1. Tính giá trị biểu thức:
    A = cos10.cos20.cos30. ... . cos70.cos80
2.Biến đổi các biểu thức thành dạng tích, thương các giá trị lượng giác.
    A= 2sin4x + $\sqrt{2}$                       B = 3 - 4cos^2 x
    C= 1-3tan^2 x                                    D = sin2x + sin4x + sin 6x
    E= 3 + 4cos4x + cos8x                       F = sin5x + sin6x + sin7x + sin8x
    G= 1 + sin2x - cos2x - tan2x     ; H = $sin^2(x+90)-3cos^2(x-90)$
    I = cos5x + cos8x + cos9x + cos12x          K = cosx + sinx + 1 
Nếu đáp án của mình đúng, bạn vui lòng nhấn V để chấp nhận VÀ nhấn mũi tên để vote up hộ mình. Cảm ơn :) –  NguyễnTốngKhánhLinh 04-08-13 08:29 PM
Có $sin(x+90)=cosx$
$cos(x-90)=cos(90-x)=sinx$
$H=cos^2x-3sin^2x$
$=cos^2x+sin^2x-4sin^2x$
$=1-4sin^2x$
$=(1+2sinx)(1-2sinx)$
$G=1+sin2x-cos2x-tan2x$

$=1-cos2x+sin2x-\frac{sin2x}{cos2x}$

$=1-cos2x+\frac{sin2xcos2x-sin2x}{cos2x}$

$=1-cos2x+\frac{sin2x(cos2x-1)}{cos2x}$

$=(1-cos2x)(1-\frac{sin2x}{cos2x})$

$=2sin^2x.(1-tan2x)$
$E=3+4cos4x+cos8x$
$=3+3cos4x+cos4x+cos8x$
$=3(cos4x+1)+2cos6x.cos2x$
$=3.2cos^22x+2cos6x.cos2x$
$=2cos2x(3cos2x+cos6x)$
$=2cos2x.4cos^32x$
$=8cos^42x$
$D=sin2x+sin6x+sin4x$

$=2sin4x.cos2x+sin4x$

$=sin4x(2cos2x+1)$

$C= 1-3tan^2x=4-3(1+tan^2x)$

$=4-\frac3{cos^2x}$

$=\frac{4cos^2x-3}{cos^2x}$

$=\frac{-4sin^2x+1}{cos^2x}$

$B=3-4cos^2x$
$=4-4cos^2x-1$
$=4sin^2x-1$
$=(2sinx-1)(2sinx+1)$

$A=2sin4x+\sqrt2$
$=2(sin4x+\frac{\sqrt2}2)$
$=2(sin4x+sin\frac{\pi}4)$
$=4sin\frac{4x+\frac{\pi}4}2.cos{\frac{4x-\frac{\pi}4}2}$
$F = sin5x + sin6x + sin7x + sin8x$

$= 2\sin \dfrac{13x}{2}\cos \dfrac{3x}{2} + 2\sin \dfrac{13x}{2}\cos \dfrac{x}{2}$

$=2\sin \dfrac{13x}{2} (\cos \dfrac{3x}{2} + \cos \dfrac{x}{2})$

 $=4\sin \dfrac{13x}{2} \cos x \cos \dfrac{x}{2}$

có mấy cái kia tương tự, dễ quá ngại làm
1)
Ta tính $\cos 10 \cos 30 \cos 50 \cos 70$

$= \cos 10 \dfrac{\sqrt 3}{2} \cos 50 \cos 70$


$=\dfrac{\sqrt 3}{4}\cos 10 (2\cos 50 \cos 70)$


$=\dfrac{\sqrt 3}{4}\cos 10(\cos 120 + \cos 20)$


$=\dfrac{\sqrt 3}{4}\cos 10( -\dfrac{1}{2} + \cos 20)$


$=\dfrac{\sqrt 3}{8}(-\cos 10 + 2\cos 20 \cos 10)$


$=\dfrac{\sqrt 3}{8}(-\cos 10 + \cos 30 + \cos 10)$


$=\dfrac{\sqrt 3}{8}\cos 30= \dfrac{\sqrt 3}{8} \dfrac{\sqrt 3}{2} = \dfrac{3}{16}$


Cái còn lại làm tương tự $\cos 20 \cos 40 \cos 60 \cos 80$ ra $\dfrac{1}{16}$


Đáp số $= \dfrac{3}{256}$

Bạn cần đăng nhập để có thể gửi đáp án

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