Question 960796
Let t be a real number with {{{cos(t)=3/7}}},  {{{3pi/2 < t < 2pi}}} Find the exact value for each of the following: 
<pre>
therefore angle t is in the 4th quadrant.
</pre>
a. sin(t+3pi/2) 
<pre>
Use the identity {{{sin(A+B)=sin(A)cos(B)+cos(A)sin(B)}}}

{{{sin(t+3pi/2)}}}{{{""=""}}}{{{sin(t)cos(3pi/2)+cos(t)sin(3pi/2)}}}
{{{sin(t+3pi/2)}}}{{{""=""}}}{{{sin(t)(0)+cos(t)(-1)}}}
{{{sin(t+3pi/2)}}}{{{""=""}}}{{{-cos(t)}}}
{{{sin(t+3pi/2)}}}{{{""=""}}}{{{-3/7}}}
</pre>
b. sec(-t)
<pre>
Multiplying an angle by -1 moves the angle t from 4th quadrant to 1st quadrant,
and the secant is the reciprocal of the cosine and is positive in 1st and 4th
quadrants. So

{{{sec(-t)}}}{{{""=""}}}{{{sec(t)}}}{{{""=""}}}{{{1/cos(t)}}}{{{""=""}}}{{{1/(3/7)}}}{{{""=""}}}{{{7/3}}}

</pre>
c. cos(2pi)
<pre>
That's just 1.  It has nothing to do with t.
</pre>
d.cos(4pi/3 - t) 
<pre>
Use the identity {{{cos(A-B)=cos(A)cos(B)+sin(A)sin(B)}}}

{{{cos(4pi/3-t)=cos(4pi/3)cos(t)+sin(4pi/3)sin(t)}}}
{{{cos(4pi/3-t)=(-1/2)(3/7)+(-sqrt(3)/2)sin(t)}}}
{{{cos(4pi/3-t)=(-3/14)+(-sqrt(3)/2)sin(t)}}}

Use identity {{{sin(A)= "" +- sqrt(1-cos^2(A))}}}
since t is in quadrant 4 where sine is negative we use 
the negative:

{{{cos(4pi/3-t)=(-3/14)+(-sqrt(3)/2)(-sqrt(1-cos^2(t)))}}}
{{{cos(4pi/3-t)=(-3/14)+(sqrt(3)/2)(sqrt(1-cos^2(t)))}}}
{{{cos(4pi/3-t)=(-3/14)+(sqrt(3)/2)(sqrt(1-(3/7)^2))}}}
{{{cos(4pi/3-t)=(-3/14)+(sqrt(3)/2)(sqrt(1-9/49))}}}
{{{cos(4pi/3-t)=(-3/14)+(sqrt(3)/2)(sqrt(49/49-9/49))}}}
{{{cos(4pi/3-t)=(-3/14)+(sqrt(3)/2)(sqrt(40/49))}}}
{{{cos(4pi/3-t)=(-3/14)+(sqrt(3)/2)(sqrt(40)/7)}}}
{{{cos(4pi/3-t)=(-3/14)+(sqrt(3)/2)(sqrt(4*10)/7)}}}
{{{cos(4pi/3-t)=(-3/14)+(sqrt(3)/2)(2sqrt(10)/7)}}}
{{{cos(4pi/3-t)=(-3/14)+(2sqrt(30)/14)}}}
{{{cos(4pi/3-t)=(-3+2sqrt(30))/14}}}
  


Edwin</pre>