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sec[sin^-1(4/7)]

Guest Mar 9, 2017

Best Answer 

 #1
avatar+86935 
+5

We're looking for the secant of an angle whose sin = 4/7

 

Sec =  r /x    and x  =  sqrt (7^2 - 4^2)   = sqrt (49 - 16)  = sqrt (33)

 

So.....sec  (theta)   = 7 / sqrt(33)   =  (7/33)*sqrt(33)

 

[Since the sine inverse only has a range of -pi/2 to pi/2, theta must fall in either the 1st or 4th quadrant......so......the secant is positive in both of these ]

 

 

cool cool cool

CPhill  Mar 9, 2017
 #1
avatar+86935 
+5
Best Answer

We're looking for the secant of an angle whose sin = 4/7

 

Sec =  r /x    and x  =  sqrt (7^2 - 4^2)   = sqrt (49 - 16)  = sqrt (33)

 

So.....sec  (theta)   = 7 / sqrt(33)   =  (7/33)*sqrt(33)

 

[Since the sine inverse only has a range of -pi/2 to pi/2, theta must fall in either the 1st or 4th quadrant......so......the secant is positive in both of these ]

 

 

cool cool cool

CPhill  Mar 9, 2017
 #2
avatar+7096 
+6

Hmmmmm! I think I understand these better now after reading this! :)

hectictar  Mar 9, 2017

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