Cho sin18 = \(\frac{\sqrt{5}-1}{4}\). Tính cos18, sin72, sin162, cos162, sin108, cos108, tan72
MK
Những câu hỏi liên quan
Chứng minh \(sin18=\frac{\sqrt{5}-1}{4}\)
CMR: \(\sin18^0=\frac{\sqrt{5}-1}{4}\)
Ta có: \(\sin18^0\approx0,3090169944\)
\(\frac{\sqrt{5}-1}{4}\approx0,3090169944\)
\(\Rightarrow\)\(\sin18^0=\frac{\sqrt{5}-1}{4}\)
1a) 6sqrt{3} - 5sqrt{12} + 3sqrt{75} b) 2sqrt{5} - dfrac{1}{4} sqrt{80} + 7sqrt{500} c) dfrac{sin43^o}{cos47^o} + tan45o d) dfrac{tan32^o}{tan68^o} - cos30o - dfrac{sin18^o}{sin82^o}
Đọc tiếp
1a) 6\(\sqrt{3}\) - 5\(\sqrt{12}\) + 3\(\sqrt{75}\)
b) 2\(\sqrt{5}\) - \(\dfrac{1}{4}\) \(\sqrt{80}\) + 7\(\sqrt{500}\)
c) \(\dfrac{sin43^o}{cos47^o}\) + tan45o
d) \(\dfrac{tan32^o}{tan68^o}\) - cos30o - \(\dfrac{sin18^o}{sin82^o}\)
\(a,=6\sqrt{3}-10\sqrt{3}+15\sqrt{3}=11\sqrt{3}\\ b,=2\sqrt{5}-\sqrt{5}+70\sqrt{5}=71\sqrt{5}\\ c,=\dfrac{\sin43^0}{\sin43^0}+1=1+1=2\\ d,Sửa:\dfrac{\tan32^0}{\cot68^0}-\cos30^0-\dfrac{\sin18^0}{\sin82^0}=\dfrac{\tan32^0}{\tan32^0}-\dfrac{\sqrt{3}}{2}-\dfrac{\sin18^0}{\cos18^0}=1-1-\dfrac{\sqrt{3}}{2}=-\dfrac{\sqrt{3}}{2}\)
Đúng 2
Bình luận (0)
cm : t\(\tan36.tan72=\sqrt{5}\)
Tính
A=\(\left(\cos30.cos18\right)^2+\frac{\tan\left(13\right)}{cot\left(77\right)}+\frac{3}{4}.cos^272\)
\(\frac{\sin^390^o-\cot^330^o-\cos^245^o+\tan20^o}{2\sqrt{7}+\sin108^o\cos32^o\tan64^o}\)
Tính \(\frac{2\sqrt{3}-4}{\sqrt{3}-1}+\frac{2\sqrt{2}-1}{\sqrt{2}-1}-\frac{1+\sqrt{6}}{\sqrt{2}+3}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+2\sqrt{12}}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-2\sqrt{75}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)
\(C=\sqrt{4+5}\)
\(C=3\)
Đúng 1
Bình luận (0)
tính:\(\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-\frac{1}{\sqrt{4}-\sqrt{5}}+\frac{1}{\sqrt{5}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{9}}\)
\(K=\frac{\sqrt{4+\sqrt{5}}+\sqrt{4-\sqrt{5}}}{\sqrt{4+\sqrt{11}}}-\sqrt{9-4\sqrt{5}}-\frac{4}{\sqrt{5}-1}\)
TÍNH