3 replies to this topic

[thanhhuyen98]

Rút gọn biểu thức:

A=sin(a-\frac{\pi }{4})cos(\frac{\pi }{3}-a)-cos(\frac{\pi }{4}-a)sin(\frac{\pi }{3}-a)

B=cos(\frac{\pi }{3}-a)cos(a+\frac{\pi }{4})+cos(a+\frac{\pi }{6})cos(a+\frac{3\pi }{4}

 

[Dahitotn94]

$A=sin(a-\frac{\pi }{4})cos(\frac{\pi }{3}-a)-cos(\frac{\pi }{4}-a)sin(\frac{\pi }{3}-a)$

 

 $\Leftrightarrow A=sin(a-\frac{\pi }{4})cos(\frac{\pi }{3}-a)+cos(a-\frac{\pi }{4})sin(\frac{\pi }{3}-a)$

 

$=sin\left [ (a-\frac{\pi }{4})+(\frac{\pi }{3})-a \right ]$

 

$=sin(2a-\frac{\pi }{12})$

Edited by Dahitotn94, 11-02-2014 - 02:29.

[Dahitotn94]

 

Rút gọn biểu thức:

A=sin(a-\frac{\pi }{4})cos(\frac{\pi }{3}-a)-cos(\frac{\pi }{4}-a)sin(\frac{\pi }{3}-a)

B=cos(\frac{\pi }{3}-a)cos(a+\frac{\pi }{4})+cos(a+\frac{\pi }{6})cos(a+\frac{3\pi }{4}

 

B$=cos(\frac{\pi }{3}-a)cos(a+\frac{\pi }{4})+cos(a+\frac{\pi }{6})cos(a+\frac{3\pi }{4}$

Edited by Dahitotn94, 11-02-2014 - 02:12.

[Dahitotn94]

$B=cos(\frac{\pi }{3}-a)cos(a+\frac{\pi }{4})+cos(a+\frac{\pi }{6})cos(a+\frac{3\pi }{4})$

 

$=sin[\frac{\pi }{2}-(\frac{\pi }{3}-a)]cos(a+\frac{\pi }{4})+cos(a+\frac{\pi }{6})sin[\frac{\pi }{2}-(a+\frac{3\pi }{4})]$

 

$=sin(a+\frac{\pi }{6})cos(a+\frac{\pi }{4})-cos(a+\frac{\pi }{6})sin(a+\frac{\pi }{4})$

 

$=sin[(a+\frac{\pi }{6})-(a+\frac{\pi }{4})]$

 

$=sin\frac{5\pi }{12}$

Edited by Dahitotn94, 11-02-2014 - 02:25.