SOLUTION: Given that cos 𝛼 =3/√10 and sin 𝛽 = −5/13, where −90° < 𝛼 < 0° and 180° < 𝛽 < 270°. Without using calculators, (a) find the values of sec(2𝛼 + 𝛽) and

Algebra ->  Trigonometry-basics -> SOLUTION: Given that cos 𝛼 =3/√10 and sin 𝛽 = −5/13, where −90° < 𝛼 < 0° and 180° < 𝛽 < 270°. Without using calculators, (a) find the values of sec(2𝛼 + 𝛽) and       Log On


   

Question 1204248: Given that cos 𝛼 =3/√10 and sin 𝛽 = −5/13, where −90° < 𝛼 < 0° and 180° < 𝛽 < 270°.
Without using calculators,
(a) find the values of sec(2𝛼 + 𝛽) and tan(2𝛼 + 𝛽);
(b) determine the quadrant which contains 2𝛼 + 𝛽. Give reasons
Found 2 solutions by math_tutor2020, Edwin McCravy:
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

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%28A%29+=+3%2Fsqrt%2810%29+=+3%2Asqrt%2810%29%2F10
sin%28B%29+=+-5%2F13

-90+%3C+A+%3C+0
180+%3C+B+%3C+270

Use the pythagorean trig identity sin%5E2%28x%29+%2B+cos%5E2%28x%29+=+1 to go from cos%28A%29+=+3%2Asqrt%2810%29%2F10 to sin%28A%29+=+-sqrt%2810%29%2F10 or sin%28A%29+=+sqrt%2810%29%2F10
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%28A%29+=+-sqrt%2810%29%2F10

If sin%28B%29+=+-5%2F13 then cos%28B%29+=+12%2F13 or cos%28B%29+=+-12%2F13 when using the pythagorean trig identity.
Angle B is in quadrant 3 where cosine is negative, which means we'll go for cos%28B%29+=+-12%2F13


Here's a short recap so far
cos%28A%29+=+3%2Fsqrt%2810%29+=+3%2Asqrt%2810%29%2F10
sin%28A%29+=+-sqrt%2810%29%2F10
sin%28B%29+=+-5%2F13
cos%28B%29+=+-12%2F13

Then,
cos%282A%2BB%29+=+cos%282A%29cos%28B%29+-+sin%282A%29sin%28B%29

cos%282A%2BB%29+=+%282%2Acos%5E2%28A%29-1%29cos%28B%29+-+2%2Asin%28A%29cos%28A%29sin%28B%29



cos%282A%2BB%29+=+-63%2F65 I skipped a few steps to get to this point. I'll let the student fill in the gaps.

sec%282A%2BB%29+=+1%2F%28cos%282A%2BB%29%29

sec%282A%2BB%29+=+-65%2F63

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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 
-0.000 000 000 000 001 998 401

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) = -65/63

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
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%2Bbeta is in.

(a) find the value of sec%282alpha+%2B+beta%29 ...

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%28alpha%29=3%2Fsqrt%2810%29 we take xα=+3. We know it's POSITIVE because it goes to 
the right. We always take rα as positive.  
r=sqrt%2810%29. Then, we find yα by 


and we know to use the NEGATIVE sign because yα 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%28beta%29=-5%2F13 we take yβ=-5, rβ=13. Then,
we find xβ by 


and again, we know to use the NEGATIVE sign because xβ goes LEFT of the y-axis.
So xβ = -12.  So we can plot both angles and their triangles on the same graph,
without one overlapping the other:

(a) find the value of ... cos%282alpha+%2B+beta%29
 
First find cos%282alpha%29%22%22=%22%222cos%5E2%28alpha%29-1%22%22=%22%222%28x%5Balpha%5D%2Fr%5Balpha%5D%29%5E2-1%22%22=%22%222%283%2Fsqrt%2810%29%29%5E2-1%22%22=%22%222%289%2F10%29-1%22%22=%22%229%2F5-1%22%22=%22%224%2F5

Next find sin%282alpha%29%22%22=%22%222sin%28alpha%29cos%28alpha%29%22%22=%22%222%28y%5Balpha%5D%2Fr%5Balpha%5D%29%28x%5Balpha%5D%2Fr%5Balpha%5D%29%22%22=%22%222%28%28-1%29%2Fsqrt%2810%29%29%283%2Fsqrt%2810%29%29%22%22=%22%22-6%2F10%22%22=%22%22-3%2F5

Next find cos%282alpha%2Bbeta%29%22%22=%22%22cos%282alpha%29cos%28beta%29-sin%282alpha%29sin%28beta%29%22%22=%22%22%284%2F5%29%28x%5Bbeta%5D%2Fr%5Bbeta%5D%29-%28-3%2F5%29%28y%5Bbeta%5D%2Fr%5Bbeta%5D%29%22%22=%22%22
%284%2F5%29%28%28-12%29%2F13%29%2B%283%2F5%29%28%28-5%29%2F13%29%22%22=%22%22-48%2F65%2B%28%28-15%29%2F65%29%22%22=%22%22-63%2F65 

Finally, sec%282alpha%2Bbeta%29%22%22=%22%221%2Fcos%282alpha%2Bbeta%29%22%22=%22%221%5E%22%22%2F%28-63%2F65%29%22%22=%22%22-65%2F63

(a) find the values of... tan%282alpha+%2B+beta%29
We already have cos(2a+b), and we know that tan%28theta%29=sin%28theta%29%2Fcos%28theta%29.

We find sin%282alpha%2Bbeta%29%22%22=%22%22sin%282alpha%29cos%28beta%29%2Bcos%282alpha%29sin%28beta%29%22%22=%22%22%28-3%2F5%29%28-12%2F13%29%2B%284%2F5%29%28-5%2F13%29%22%22=%22%2236%2F65%2B%28%28-20%29%2F65%29%22%22=%22%2216%2F65

tan%282alpha%2Bbeta%29%22%22=%22%22sin%282alpha%2Bbeta%29%2Fcos%282alpha%2Bbeta%29%22%22=%22%22%2816%2F65%29%2F%28-63%2F65%29%22%22=%22%22-16%2F63

(b) determine the quadrant which contains 2alpha%2Bbeta. Give reasons
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%282alpha%2Bbeta%29 is negative and tan%282alpha%2Bbeta%29 is also negative, then 2alpha%2Bbeta is in QII.
 
Edwin