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trig function of the rest from sec(delta) = -6/5 tan(delta)>0

 Aug 19, 2015

Best Answer 

 #1
avatar+118587 
+5

 

$$\\sec(\delta) = -6/5 tan(\delta)\\\\\\
\frac{1}{cos(\delta)}}=\frac{-6Sin(\delta)}{5cos(\delta)}\\\\
so\qquad \delta \ne n\pi\;\;radians \quad n\in Z\qquad and\\\\
1=\frac{-6Sin(\delta)}{5}\\\\
\frac{5}{-6}=sin(\delta)\\\\
\delta $must be in 3rd or 4th quadrant$\\
$but tan has to be neg so $\delta $ must be in the 4th quad $\\\\
\delta=asin(\frac{-5}{6})$$

 

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\mathtt{\,-\,}}{\frac{{\mathtt{5}}}{{\mathtt{6}}}}\right)} = -{\mathtt{56.442\: \!690\: \!238\: \!079^{\circ}}}$$

 

$$\delta \approx -56^0\pm 360N\qquad where \quad N\in Z$$

.
 Aug 19, 2015
 #1
avatar+118587 
+5
Best Answer

 

$$\\sec(\delta) = -6/5 tan(\delta)\\\\\\
\frac{1}{cos(\delta)}}=\frac{-6Sin(\delta)}{5cos(\delta)}\\\\
so\qquad \delta \ne n\pi\;\;radians \quad n\in Z\qquad and\\\\
1=\frac{-6Sin(\delta)}{5}\\\\
\frac{5}{-6}=sin(\delta)\\\\
\delta $must be in 3rd or 4th quadrant$\\
$but tan has to be neg so $\delta $ must be in the 4th quad $\\\\
\delta=asin(\frac{-5}{6})$$

 

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\mathtt{\,-\,}}{\frac{{\mathtt{5}}}{{\mathtt{6}}}}\right)} = -{\mathtt{56.442\: \!690\: \!238\: \!079^{\circ}}}$$

 

$$\delta \approx -56^0\pm 360N\qquad where \quad N\in Z$$

Melody Aug 19, 2015

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