Bài tập 1.56 trang 28 SBT toán 11 tập 1 kết nối: Chứng minh các biểu thức sau không phụ thuộc vào x
a) $A = sin(\frac{\pi}{4}+x)-cos(\frac{\pi}{4}-x)$
b) $B=cos(\frac{\pi}{6}-x)-sin(\frac{\pi}{3}+x)$
c) $C=sin^{2}x+cos(\frac{\pi}{3}-x)cos(\frac{\pi}{3}+x)$
d) $D=\frac{1-cos2x+sin2x}{1+cos2x+sin2x}.cotx$
Bài Làm:
a) $A = sin(\frac{\pi}{4}+x)-cos(\frac{\pi}{4}-x)$
$=sin(\frac{\pi}{4}+x)-sin[\frac{\pi}{2}-(\frac{\pi}{4}-x)]$
$=sin(\frac{\pi}{4}+x)-sin(\frac{\pi}{4}+x=0$ $\forall x$
b) $B=cos(\frac{\pi}{6}-x)-sin(\frac{\pi}{3}+x)$
$=cos(\frac{\pi}{6}-x)-cos[\frac{\pi}{2}-(\frac{\pi}{3}+x)]$
$=cos(\frac{\pi}{6}-x)-cos(\frac{\pi}{6}-x)=0$ $\forall x$
c) $C=sin^{2}x+cos(\frac{\pi}{3}-x)cos(\frac{\pi}{3}+x)$
$=sin^{2}+\frac{1}{2}[cos(\frac{\pi}{3}-x+\frac{\pi}{3}+x)+cos(\frac{\pi}{3}-x-\frac{\pi}{3}-x)]$
$=sin^{2}+\frac{1}{2}(cos\frac{2\pi}{3}+cos(-2x))=sin^{2}x+\frac{1}{2}(-\frac{1}{2}+cos(-2x))$
$=sin^{2}+\frac{1}{2}[-\frac{1}{2}+(1-2sin^{2})]$
$=sin^{2}x+\frac{1}{2}(\frac{1}{2}-2sin^{2}x)$
$=sin^{2}x+\frac{1}{4}-sin^{2}x=\frac{1}{4}$ $\forall x$
d) $D=\frac{1-cos2x+sin2x}{1+cos2x+sin2x}.cotx$
$=\frac{1-(1-2sin^{2}x)+2sinxcosx}{1+(2cos^{2}x-1)+2sinxcosx}.cotx$
$=\frac{2sin^{2}x+2sinxcosx}{2cos^{2}x+2sinxcosx}cotx$
$=\frac{2sinx(sinx+cosx)}{2cosx(cosx+sinx)}cotx$
$=\frac{sinx}{cosx}.cotx=tanx.cotx=1$ $\forall x$
