[Toán 10] Lượng giác

M

min.hb

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L

lan_phuong_000

1. $VT = cos^2x + cos^2(\dfrac{2\pi}{3} + x) + cos^2(\dfrac{2\pi}{3} - x)$
$= \dfrac{1}{2}.(cos2x + 1 + cos(\dfrac{4\pi}{3} + 2x) + 1 + cos(\dfrac{2\pi}{3} - x) + 1$
$= \dfrac{3}{2} + \dfrac{1}{2}.(cos2x + cos\dfrac{4\pi}{3}.cos2x - sin\dfrac{4\pi}{3}.sin2x + cos\dfrac{4\pi}{3}.cos2x + sin\dfrac{4\pi}{3}.sin2x$
$= \dfrac{3}{2} + \dfrac{1}{2}.(cos2x - 2.cos\dfrac{\pi}{3}.cos2x)$
$= \dfrac{3}{2} + \dfrac{1}{2}.cos2x.(1 - 1)$
$= \dfrac{3}{2}$ (đpcm)
 
N

nguyenbahiep1

Chứng minh
1. [TEX]cos^2 x + cos^2 (\frac{2\pi}{3} + x) + cos^2 (\frac{2\pi}{3} - x) = \frac{3}{2}[/TEX]

[laTEX]VT = cos^2x + (-\frac{1}{2}cosx - \frac{\sqrt{3}}{2}sinx)^2 + (-\frac{1}{2}cosx - \frac{\sqrt{3}}{2}sinx)^2 \\ \\ cos^2x + \frac{1}{4}( cos^2x+cos^2x +3(sin^2x+sin^2x) ) \\ \\ cos^2x + \frac{cos^2x}{2} + \frac{3sin^2x}{2} = \frac{3}{2}(sin^2x+cos^2x) = \frac{3}{2} \Rightarrow dpcm[/laTEX]
 
L

lan_phuong_000

$ cos2A + cos2B + cos2C + 4cosAcosBcosC +1 $
$= 2.cos(A+B).cos(A-B) + cos2C + 4cosAcosBcosC +1$
$= 2.cos(\pi - C).cos(A-B) + 2.cos^C + 4cosAcosBcosC$
$= -2cosC(cos(A-B) - cosC) + 4cosAcosBcosC$
$= -2.cosC.(cos(A-B) + cos(A+B)) + 4cosAcosBcosC$
$= -4.cosC.cosB.cosc + 4cosAcosBcosC$
=0 (đpcm)
 
N

nguyenbahiep1

min.hb said:
Chứng minh:

[TEX] sin^8 x + cos^8 x = \frac{1}{64} (35+28cos4x+cos8x)[/TEX]
Bấm để xem đầy đủ nội dung ...

[laTEX](cos^4x-sin^4x)^2 + 2sin^4x.cos^4x \\ \\ cos^22x + \frac{1}{8}sin^42x = \frac{1+cos4x}{2} + \frac{(1-cos4x)^2}{32} \\ \\ \frac{1}{32}( 16 + 16cos4x + 1- 2cos4x + cos^24x) \\ \\ \frac{1}{32}( 17 + 14cos4x + \frac{1+cos8x}{2}) \\ \\ \frac{1}{64} (35+28cos4x+cos8x)[/laTEX]
 
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