Giải giúp mình với ạ

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Giải giúp mình với ạ
giai-giup-minh-voi-a

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Kiệt Gia 4 years 2021-05-02T17:23:17+00:00 2 Answers 52 views 0

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  1. mocmien2
    0
    2021-05-02T17:25:05+00:00
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    g) `tan4x-\frac{1}{cos4x}=\frac{sin4x}{cos4x}-\frac{1}{cos4x}`

    `=\frac{sin4x-1}{cos4x}=\frac{2sin2x.cos2x-1}{cos^{2}2x-sin^{2}2x}`

    `=\frac{-(1-2sin2x.cos2x)}{cos^{2}2x-sin^{2}2x}`

    `=\frac{-(sin^{2}2x+cos^{2}2x-2sin2x.cos2x)}{cos^{2}2x-sin^{2}2x}`

    `=\frac{-(sin2x-cos2x)^2}{(cos2x+sin2x)(cos2x-sin2x)}`

    `=\frac{sin2x-cos2x}{sin2x+cos2x} \ (dpcm)`

    h) `tan6x-tan4x-tan2x=\frac{sin6x}{cos6x}-\frac{sin4x}{cos4x}-\frac{sin2x}{cos2x}`

    `=\frac{sin6x.cos4x-sin4x.cos6x}{cos6x.cos4x}-\frac{sin2x}{cos2x}`

    `=\frac{\frac{1}{2}.(sin2x+sin10x)-\frac{1}{2}.(-sin2x+sin10x)}{cos6x.cos4x}-\frac{sin2x}{cos2x}`

    `=\frac{\frac{1}{2}sin2x+\frac{1}{2}sin2x}{cos6x.cos4x}-\frac{sin2x}{cos2x}`

    `=\frac{sin2x}{cos6x.cos4x}-\frac{sin2x}{cos2x}`

    `=sin2x.(\frac{1}{cos6x.cos4x}-\frac{1}{cos2x})`

    `=sin2x.\frac{cos2x-cos6x.cos4x}{cos6x.cos4x.cos2x}`

    `=\frac{sin2x}{cos2x}.\frac{cos2x-cos6x.cos4x-sin6x.sin4x+sin6x.sin4x}{cos6x.cos4x}`

    `=tan2x.\frac{cos2x-\frac{1}{2}.(cos2x+cos10x)-\frac{1}{2}.(cos2x-cos10x)+sin6x.sin4x}{cos4x.cos6x}`

    `=tan2x.\frac{cos2x-cos2x+sin6x.sin4x}{cos4x.cos6x}=tan2x.\frac{sin6x}{cos6x}.\frac{sin4x}{cos4x}`

    `=tan2x.tan4x.tan6x \ (dpcm)`

    i) `\frac{sin7x}{sinx}=1+2cos2x+2cos4x+2cos6x`

    `⇔ sin7x=sinx+2cos2xsinx+2cos4xsinx+2cos6xsinx`

    `⇔ sin7x=sinx+2.\frac{1}{2}.[sin(-x)+sin3x]+2.\frac{1}{2}.[sin(-3x)+sin5x]+2.\frac{1}{2}.[sin(-5x)+sin7x]`

    `⇔ sin7x=sin7x \ (dung) ⇒ dpcm`

  2. Acacia
    0
    2021-05-02T17:25:06+00:00
    Reply

    $\tan 4x – \dfrac{1}{{\cos 4x}} = \dfrac{{\sin 4x}}{{\cos 4x}} – \dfrac{1}{{\cos 4x}} = \dfrac{{2\sin 2x\cos 2x – {{\sin }^2}2x – {{\cos }^2}2x}}{{{{\cos }^2}2x – {{\sin }^2}2x}} = \dfrac{{ – {{\left( {\cos 2x – \sin 2x} \right)}^2}}}{{\left( {\cos 2x – \sin 2x} \right)\left( {\cos 2x + \sin 2x} \right)}} = \dfrac{{\sin 2x – \cos 2x}}{{\cos 2x + \sin 2x}}$

    $\begin{array}{l}
    \tan 6x = \tan \left( {4x + 2x} \right) \Leftrightarrow \tan 6x = \dfrac{{\tan 4x + \tan 2x}}{{1 – \tan 4x.\tan 2x}} = \tan 6x – \tan 2x.\tan 4x.\tan 6x = \tan 4x + \tan 2x\\
     \Leftrightarrow \tan 6x – \tan 4x – \tan 2x = \tan 2x.\tan 4x.\tan 6x
    \end{array}$

    $\begin{array}{l}
    \sin x\left( {1 + 2\cos 2x + 2\cos 4x + 2\cos 6x} \right) = \sin x + 2\sin x.\cos 2x + 2\sin x.\cos 4x + 2\sin x.\cos 6x = \sin x + \sin 3x + \sin ( – x) + \sin 5x + \sin ( – 3x) + \sin 7x + \sin ( – 5x)\\
     = \sin x + \sin 3x – \sin x + \sin 5x – \sin 3x + \sin 7x = \sin 7x \Rightarrow \dfrac{{\sin 7x}}{{\sin x}} = 1 + 2\cos 2x + 2\cos 4x + 2\cos 6x
    \end{array}$

     

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