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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 10: `CM:tan20^{o} + tan40^{o} + tan80^{o} – tan60^{o} = 8sin40^{o}`

Tháng Chín 14, 2022 Toán học 1 Comment 0 views

Toán Lớp 10: CM:tan20^{o} + tan40^{o} + tan80^{o} – tan60^{o} = 8sin40^{o}

Comments ( 1 )

  1. daithanh
    daithanh
    14 Tháng Chín, 2022 at 5:42 chiều
    Reply
    $VT=\tan20^o+\tan40^o+\tan80^o-\tan60^o$
    $=\dfrac{\sin20^o}{\cos20^o}+\dfrac{\sin40^o}{\cos40^o}+\dfrac{\sin80^o}{\cos80^o}-\dfrac{\sin60^o}{\cos60^o}$
    $=\dfrac{\sin20^o\cos40^o+\sin40^o\cos20^o}{ \cos20^o\cos40^o}+\dfrac{\sin80^o\cos60^o-\sin60^o\cos80^o}{\cos60^o\cos80^o}$
    $=\dfrac{\sin60^o}{\cos20^o\cos40^o}+\dfrac{\sin20^o}{\cos60^o\cos80^o}$
    $=\dfrac{\sin60^o(\cos60^o\cos80^o)+\sin20^o(\cos20^o\cos40^o) }{\cos20^o\cos40^o\cos60^o\cos80^o}$
    $=\dfrac{\dfrac{1}{2}\sin120^o\cos80^o+\dfrac{1}{4}\sin80^o}{ \cos20^o\cos40^o\cos60^o\cos80^o}$
    $=\dfrac{\dfrac{1}{4}(\sqrt3\cos80^o+\sin80^o) }{\cos20^o\cos40^o\cos60^o\cos80^o}$
    $=\dfrac{\dfrac{1}{2}\left( \cos80^o\cos 30^o+\sin80^o\sin30^o\right)}{\cos20^o\cos40^o\cos60^o\cos80^o}$
    $=\dfrac{\cos50^o}{2\cos20^o\cos40^o\cos60^o\cos80^o}$
    $=\dfrac{\sin20^o\cos50^o}{ \cos60^o.\dfrac{1}{4}\sin160^o}$
    $=\dfrac{\sin20^o\cos50^o}{\dfrac{1}{4}\cos60^o.\sin20^o}$
    $=\dfrac{\sin40^o}{\dfrac{1}{8}}$
    $=8\sin40^o$
    $=VP$ (đpcm)

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222-9+11+12:2*14+14 = ? ( )

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