Question

If $\cos A = \dfrac{4}{5},\cos B = \dfrac{{12}}{{13}},\dfrac{{3\pi }}{2} < A,B < 2\pi$, find the value of the following:
A. Cos (A + B)
B. Sin (A – B)

Answer 1

We will first find the cosine of A and sine of B using the given data and then, using the formula of expansion of parts in A and B, we will then use those values to find the answer.

Complete step-by-step answer:
Since, we are given that $\cos A = \dfrac{4}{5},\dfrac{{3\pi }}{2} < A < 2\pi$.
We know that the cosine is positive in the first or fourth quadrant and since we are given in condition that $\dfrac{{3\pi }}{2} < A$, therefore, A lies in the fourth quadrant.
We know that cosine is the ratio of cosine is given by base and hypotenuse. We will use the Pythagorean Theorem which is given by: ${H^2} = {P^2} + {B^2}$
$\Rightarrow {5^2} = {P^2} + {4^2}$
Solving the above equation, we get: P = 3
And, we know that sine of an angle is given by the ratio of perpendicular and hypotenuse.
So, $\sin A = \dfrac{3}{5}$. Since, it lies in the fourth quadrant and sine is negative in the fourth quadrant.
Therefore, $\sin A = - \dfrac{3}{5}$ ………………(1)
We are also given that $\cos B = \dfrac{{12}}{{13}},\dfrac{{3\pi }}{2} < B < 2\pi$
We know that the cosine is positive in the first or fourth quadrant and since we are given in condition that $\dfrac{{3\pi }}{2} < B < 2\pi$, therefore, B lies in the fourth quadrant.
We know that cosine is the ratio of cosine is given by base and hypotenuse. We will use the Pythagorean Theorem which is given by: ${H^2} = {P^2} + {B^2}$
$\Rightarrow {13^2} = {P^2} + {12^2}$
Solving the above equation, we get: P = 5
And, we know that sine of an angle is given by the ratio of perpendicular and hypotenuse.
So, $\sin B = \dfrac{5}{{13}}$. Since, it lies in the fourth quadrant and sine is negative in the fourth quadrant.
Therefore, $\sin B = - \dfrac{5}{{13}}$ ………………(2)
Part (A):
Here, we have to find the value of cos(A + B).
We know that cos(A + B) = cosA.cosB – sinA.sinB
Using the given values and the values we found in the equations (1) and (2), we will get:-
$\Rightarrow cos\left( {A + B} \right) = \dfrac{4}{5} \times \dfrac{{12}}{{13}}-\left( { - \dfrac{3}{5}} \right) \times \left( { - \dfrac{5}{{13}}} \right)$
Simplifying a bit to get:-
$\Rightarrow cos\left( {A + B} \right) = \dfrac{{48}}{{65}}-\dfrac{{15}}{{65}}$
Simplifying it further to get:-
$\Rightarrow cos\left( {A + B} \right) = \dfrac{{33}}{{65}}$
Part (B):
Here, we have to find the value of sin(A - B).
We know that sin(A - B) = sinA.cosB – cosA.sinB
Using the given values and the values we found in the equations (1) and (2), we will get:-
$\Rightarrow \sin \left( {A - B} \right) = \left( { - \dfrac{3}{5}} \right) \times \dfrac{{12}}{{13}}-\left( { - \dfrac{4}{5}} \right) \times \dfrac{5}{{13}}$
Simplifying a bit to get:-
$\Rightarrow \sin \left( {A - B} \right) = - \dfrac{{36}}{{65}} + \dfrac{{20}}{{65}}$
Simplifying it further to get,

$\Rightarrow \sin \left( {A - B} \right) = - \dfrac{{16}}{{65}}$

Note:
The students might make the mistake of not considering the intervals for both the angles and just assume generally the sine values we got as positive values only and end up getting the wrong answer for the second part of the question. Always take in consideration of the interval and take values as per the norms of that quadrant.
The students must note that finding angles in this question is not the easiest task. Therefore, we found out the easy way by elaborating the formulas and thus finding the required values.