sin \(32^o\) - cos \(58^o\) + 5 (\(sin^210^o\) + \(sin^280^o\)) + 2 . \(\dfrac{tan43^o}{cot47^o}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\cos^215+\cos^225+\cos^235+\cos^245+\sin^235+\sin^225+\sin^215\)
=\(\left(\cos^215+\sin^215\right)+\left(\cos^225+\sin^225\right)+\left(\cos^235+\sin^235\right)+\cos^245\)
=\(1+1+1+\frac{1}{2}=\frac{7}{2}\)
b.\(\sin^210-\sin^220-\sin^230-\sin^240-\cos^240-\cos^220+\cos^210\)
=\(\left(\sin^210+\cos^210\right)-\left(\sin^220+\cos^220\right)-\left(\sin^240+\cos^240\right)-\sin^230\)
=\(1-1-1-\frac{1}{4}=-\frac{5}{4}\)
c,\(\sin15+\sin75-\sin75-\cos15+\sin30=\sin30=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 : \(sin^210-sin^220+sin^230-sin^240-sin^250-sin^270+sin^280\)
\(=sin^210+sin^280-sin^220-sin^270-sin^240-sin^250+sin^230\) \(=sin^210+sin^2\left(90-10\right)-sin^220-sin^2\left(90-20\right)-sin^240-sin^2\left(90-40\right)+sin^230\) \(=sin^210+cos^210-sin^220-cos^220-sin^240-cos^240+sin^230\) \(=1-1-1+\dfrac{1}{4}=\dfrac{-3}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
Mà\(\left\{{}\begin{matrix}sin32^o=cos58^o\\sin^210^o=cos^280^o\\tan43^o=cot47^0\end{matrix}\right.\)
=> sin 32o - cos 58o + 5(sin210o + sin280o) + 2.\(\dfrac{tan43^o}{cot47^o}\)
= cos 58o - cos 58o + 5(cos280o + sin280o) + 2. \(\dfrac{cot47^o}{cot47^o}\)
Mà cos280o + sin280o = 1
=> cos 58o - cos 58o + 5(cos280o + sin280o) + 2. \(\dfrac{cot47^o}{cot47^o}\)
= 5.1 + 2.1
= 5 + 2
= 7
Vậy sin 32o - cos 58o + 5(sin210o + sin280o) + 2.\(\dfrac{tan43^o}{cot47^o}\) = 7
có cách nào làm khác không ạ??