SOLUTION: A is an angle in quadrant 2. TanA = -3/root5. Solve the following and rationalize the denominator to the exact value. a) Sin2A b) Cos 2A c) Cos(A + pi/2) d) Sin(A + pi/2)

Algebra ->  Trigonometry-basics -> SOLUTION: A is an angle in quadrant 2. TanA = -3/root5. Solve the following and rationalize the denominator to the exact value. a) Sin2A b) Cos 2A c) Cos(A + pi/2) d) Sin(A + pi/2)      Log On


   

Question 1030548: A is an angle in quadrant 2. TanA = -3/root5. Solve the following and rationalize the denominator to the exact value.
a) Sin2A
b) Cos 2A
c) Cos(A + pi/2)
d) Sin(A + pi/2)
Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
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A is an angle in quadrant 2. TanA = -3/root5. Solve the following and rationalize the denominator to the exact value.
a) Sin2A
b) Cos 2A
c) Cos(A + pi/2)
d) Sin(A + pi/2)
~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

First of all, we need to find  sin(A)  and  cos(A)  based on the given value of tan(A) and the fact that angle A is in quadrant 2.

For it, we will use well known formulas  of Trigonometry 

sin%5E2%28A%29 = tan%5E2%28A%29%2F%281+%2B+tan%5E2%28A%29%29  and  cos%5E2%28A%29 = 1%2F%281+%2B+tan%5E2%28A%29%29.


Since tan(A) = -3%2Fsqrt%285%29, we have 
sin(A) = sqrt%28tan%5E2%28A%29%2F%281+%2B+tan%5E2%28A%29%29%29 = sqrt%28%28-3%2Fsqrt%285%29%29%5E2%2F%281+%2B+%28-3%2Fsqrt%285%29%29%5E2%29%29 = sqrt%28%28%289%2F5%29%29%2F%281+%2B+9%2F5%29%29 = sqrt%28%28%289%2F5%29%29%2F%28%2814%2F5%29%29%29 = sqrt%289%2F14%29     and 


cos(A) = -1%2Fsqrt%281+%2B+tan%5E2%28A%29%29 = -1%2Fsqrt%281+%2B+%28-3%2Fsqrt%285%29%29%5E2%29 = -1%2Fsqrt%281+%2B+9%2F5%29 = -1%2Fsqrt%2814%2F5%29 = -sqrt%285%2F14%29.


The sign for sin(A) is "+" since the angle A is in Q2.  The sign for cos(A) is "-" due to the same reason.


Now

a)  sin(2A) = 2*sin(a)*cos(A) = 2%2Asqrt%289%2F14%29%2A%28-sqrt%285%2F14%29%29 = -2%2Asqrt%28%289%2A5%29%2F14%5E2%29 = -2%2Asqrt%2845%29%2F14 = -sqrt%2845%29%2F7.

b)  cos(2A) = 2%2Acos%5E2%28A%29+-+1 = 2%2A%285%2F14%29+-+1 = 5%2F7+-+1 = -2%2F7.

    (By the way, the fact that cos(2A) is negative means that the angle 2A is in Q3).


c)  cos(A + pi/2) = -sin(A) = -sqrt%289%2F14%29.

d)  sin(A + pi/2) = cos(A) = -sqrt%285%2F14%29.

The problem is solved.