+0

# http://prntscr.com/6le9v0

0
679
2

http://prntscr.com/6le9v0

Guest Mar 25, 2015

#2
+94106
+10

Mmm,

first  $$\frac{\alpha}{2}$$   is between  45 and 67.5 degrees.   so that is the first quadrant.   All trig ratios are positive

so   $$\alpha$$    must be between 90 and 135 degrees so that is the second quadrant.  Only sine and cosec are positive all other ratios will be negative,

Now

$$\\cos \left(\frac{\alpha}{2}\right)=\frac{5}{8}\\\\ cos^2 \left(\frac{\alpha}{2}\right)=\frac{25}{64}\\\\ cos^2 \left(\frac{\alpha}{2}\right)+sin^2 \left(\frac{\alpha}{2}\right)=1}\\\\ \frac{25}{64}+sin^2 \left(\frac{\alpha}{2}\right)=1}\\\\ sin^2 \left(\frac{\alpha}{2}\right)=1-\frac{25}{64}\\\\ sin^2 \left(\frac{\alpha}{2}\right)=\frac{39}{64}\\\\ sin \left(\frac{\alpha}{2}\right)=\frac{\sqrt{39}}{8}\\\\\\$$

$$\\sin(\alpha) = sin(\frac{\alpha}{2}+\frac{\alpha}{2})\\\\ sin(\alpha)=2sin\frac{\alpha}{2}cos\frac{\alpha}{2}\\\\ sin(\alpha)=2*\frac{\sqrt{39}}{8}*\frac{5}{8}\\\\ sin(\alpha)=\frac{\sqrt{39}}{4}*\frac{5}{8}\\\\ sin(\alpha)=\frac{5\sqrt{39}}{32}\\\\\\ Cosec(\alpha)=\frac{32}{5\sqrt{39}}\\\\ Cosec(\alpha)=\frac{32\sqrt{39}}{5*39}\\\\ Cosec(\alpha)=\frac{32\sqrt{39}}{195}\\\\$$

$$\\cos(\alpha) = cos(\frac{\alpha}{2}+\frac{\alpha}{2})\\\\ cos(\alpha)=cos^2(\frac{\alpha}{2})-sin^2(\frac{\alpha}{2})\\\\ cos(\alpha)=\frac{25}{64}-\frac{39}{64}\\\\ cos(\alpha)=-\frac{14}{64}\\\\ cos(\alpha)=-\frac{7}{32}\\\\\\ sec(\alpha)=-\frac{32}{7}\\\\\\$$

$$\\tan(\alpha)=sin(\alpha)}\div{cos(\alpha)}\\\\ tan(\alpha)=\frac{5\sqrt{39}}{32}\div\frac{-7}{32}\\\\ tan(\alpha)=\frac{5\sqrt{39}}{32}\times\frac{32}{-7}\\\\ tan(\alpha)=-\frac{5\sqrt{39}}{7}\\\\\\ cot(\alpha)=-\frac{7}{5\sqrt{39}}\\\\ cot(\alpha)=-\frac{7\sqrt{39}}{5*39}\\\\ cot(\alpha)=-\frac{7\sqrt{39}}{195}\\\\$$

Melody  Mar 26, 2015
#1
+94106
+5

Melody  Mar 26, 2015
#2
+94106
+10

Mmm,

first  $$\frac{\alpha}{2}$$   is between  45 and 67.5 degrees.   so that is the first quadrant.   All trig ratios are positive

so   $$\alpha$$    must be between 90 and 135 degrees so that is the second quadrant.  Only sine and cosec are positive all other ratios will be negative,

Now

$$\\cos \left(\frac{\alpha}{2}\right)=\frac{5}{8}\\\\ cos^2 \left(\frac{\alpha}{2}\right)=\frac{25}{64}\\\\ cos^2 \left(\frac{\alpha}{2}\right)+sin^2 \left(\frac{\alpha}{2}\right)=1}\\\\ \frac{25}{64}+sin^2 \left(\frac{\alpha}{2}\right)=1}\\\\ sin^2 \left(\frac{\alpha}{2}\right)=1-\frac{25}{64}\\\\ sin^2 \left(\frac{\alpha}{2}\right)=\frac{39}{64}\\\\ sin \left(\frac{\alpha}{2}\right)=\frac{\sqrt{39}}{8}\\\\\\$$

$$\\sin(\alpha) = sin(\frac{\alpha}{2}+\frac{\alpha}{2})\\\\ sin(\alpha)=2sin\frac{\alpha}{2}cos\frac{\alpha}{2}\\\\ sin(\alpha)=2*\frac{\sqrt{39}}{8}*\frac{5}{8}\\\\ sin(\alpha)=\frac{\sqrt{39}}{4}*\frac{5}{8}\\\\ sin(\alpha)=\frac{5\sqrt{39}}{32}\\\\\\ Cosec(\alpha)=\frac{32}{5\sqrt{39}}\\\\ Cosec(\alpha)=\frac{32\sqrt{39}}{5*39}\\\\ Cosec(\alpha)=\frac{32\sqrt{39}}{195}\\\\$$

$$\\cos(\alpha) = cos(\frac{\alpha}{2}+\frac{\alpha}{2})\\\\ cos(\alpha)=cos^2(\frac{\alpha}{2})-sin^2(\frac{\alpha}{2})\\\\ cos(\alpha)=\frac{25}{64}-\frac{39}{64}\\\\ cos(\alpha)=-\frac{14}{64}\\\\ cos(\alpha)=-\frac{7}{32}\\\\\\ sec(\alpha)=-\frac{32}{7}\\\\\\$$

$$\\tan(\alpha)=sin(\alpha)}\div{cos(\alpha)}\\\\ tan(\alpha)=\frac{5\sqrt{39}}{32}\div\frac{-7}{32}\\\\ tan(\alpha)=\frac{5\sqrt{39}}{32}\times\frac{32}{-7}\\\\ tan(\alpha)=-\frac{5\sqrt{39}}{7}\\\\\\ cot(\alpha)=-\frac{7}{5\sqrt{39}}\\\\ cot(\alpha)=-\frac{7\sqrt{39}}{5*39}\\\\ cot(\alpha)=-\frac{7\sqrt{39}}{195}\\\\$$

Melody  Mar 26, 2015