pi }{4}} \right) = \cos \left( {\frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2}\\\sin \left( { - \frac{{159\pi }}{4}} \right) = - \sin \left( {\frac{{159\pi }}{4}} \right) = - \sin \left( {40.\pi - \frac{\pi }{4}} \right) = - \sin \left( { - \frac{\pi }{4}} \right) = \sin \left( {\frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2}\\\tan \left( { - \frac{{159\pi }}{4}} \right) = \frac{{\cos \left( { - \frac{{159\pi }}{4}} \right)}}{{\sin \left( { - \frac{{159\pi }}{4}} \right)}} = 1\\\cot \left( { - \frac{{159\pi }}{4}} \right) = \frac{1}{{\tan \left( { - \frac{{159\pi }}{4}} \right)}} = 1\end{array}\)
H9
HT.Phong (9A5)
CTVHS
22 tháng 7 2023
Ta có tổng 4 góc trong tứ giác là: \(360^o\)
\(\Rightarrow\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\)
Hay: \(60^o+110^o+\widehat{C}+70^o=360^o\)
\(\Rightarrow\widehat{C}=360^o-\left(110^o+60^o+70^o\right)120^o\)
Vậy chọn đáp án A
29 tháng 10 2022
a ) \(2\sqrt{45}+\sqrt{5}-3\sqrt{80}\)
= \(2\sqrt{9.5}+\sqrt{5}-3\sqrt{16.5}\) \
= \(2.3\sqrt{5}+\sqrt{5}-3.4\sqrt{5}\)
= \(6\sqrt{5}+\sqrt{5}-12\sqrt{5}\)
= \(\left(6+1-12\right)\sqrt{5}\)
= \(-5\sqrt{5}\)
b ) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\dfrac{2}{\sqrt{3}+1}-6\sqrt{\dfrac{16}{3}}\)
= / \(2-\sqrt{3}\) / \(+\dfrac{2.\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right).\left(\sqrt{3}-1\right)}-6\sqrt{\dfrac{48}{3^2}}\)
= \(2-\sqrt{3}+\dfrac{2.\left(\sqrt{3}-1\right)}{\sqrt{3}^2-1^2}-\dfrac{6}{3}\sqrt{48}\)
= \(2-\sqrt{3}+\dfrac{2.\left(\sqrt{3}-1\right)}{3-1}-2\sqrt{48}\)
=\(2-\sqrt{3}+\sqrt{3}-1-2\sqrt{16.3}\)
= \(2-\sqrt{3}+\sqrt{3}-1-8\sqrt{3}\)
= \(1-8\sqrt{3}\)
ý c ) em không biết làm ☹
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
a) \({\cos ^2}\frac{\pi }{8} + {\cos ^2}\frac{{3\pi }}{8} = {\cos ^2}\frac{\pi }{8} + {\cos ^2}\left( {\frac{\pi }{2} - \frac{\pi }{8}} \right) = {\cos ^2}\frac{\pi }{8} + {\sin ^2}\frac{\pi }{8} = 1\)
b)
\(\begin{array}{l}\tan {1^ \circ }.\tan {2^ \circ }.\tan {45^ \circ }.\tan {88^ \circ }.\tan {89^ \circ }\\ = (\tan {1^ \circ }.\tan {89^ \circ }).(\tan {2^ \circ }.\tan {88^ \circ }).\tan {45^ \circ }\\ = (\tan {1^ \circ }.\cot {1^ \circ }).(\tan {2^ \circ }.\cot {2^ \circ }).\tan {45^ \circ }\\ = 1\end{array}\)
H9
HT.Phong (9A5)
CTVHS
22 tháng 7 2023
Vì ABCD là hình thang cân
\(\Rightarrow\widehat{A}+\widehat{D}=180^o\)
Nên: \(\widehat{D}=180^o-\widehat{A}=180^o-65^o=115^o\)
Mặt khác ta có ABCD là hình thang cân nên:
\(\widehat{C}=\widehat{D}=115^o\)
Vậy chọn đáp án A
Ta có \(M=\dfrac{sin48^{\text{o}}}{cos42^{\text{o}}}-cos60^{\text{o}}+tan27^{\text{o}}.tan63^{\text{o}}\)
\(=\dfrac{sin48^{\text{o}}}{sin48^o}-cos60^{\text{o}}+tan27^o.cot27^o=1-cos60^{\text{o}}+1\)
\(=2-cos60^{\text{o}}=2-\dfrac{1}{2}=\dfrac{3}{2}\)
N = \(cot27^{\text{o}}.cot60^{\text{o}}.cot63^{\text{o}}+sin^244^{\text{o}}+sin^246^{\text{o}}\)
\(=cot27^{\text{o}}.tan27^{\text{o}}.cot60^{\text{o}}+sin^244^{\text{o}}+cos^244^{\text{o}}=cot60^{\text{o}}+1=\dfrac{1}{\sqrt{3}}+1\)
\(=\dfrac{\sqrt{3}+3}{3}\)
Ta có M=sin48ocos42o−cos60o+tan27o.tan63oM=cos42osin48o−cos60o+tan27o.tan63o
=sin48osin48o−cos60o+tan27o.cot27o=1−cos60o+1=sin48osin48o−cos60o+tan27o.cot27o=1−cos60o+1
=2−cos60o=2−12=32=2−cos60o=2−21=23
N = cot27o.cot60o.cot63o+sin244o+sin246ocot27o.cot60o.cot63o+sin244o+sin246o
=cot27o.tan27o.cot60o+sin244o+cos244o=cot60o+1=13+1=cot27o.tan27o.cot60o+sin244o+cos244o=cot60o+1=31+1
=3+33=33+3