Question

Să se arate că are loc identitatea:


(ctg10° - ctg80°)cos70° = 2sin110°


Mulțumesc mult !

Answer 1

sin110=sin(180-70)= sin70
patanteza din primul membru
(cos10sin80-sin10cos80)/sin10sin70=sin(80-10)/[cos(80+10)+cos(80-10)]/2= 2sin70/cos70
deci avem că:
2sin70*cos70/cos70=2sin70
relație adevărată!

Answer 2

Explicație pas cu pas:

Prelucram membrul stang si folosim formulele:

• $ctg~x=\frac{cos~x}{sin~x}$
• $sin~(x-y)=sin~x*cos~y-sin~y*cos~x$
• $sin~x*sin~y=\frac{1}{2}*[cos~(x-y)-cos~(x+y)]$
• $cos~90^{\circ}=0$
• $cos~(-x)=cos~x$

$(ctg10^{\circ}-ctg80^{\circ})*cos70^{\circ}=\\=(\frac{cos10^{\circ}}{sin10^{\circ}}-\frac{cos80^{\circ}}{sin80^{\circ}})*cos70^{\circ}=\\=(\frac{cos10^{\circ}*sin80^{\circ}}{sin10^{\circ}*sin80^{\circ}}-\frac{cos80^{\circ}*sin10^{\circ}}{sin10^{\circ}*sin80^{\circ}})*cos70^{\circ}=\\=\frac{sin80^{\circ}*cos10^{\circ}-cos80^{\circ}*sin10^{\circ}}{sin10^{\circ}*sin80^{\circ}}*cos70^{\circ}=\\=\frac{sin(80^{\circ}-10^{\circ})}{sin10^{\circ}*sin80^{\circ}}*cos70^{\circ}=$

$=\frac{sin70^{\circ}*cos70^{\circ}}{sin10^{\circ}*sin80^{\circ}}=\\=\frac{sin70^{\circ}*cos70^{\circ}}{\frac{1}{2}*[cos(10^{\circ}-80^{\circ})-cos(10^{\circ}+80^{\circ})]}=\\=\frac{2*sin70^{\circ}*cos70^{\circ}}{cos(-70^{\circ})-cos90^{\circ}}=\\=\frac{2*sin70^{\circ}*cos70^{\circ}}{cos70^{\circ}-0}=\\=\frac{2*sin70^{\circ}*cos70^{\circ}}{cos70^{\circ}}=\\=2sin70^{\circ}~(relatia~1)$

Prelucram membrul drept si folosim formula:

• $sin~(180^{\circ}-x)=sin~x$

$2*sin110^{\circ}=2*sin(180^{\circ}-110^{\circ})=2*sin70^{\circ}~(relatia~2)$

Obervam ca relatia 1 coincide cu relatia 2 si, deci, are loc egalitatea:

$(ctg10^{\circ}-ctg80^{\circ})*cos70^{\circ}=2*sin110^{\circ}$