Biến đổi tổng thành tích biểu thức A= sin²2x + sin²3x + sin²4x + sin²5x – 2 Giúp mình với 03/07/2021 Bởi Arianna Biến đổi tổng thành tích biểu thức A= sin²2x + sin²3x + sin²4x + sin²5x – 2 Giúp mình với
$\begin{array}{l} {\sin ^2}2x + {\sin ^2}3x + {\sin ^2}4x + {\sin ^2}5x – 2\\ = \dfrac{{1 – \cos 4x}}{2} + \dfrac{{1 – \cos 6x}}{2} + \dfrac{{1 – \cos 8x}}{2} + \dfrac{{1 – \cos 10x}}{2} – 2\\ = \dfrac{{ – \left( {\cos 4x + \cos 6x + \cos 8x + \cos 10x} \right) + 4}}{2} – 2\\ = – \left( {\cos 4x + \cos 6x + \cos 8x + \cos 10x} \right) + 2 – 2\\ = – \left[ {\left( {\cos 4x + 10x} \right) + \left( {\cos 6x + \cos 8x} \right)} \right]\\ = – \left[ {2\cos 7x\cos 3x + 2\cos 7x\cos x} \right]\\ = – 2\cos 7x\left( {\cos 3x + \cos x} \right)\\ = – 2\cos 7x.2\cos 2x.\cos x\\ = – 4\cos x\cos 2x\cos 7x \end{array}$ Bình luận
Đáp án: $\sin^22x+\sin623x+\sin^24x+\sin^25x-2=-2\cos x.\cos2x.\cos7x$ Giải thích các bước giải: $\sin^22x+\sin623x+\sin^24x+\sin^25x-2$ $=\dfrac{1-\cos4x}{2}+\dfrac{1-\cos6x}{2}+\dfrac{1-\cos8x}{2}+\dfrac{1-\cos10x}{2}-2$ $=\dfrac{4-\cos4x-\cos6x-\cos8x-\cos10x}{2}-2$ $=2-\dfrac{\cos4x+\cos6x+\cos8x+\cos10x}{2}-2$ $=-\dfrac{\cos4x+\cos6x+\cos8x+\cos10x}{2}$ $=-\dfrac{(\cos4x+\cos10x)+(\cos6x+\cos8x)}{2}$ $=-\dfrac{2\cos7x.\cos3x+2\cos7x.\cos x}{2}$ $=-\cos7x(\cos3x+\cos x)$ $=-\cos7x.2.\cos2x.\cos x$ $=-2\cos x.\cos2x.\cos7x$. Bình luận
$\begin{array}{l} {\sin ^2}2x + {\sin ^2}3x + {\sin ^2}4x + {\sin ^2}5x – 2\\ = \dfrac{{1 – \cos 4x}}{2} + \dfrac{{1 – \cos 6x}}{2} + \dfrac{{1 – \cos 8x}}{2} + \dfrac{{1 – \cos 10x}}{2} – 2\\ = \dfrac{{ – \left( {\cos 4x + \cos 6x + \cos 8x + \cos 10x} \right) + 4}}{2} – 2\\ = – \left( {\cos 4x + \cos 6x + \cos 8x + \cos 10x} \right) + 2 – 2\\ = – \left[ {\left( {\cos 4x + 10x} \right) + \left( {\cos 6x + \cos 8x} \right)} \right]\\ = – \left[ {2\cos 7x\cos 3x + 2\cos 7x\cos x} \right]\\ = – 2\cos 7x\left( {\cos 3x + \cos x} \right)\\ = – 2\cos 7x.2\cos 2x.\cos x\\ = – 4\cos x\cos 2x\cos 7x \end{array}$
Đáp án:
$\sin^22x+\sin623x+\sin^24x+\sin^25x-2=-2\cos x.\cos2x.\cos7x$
Giải thích các bước giải:
$\sin^22x+\sin623x+\sin^24x+\sin^25x-2$
$=\dfrac{1-\cos4x}{2}+\dfrac{1-\cos6x}{2}+\dfrac{1-\cos8x}{2}+\dfrac{1-\cos10x}{2}-2$
$=\dfrac{4-\cos4x-\cos6x-\cos8x-\cos10x}{2}-2$
$=2-\dfrac{\cos4x+\cos6x+\cos8x+\cos10x}{2}-2$
$=-\dfrac{\cos4x+\cos6x+\cos8x+\cos10x}{2}$
$=-\dfrac{(\cos4x+\cos10x)+(\cos6x+\cos8x)}{2}$
$=-\dfrac{2\cos7x.\cos3x+2\cos7x.\cos x}{2}$
$=-\cos7x(\cos3x+\cos x)$
$=-\cos7x.2.\cos2x.\cos x$
$=-2\cos x.\cos2x.\cos7x$.