Rut gon sin4x+sin5x+sin6x/cos4x+cos5x+cos6x Chứng minh tanx.tan3x=tan²2x-tanx/1-tan²2x.tan²x 09/11/2021 Bởi Skylar Rut gon sin4x+sin5x+sin6x/cos4x+cos5x+cos6x Chứng minh tanx.tan3x=tan²2x-tanx/1-tan²2x.tan²x
`D=\frac{sin4x+sin5x+sin6x}{cos4x+cos5x+cos6x}` `D=\frac{(sin4x+sin6x)+sin5x}{(cos4x+cos6x)+cos5x}` `D=\frac{2sin5xcosx+sin5x}{2cos5xcosx+cos5x}` `D=\frac{sin5x(2cosx+1)}{cos5x(2cosx+1)}` `D=\frac{sin5x}{cos5x}` `D=tan5x` `F=\frac{tan^{2}2x-tanx}{1-tan^{2}2x.tan^2x}` `F=\frac{(tan2x+tanx).(tan2x-x)}{(1-tan2x.tanx).(1+tan2x.tanx)}` `F=\tanx.tan3x` (ĐPCM) Bình luận
Giải thích các bước giải: 1) $\dfrac{sin4x + sin5x + sin6x}{cos4x + cos5x + cos6x}$ $= \dfrac{\left ( sin4x + sin6x \right ) + sin5x}{\left ( cos4x + cos6x \right ) + cos5x}$ $= \dfrac{2sin5xcosx + sin5x}{2cos5xcosx + cos5x}$ $= \dfrac{sin5x\left ( 2cosx + 1 \right )}{cos5x\left ( 2cosx + 1 \right )}$ $= \dfrac{sin5x}{cos5x}$ $= tan5x$ 2) $tanx.tan3x = \dfrac{\left ( tan2x + tanx \right )\left ( tan2x – tanx \right )}{\left ( 1 – tanx.tan2x \right )\left ( 1 + tanx.tanx \right )} = \dfrac{tan^{2}2x – tan^{2}x}{1 – tan^{2}2x.tan^{2}x}$ Bình luận
`D=\frac{sin4x+sin5x+sin6x}{cos4x+cos5x+cos6x}`
`D=\frac{(sin4x+sin6x)+sin5x}{(cos4x+cos6x)+cos5x}`
`D=\frac{2sin5xcosx+sin5x}{2cos5xcosx+cos5x}`
`D=\frac{sin5x(2cosx+1)}{cos5x(2cosx+1)}`
`D=\frac{sin5x}{cos5x}`
`D=tan5x`
`F=\frac{tan^{2}2x-tanx}{1-tan^{2}2x.tan^2x}`
`F=\frac{(tan2x+tanx).(tan2x-x)}{(1-tan2x.tanx).(1+tan2x.tanx)}`
`F=\tanx.tan3x` (ĐPCM)
Giải thích các bước giải:
1) $\dfrac{sin4x + sin5x + sin6x}{cos4x + cos5x + cos6x}$
$= \dfrac{\left ( sin4x + sin6x \right ) + sin5x}{\left ( cos4x + cos6x \right ) + cos5x}$
$= \dfrac{2sin5xcosx + sin5x}{2cos5xcosx + cos5x}$
$= \dfrac{sin5x\left ( 2cosx + 1 \right )}{cos5x\left ( 2cosx + 1 \right )}$
$= \dfrac{sin5x}{cos5x}$
$= tan5x$
2) $tanx.tan3x = \dfrac{\left ( tan2x + tanx \right )\left ( tan2x – tanx \right )}{\left ( 1 – tanx.tan2x \right )\left ( 1 + tanx.tanx \right )} = \dfrac{tan^{2}2x – tan^{2}x}{1 – tan^{2}2x.tan^{2}x}$