Question 1204248
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I'll do the first portion of part (a) to get you started.


I'll use letters A and B in place of alpha and beta. 


{{{cos(A) = 3/sqrt(10) = 3*sqrt(10)/10}}}
{{{sin(B) = -5/13}}}


{{{-90 < A < 0}}}
{{{180 < B < 270}}}


Use the pythagorean trig identity {{{sin^2(x) + cos^2(x) = 1}}} to go from {{{cos(A) = 3*sqrt(10)/10}}} to {{{sin(A) = -sqrt(10)/10}}} or {{{sin(A) = sqrt(10)/10}}}
I'll leave the scratch work steps for the student to do.


Which sine do we go for? The angle A is between -90 and 0, i.e. between 270 and 360, which places the angle in quadrant 4. This is where sine is negative.
Therefore, we go with {{{sin(A) = -sqrt(10)/10}}}


If {{{sin(B) = -5/13}}} then {{{cos(B) = 12/13}}} or {{{cos(B) = -12/13}}} when using the pythagorean trig identity. 
Angle B is in quadrant 3 where cosine is negative, which means we'll go for {{{cos(B) = -12/13}}}



Here's a short recap so far
{{{cos(A) = 3/sqrt(10) = 3*sqrt(10)/10}}}
{{{sin(A) = -sqrt(10)/10}}}
{{{sin(B) = -5/13}}}
{{{cos(B) = -12/13}}}


Then,
{{{cos(2A+B) = cos(2A)cos(B) - sin(2A)sin(B)}}}


{{{cos(2A+B) = (2*cos^2(A)-1)cos(B) - 2*sin(A)cos(A)sin(B)}}}


{{{cos(2A+B) = (2*(3/sqrt(10))^2-1)*(-12/13) - 2*(-sqrt(10)/10)*(3/sqrt(10))*(-5/13)}}}


{{{cos(2A+B) = -63/65}}} I skipped a few steps to get to this point. I'll let the student fill in the gaps.


{{{sec(2A+B) = 1/(cos(2A+B))}}}


{{{sec(2A+B) = -65/63}}}


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How can we help to confirm this?


cos(A) = 3/sqrt(10) leads to either 
A = arccos(3/sqrt(10)) = 18.434948822922
or
A = -arccos(3/sqrt(10)) = -18.434948822922


We go with the negative value to place angle A in quadrant 4.


Make sure that your calculator is set to degree mode.
Then add on 360 to get a coterminal angle between 0 and 360.
-18.434948822922 + 360 = 341.565051177078
Notice how that value satisfies the interval 270 < A < 360.


sin(B) = -5/13 leads to 
B = arcsin(-5/13) = -22.6198649480404
or
B = 180-arcsin(-5/13) = 202.61986494804
We go with the second result to place angle B in quadrant 3.


We have these approximate angle values
A = 341.565051177078
B = 202.61986494804


Then we can type this into a calculator:
1/(cos(2*341.565051177078+202.61986494804)) - (-65/63)


The 1/(cos(2*341.565051177078+202.61986494804)) portion is us computing 1/(cos(2A+B)), i.e. sec(2A+B).
We then subtract off (-65/33) to finish off the calculation.
I'm using the idea that if x = y, then x-y = 0.
So for example, 2+3 = 5 leads to 2+3-5 = 0. For such trivial examples like this, we won't need to use this rule. But it's useful for more complicated algebraic and trig expressions.


The result of this calculation should be very close to zero. It won't be zero itself because those A,B values we found were approximate. 


Despite the difference of those values not being zero itself, it's quite possible your calculator rounds the result to 0.


My calculator shows the difference is the very small value of -1.998401 * 10^(-15) which is the same as writing 
<pre>-0.000 000 000 000 001 998 401</pre>
I have separated the decimal digits with a space every 3 digits to help make the number more readable. There are 14 zeros between the decimal point and the first copy of "1".


This all but confirms that we have the correct value of sec(2A+B).


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Answer: sec(2A+B) = <font color=red size=4>-65/63</font>
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