Question

If $ \sin A = \sin B\,\,and\,\,\cos A = \cos B $ , then prove that \[A = 2n\pi + B\] for some integer n.

Answer 1

Hint: Whenever two same trigonometric functions are given equal then we use the general solution formula to find the answer of the given problem as we know that trigonometric ratios are periodic in nature. So, writing their general solution of both equations and solving them for odd and even integers. We can find required proof.
General formulas for sin and cosine if: $ \sin A = \sin B\,\,then\,\,A = n\pi \pm {( - 1)^n}B.\, $ Also, if $ \cos A = \cos B\,\,then\,\,A = (2n\pi + 1)B $ for all value of $ n \in N $ .

Complete step-by-step answer:
From statement we have
 $ \operatorname{Sin} A = \sin B $
We will use the general formula to find relation between A and B as sine is a periodic function.
Hence, from general formula we have,
 $ \,\,A = n\pi \pm {( - 1)^n}B $
There are two cases if n is an even number or if n is an odd number.
So, for Ist case we take n is an odd number.
Hence, using n as odd from above we have,
 $ A = \left( {2n + 1} \right)\pi - B $ ……………….(i)
And for n is even we have
$\Rightarrow A = 2n\pi + B $ ……………..(ii)
Also, it is given that $ \operatorname{Cos} A = \cos B $
Then again using general formula for cosine we have
$\Rightarrow A = 2n\pi \pm B $
Or we can write above equation as
 $
\Rightarrow A = 2n\pi + B..............(iii) \\
  and \\
\Rightarrow A = 2n\pi - B..............(iv) \;
  $
Therefore, from above four equation we see that common solution is
 $ A = 2n\pi + B $
Which is the required proof of the problem.

Note: If in a problem two T-ratio are equated with each other. We can find a solution to the problem by using general formulas but one must see that general formulas are only applicable when T-ratios of the same functions are equated for different functions. We first convert them to make them the same first and then use a general formula