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Guest Mar 25, 2015

Best Answer 

 #2
avatar+91436 
+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
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2+0 Answers

 #1
avatar+91436 
+5

Melody  Mar 26, 2015
 #2
avatar+91436 
+10
Best Answer

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

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