Question 1030548
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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)
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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^2(A)}}} = {{{tan^2(A)/(1 + tan^2(A))}}}  and  {{{cos^2(A)}}} = {{{1/(1 + tan^2(A))}}}.


Since tan(A) = {{{-3/sqrt(5)}}}, we have 
sin(A) = {{{sqrt(tan^2(A)/(1 + tan^2(A)))}}} = {{{sqrt((-3/sqrt(5))^2/(1 + (-3/sqrt(5))^2))}}} = {{{sqrt(((9/5))/(1 + 9/5))}}} = {{{sqrt(((9/5))/((14/5)))}}} = {{{sqrt(9/14)}}}     and 


cos(A) = -{{{1/sqrt(1 + tan^2(A))}}} = -{{{1/sqrt(1 + (-3/sqrt(5))^2)}}} = -{{{1/sqrt(1 + 9/5)}}} = -{{{1/sqrt(14/5)}}} = {{{-sqrt(5/14)}}}.


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*sqrt(9/14)*(-sqrt(5/14))}}} = {{{-2*sqrt((9*5)/14^2)}}} = {{{-2*sqrt(45)/14}}} = {{{-sqrt(45)/7}}}.

b)  cos(2A) = {{{2*cos^2(A) - 1}}} = {{{2*(5/14) - 1}}} = {{{5/7 - 1}}} = {{{-2/7}}}.

    (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(9/14)}}}.

d)  sin(A + pi/2) = cos(A) = {{{-sqrt(5/14)}}}.

The problem is solved. 
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