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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 : \(sin^210-sin^220+sin^230-sin^240-sin^250-sin^270+sin^280\)
\(=sin^210+sin^280-sin^220-sin^270-sin^240-sin^250+sin^230\) \(=sin^210+sin^2\left(90-10\right)-sin^220-sin^2\left(90-20\right)-sin^240-sin^2\left(90-40\right)+sin^230\) \(=sin^210+cos^210-sin^220-cos^220-sin^240-cos^240+sin^230\) \(=1-1-1+\dfrac{1}{4}=\dfrac{-3}{4}\)
a) Ta có : sin\(^2\)12o=cos278o=> sin212o+sin278o=1.
tương tự => A=3
b) tương tự câu (a) ta có: cos215o=sin275o ( do 15+75=90 nha bạn ) => cos215o+cos275o=1. Tương tự => B=0
sin20<sin70
cos25 > cos65*15'
tan73*20' >tan45
cotg2 >cotg73*40'
tan25>sin25
cotg32 >cos32
\(A=sin23^0-cos67^0=cos67^0-cos67^0=0\)
Vậy ...
\(B=\dfrac{tan70^0.tan45^0.tan20^0}{cos70^0.cos45^0.cos20^0}\)
\(\Leftrightarrow B=\dfrac{tan70^0.tan45^0.tan20^0}{tan70^0.cos45^0.tan20^0}=1\)
Vậy ...
Ta có:
\(sin^4a+cos^4a=\left(sin^2a+cos^2a\right)^2-2sin^2acos^2a=1-2sin^2a.cos^2a\)
Và:
\(sin^6a+cos^6a=\left(sin^2a+cos^2a\right)^3-3sin^2a.cos^2a.\left(sin^2a+cos^2a\right)\)
\(=1-3sin^2a.cos^2a\)
Do đó:
\(A=3\left(1-2sin^2a.cos^2a\right)-2\left(1-3sin^2a.cos^2a\right)=1\)
\(B=1-3sin^2.cos^2a+3sin^2a.cos^2a=1\)
2. \(\left(\sin a+\cos a\right)^2+\left(\sin a-\cos a\right)^2+2\)
\(=\sin^2a+2.\sin a.\cos a+\cos^2a+\sin^2a\cdot2.\sin a.\cos a+\cos^2a+2\)
\(=2\sin^2a+2\cos^2a+2\)
\(=2\left(\sin^2a+\cos^2a\right)+2\)
\(=2.1+2=4\)
=> biểu thức trên ko phụ thuộc vào a
1. a.) \(\cot a=\dfrac{1}{\tan a}=\dfrac{1}{\sqrt{3}}\)
\(\tan\sqrt{3}=60\Rightarrow a=60^o\)
\(\sin60=\dfrac{\sqrt{3}}{2}\)
\(\cos60=\dfrac{1}{2}\)
b.) \(\cos^2a=1-\left(\dfrac{15}{17}\right)^2=\dfrac{64}{289}\Rightarrow\cos a=\dfrac{8}{17}\)
\(\tan a=\dfrac{\sin a}{\cos a}=\dfrac{\dfrac{15}{17}}{\dfrac{8}{17}}=\dfrac{15}{17}.\dfrac{17}{8}=\dfrac{15}{8}\)
ta có : \(C=sin^215+sin^235+sin^255+sin^275\)
\(=sin^215+sin^2\left(90-15\right)+sin^235+sin^2\left(90-35\right)\)
\(=sin^215+cos^215+sin^235+cos^235=1+1=2\)
cảm ơn bn Mysterious Person nha!