Trigonometry Question

Mathematics
Tutor: None Selected Time limit: 1 Day

Given that sin A=-3/5 and the terminal side of angle A lies in the third quadrant and cos B=5/13 and the terminal side of angle B lies in the fourth quadrant, find:

a) Cos A

b) Sin 2A

c) sec^2 B

d) Cos(A-B)

May 18th, 2015

Cos A =  -4/5

 Sin 2A = -6/5 * -4/5 = 24/25

 sec^2 B = 169/25

Cos (A-B) =  -4/13 + 36/65 = (-20+36)/65 = 16/65

Please let me know if you have any questions and best me if you are satisfactory.

May 18th, 2015

Can you please explain how you found sec^2 B and Cos (A-B) step by step?

May 18th, 2015


Sec B = 1/cos B = 13/5

sec^2 B =  169/25

Cos(A-B) = Cos A Cos B + Sin A  Sin B

sin A=-3/5, Cos A =  -4/5,  Sin B = -12/13  (Negative since angle is in the 4th quadrant ),

 Cos B = 5/13

Substituting the above values , we get

Cos(A-B) = -4/13 + 36/65 = (-20+36)/65 = 16/65

May 18th, 2015

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May 18th, 2015
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May 18th, 2015
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