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a) Vì 20 ° < 70 ° n ê n sin 20 ° < sin 70 ° (góc tăng, sin tăng)
b) Vì 25 ° < 63 ° 15 ' n ê n cos 25 ° > cos 63 ° 15 ' (góc tăng, cos giảm)
c) Vì 73 ° 20 ' > 45 ° n ê n t g 73 ° 20 ' > t g 45 ° (góc tăng, tg tăng)
d) Vì 2 ° < 37 ° 40 ' n ê n c o t g 2 ° > c o t g 37 ° 40 ' (góc tăng, cotg giảm )
a, cos220o + cos240o + cos250o + cos270o
= (cos220o + cos270o) + (cos240o + cos250o)
= (cos220o + sin220o) + (cos240o + sin240o)
= 1 + 1 = 2
Mình nghĩ chắc sin285o là sin255o
b, sin225o + sin245o + sin265o + sin255o
= (sin225o + sin265o) + (sin245o + sin255o)
= (sin225o + cos225o) + (sin245o + cos245o)
= 1 + 1 = 2
Chúc bn học tốt!
\(B=tan^267^0-cot^223^0+2\cdot\left(sin^216^0+cos^216^0\right)-2\)
\(=0+2\cdot1-2=0\)
\(A=cot67\cdot tan67-2\left(\dfrac{\sqrt{2}}{2}\cdot sin64\right)^2-2\cdot\dfrac{sin23}{3\cdot sin23}-sin^226^0\)
\(=1-2\cdot\dfrac{1}{2}\cdot sin^264^0-\dfrac{2}{3}-sin^226^0\)
\(=1-1-\dfrac{2}{3}=-\dfrac{2}{3}\)
Câu 1:
Ta có: \(\cos\left(90^0-\alpha\right)=\sin\alpha\)
\(\Leftrightarrow\sin\alpha=1:\sqrt{\dfrac{1^2+2^2}{1}}=1:\sqrt{5}=\dfrac{\sqrt{5}}{5}\)
Câu 2:
a) \(\cos\alpha=\sqrt{1-\sin^2\alpha}=\sqrt{1-\dfrac{16}{25}}=\dfrac{3}{5}\)
\(\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}\)
c: \(\cot50^0>\cos50^0>\cos70^0\)
a: \(\tan40^0>\cos40^0>\cos60^0\)
b: \(\cot70^0=\tan20^0>\sin20^0>\sin10^0\)
a: \(=\left(\cos^215^0+\cos^275^0\right)+\left(\cos^225^0+\cos^265^0\right)+\left(\cos^235^0+\cos^255^0\right)+\cos^245^0\)
=1+1+1+1/2
=3,5
b: \(=\left(\sin^210^0+\sin^280^0\right)-\left(\sin^220^0+\sin^270^0\right)+\left(\sin^230^0\right)-\left(\sin^240^0+\sin^250^0\right)\)
=1-1-1+1/4
=-1+1/4=-3/4
c: \(=\left(\sin15^0-\cos75^0\right)+\left(\sin75^0-\cos15^0\right)+\sin30^0\)
=1/2
mk bỏ dấu độ nha . trong toán người ta cho phép
a) ta có : \(cos^215+cos^225+cos^235+cos^245+cos^255+cos^265+cos^275\)
\(=cos^215+cos^275+cos^225+cos^265+cos^235+cos^255+cos^245\) \(=cos^215+cos^2\left(90-15\right)+cos^225+cos^2\left(90-25\right)+cos^235+cos^2\left(90-35\right)+cos^245\) \(=cos^215+sin^215+cos^225+sin^225+cos^235+sin^235+cos^245\)\(=1+1+1+\dfrac{1}{2}=\dfrac{7}{2}\)
b) ta có : \(sin^210-sin^220+sin^230-sin^240-sin^250-sin^270+sin^280\)
\(=sin^210+sin^280-sin^220-sin^270-sin^240-sin^250+sin^230\) \(=sin^210+sin^2\left(90-10\right)-sin^220-sin^2\left(90-20\right)-sin^240-sin^2\left(90-40\right)+sin^230\) \(=sin^210+cos^210-sin^220-cos^220-sin^240-cos^240+sin^230\) \(=1-1-1+\dfrac{1}{4}=\dfrac{-3}{4}\)
sin20<sin70
cos25 > cos65*15'
tan73*20' >tan45
cotg2 >cotg73*40'
tan25>sin25
cotg32 >cos32