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a, sin 27o = cos 63o
\(\Rightarrow\) \(\frac{sin27^o}{cos63^o}\) = 1
b, tan 18o = cot 72o
\(\Rightarrow\) tan 18o - cot 72o = 0
c, sin230o + cos230o = 1
d, tan 27o x cos 27o = sin 27o \(\approx\) 0,45
Chúc bn học tốt
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)sin a-sin a.cos^2 a=sin a(1-cos^2 a)=sin a(sin^2 a)=sin^3 a
b)sin^4a+cos^4a+2sin^2acos^2a=(sin^2a+cos^2a)^2=1^2=1
sin20<sin70
cos25 > cos65*15'
tan73*20' >tan45
cotg2 >cotg73*40'
tan25>sin25
cotg32 >cos32
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
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:
\(A=3\left(1-2sin^2a.cos^2a\right)-2\left(1-3sin^2a.cos^2a\right)=1\)
\(B=1-3sin^2.cos^2a+3sin^2a.cos^2a=1\)
ta có : \(C=sin^215+sin^235+sin^255+sin^275\)
\(=sin^215+sin^2\left(90-15\right)+sin^235+sin^2\left(90-35\right)\)
\(=sin^215+cos^215+sin^235+cos^235=1+1=2\)
a/ \(\left(1-cos\alpha\right)\left(1+cos\alpha\right)=1-cos^2\alpha=\left(sin^2\alpha+cos^2\alpha\right)-cos^2\alpha=sin^2\alpha\)
b/ \(1+sin^2\alpha+cos^2\alpha=1+1=2\)
c/ \(sin\alpha-sin\alpha.cos^2\alpha=sin\alpha\left(1-cos^2\alpha\right)=sin\alpha.sin^2\alpha=sin^3\alpha\)