Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ta có : \(M=2cot37.cot53+sin^228\dfrac{3tan54}{cot36}+sin^262\)
\(=2.cot37.cot\left(90-37\right)+sin^228\dfrac{3tan54}{cot\left(90-54\right)}+sin^262\)
\(=2.cot37.tan37+sin^228\dfrac{3tan54}{tan54}+sin^262\)\(=2+3sin^228+sin^262=2+2sin^228+sin^228+sin^2\left(90-28\right)\)
\(=2+2sin^228+sin^228+cos^228=3+2sin^228\)
A=(sin220°+sin270°)+(sin230°+sin260°)
+(sin240°+sin250°)-tan245°
=(sin220°+cos220°)+(sin230°+cos230°)+(sin240°+cos240°)-1
=1+1+1-1=2
Ta có:
\(C=sin^22^0+sin^24^0+...+sin^288^0\)
\(C=\left(sin^22^0+sin^288^0\right)+\left(sin^24^0+sin^286^0\right)+...+\left(sin^244^0+sin^246^0\right)\)
\(C=\left(sin^22^0+cos^22^0\right)+\left(sin^24^0+cos^24^0\right)+...+\left(sin^244^0+cos^244^0\right)\)
\(C=1+1+...+1\) \(C=22\)
A=(sin210+sin280)+(sin220+sin70)+(sin230+sin260)+(sin240+sin250)
Lại có: sin80=cos10; sin70=cos20; sin60=cos30; sin50=cos40
=> sin280=cos210; sin270=cos220; sin260=cos230; sin250=cos240
=>A=(sin210+cos210)+(sin220+cos220)+(sin230+cos230)+(sin240+cos240)
=>A=1+1+1+1=4
\(=2\cdot sin53^0\cdot cos53^0+1-3=sin106^0-2\)