anh HCT cho em hỏi cái?

a) 1 + cos2x + cos4x = 1 + cos4x + cos2x = 2cos²2x + cos2x =

= cos2x(2cos2x + 1)

= cos2x(4cos²x - 1) = cos2x(2cosx - 1)(2cosx + 1)

b) tan30 + tan40 + tan50 + tan60 =

= tan30 + tan50 + tan40 + tan60

= sin80/cos30.cos50 + sin100/cos40.cos60

= sin80(1/cos30.cos50 + 1/cos40.cos60) =

sin80(cos60.cos40+cos50.cos30) /cos30cos60cos40cos50

sin80(cos20+cos100+cos20+cos80) /2√3/2.1/2.cos40cos50

= sin80(2cos20) / (√3/2).cos40.cos50

= 4.sin80.cos20 / √3.cos40.sin40

= 8.sin40.cos40.cos20 /√3cos40.sin40

= 8cos20 /√3

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ra đến đây cũng đủ thấy em ghi đề bị thiếu 1 cái tan30

minh` giai? o? duoi 'roi` do' ban co' gi` xem qua nhe'