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can someone help me with these two problems?

Guest Mar 6, 2017

Best Answer 

 #1
avatar+75332 
+5

18. A little hard to see....but.....I think we have

 

sin (theta)  = 3/4   and cos(theta) < 0

 

cos(theta)  = - sqrt ( 1 - (3/4)^2)  = -sqrt(7)/4

 

tan(theta)   =  sin(theta) / cos(theta) = {3/4] / [-sqrt(7)/4] =  -3/sqrt(7)   =  -(3/7)sqrt(7)

 

sec(theta)  = 1/cos(theta) =  1 / [ -sqrt(7)/4]  =  -4/sqrt(7)  = -(4/7)sqrt(7)

 

 

cool cool cool

CPhill  Mar 6, 2017
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2+0 Answers

 #1
avatar+75332 
+5
Best Answer

18. A little hard to see....but.....I think we have

 

sin (theta)  = 3/4   and cos(theta) < 0

 

cos(theta)  = - sqrt ( 1 - (3/4)^2)  = -sqrt(7)/4

 

tan(theta)   =  sin(theta) / cos(theta) = {3/4] / [-sqrt(7)/4] =  -3/sqrt(7)   =  -(3/7)sqrt(7)

 

sec(theta)  = 1/cos(theta) =  1 / [ -sqrt(7)/4]  =  -4/sqrt(7)  = -(4/7)sqrt(7)

 

 

cool cool cool

CPhill  Mar 6, 2017
 #2
avatar+75332 
+5

19.   tan(theta) = -5/3  and theta in Q4

 

sin(theta)  =   -5 / sqrt [ (3)^2 + (-5)^2]  =  -5 / sqrt(34)  =  (-5/34)sqrt(34)

 

sec(theta)   =   -sqrt(34)/ -3   =  sqrt(34)/3

 

 

cool cool cool

CPhill  Mar 6, 2017

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