Question 1204248
<pre>
You were told not to use any calculators.  The other tutor evaluated angles, 
which requires a calculator so you cannot use the part where he evaluated angles
to find which quadrant {{{2alpha+beta}}} is in.</pre>
(a) find the value of {{{sec(2alpha + beta)}}} ...<pre>

Please refer to the graph below.

Since {{{alpha}}} is negative and in QIV, it is measured clockwise from the right side of the
x-axis. Since {{{cos(alpha)=3/sqrt(10)}}} we take x<sub>&alpha;</sub>=+3. We know it's POSITIVE because it goes to 
the right. We always take r<sub>&alpha;</sub> as positive.  
r={{{sqrt(10)}}}. Then, we find y<sub>&alpha;</sub> by 
{{{matrix(5,1,x[alpha]^2+y[alpha]^2=r[alpha]^2,
   3^2+y[alpha]^2=(sqrt(10))^2,
   9+y[alpha]^2=10, 
     y[alpha]^2=1,
       y[alpha]= "" +- 1)}}}

and we know to use the NEGATIVE sign because y<sub>&alpha;</sub> goes DOWN from the x-axis.

Since {{{beta}}} is positive, it is in QIII, measured counter-clockwise from the right
side of the x-axis. Since {{{sin(beta)=-5/13}}} we take y<sub>&beta;</sub>=-5, r<sub>&beta;</sub>=13. Then,
we find x<sub>&beta;</sub> by 
{{{matrix(5,1,x[beta]^2+y[beta]^2=r[beta]^2,
   x[beta]^2+(-5)^2=13^2,
   x[beta]^2+25=169, 
     x[beta]^2=144,
       x[beta]= "" +- 12)}}}

and again, we know to use the NEGATIVE sign because x<sub>&beta;</sub> goes LEFT of the y-axis.
So x<sub>&beta;</sub> = -12.  So we can plot both angles and their triangles on the same graph,
without one overlapping the other:

{{{drawing(800,3200/9,-13,5,-6,2,graph(800,3200/9,-13,5,-6,2),

line(3,0,3,-1),line(-12,0,-12,-5),

line(0,0,3,-1),line(-12,-5,0,0),

locate(1.5,.85,x[alpha]=3),locate(1.1,-.52,r[alpha]=sqrt(10)),locate(-6.4,-2.5,r[beta]=13), locate(-7,.85,x[beta]=-12),

locate(-11.9,-2,y[beta]=-5), locate(-.5,.74,beta),locate(2.51,-.3,alpha),
locate(3.1,-.16,y[alpha]=-1),

arc(0,0,1,-1,0,203), arc(0,0,5,-5,341,360)

  )}}}</pre>
(a) find the value of ... {{{cos(2alpha + beta)}}}<pre> 
First find {{{cos(2alpha)}}}{{{""=""}}}{{{2cos^2(alpha)-1}}}{{{""=""}}}{{{2(x[alpha]/r[alpha])^2-1}}}{{{""=""}}}{{{2(3/sqrt(10))^2-1}}}{{{""=""}}}{{{2(9/10)-1}}}{{{""=""}}}{{{9/5-1}}}{{{""=""}}}{{{4/5}}}

Next find {{{sin(2alpha)}}}{{{""=""}}}{{{2sin(alpha)cos(alpha)}}}{{{""=""}}}{{{2(y[alpha]/r[alpha])(x[alpha]/r[alpha])}}}{{{""=""}}}{{{2((-1)/sqrt(10))(3/sqrt(10))}}}{{{""=""}}}{{{-6/10}}}{{{""=""}}}{{{-3/5}}}

Next find {{{cos(2alpha+beta)}}}{{{""=""}}}{{{cos(2alpha)cos(beta)-sin(2alpha)sin(beta)}}}{{{""=""}}}{{{(4/5)(x[beta]/r[beta])-(-3/5)(y[beta]/r[beta])}}}{{{""=""}}}
{{{(4/5)((-12)/13)+(3/5)((-5)/13)}}}{{{""=""}}}{{{-48/65+((-15)/65)}}}{{{""=""}}}{{{-63/65}}} 

Finally, {{{sec(2alpha+beta)}}}{{{""=""}}}{{{1/cos(2alpha+beta)}}}{{{""=""}}}{{{1^""/(-63/65)}}}{{{""=""}}}{{{-65/63}}}</pre>

(a) find the values of... {{{tan(2alpha + beta)}}}<pre>
We already have cos(2a+b), and we know that {{{tan(theta)=sin(theta)/cos(theta)}}}.

We find {{{sin(2alpha+beta)}}}{{{""=""}}}{{{sin(2alpha)cos(beta)+cos(2alpha)sin(beta)}}}{{{""=""}}}{{{(-3/5)(-12/13)+(4/5)(-5/13)}}}{{{""=""}}}{{{36/65+((-20)/65)}}}{{{""=""}}}{{{16/65}}}

{{{tan(2alpha+beta)}}}{{{""=""}}}{{{sin(2alpha+beta)/cos(2alpha+beta)}}}{{{""=""}}}{{{(16/65)/(-63/65)}}}{{{""=""}}}{{{-16/63}}}</pre>
(b) determine the quadrant which contains {{{2alpha+beta}}}. Give reasons<pre>
Now if you've learned "ALL STUDENTS TAKE CALCULUS" or some other way to
determine which trig ratios are positive or negative in which quadrants,
you know that the cosine is negative only in QII and QIII, and the tangent 
is negative only in QII and QIV.

Since {{{cos(2alpha+beta)}}} is negative and {{{tan(2alpha+beta)}}} is also negative, then {{{2alpha+beta}}} is in QII.
 
Edwin</pre>