Prove that ${\sin75^{\circ}} – {\sin 15^{\circ}} = {\cos105^{\circ}} + {\cos 15^{\circ}}$?

Hint: This question can be done by converting the both the cosines in the right hand side to the sines. The formula you can use is the $\sin x = \cos {\left(90-x \right)}$ which can be rewritten as cos(x) = sin(90-x). You can write 105 as 90 + 15 and you can write 15 as 90 – 75. BY doing these substitutions and using the formulas, you will get the final answer.

Complete step-by-step solution:
This question can be done by converting the both the cosines in the right hand side to the sines. The formula you can use is the sin(x) = cos(90-x) which can be rewritten as cos(x) = sin(90-x). You can write 105 as 90 + 15 and you can write 15 as 90 – 75. BY doing these substitutions and using the formulas, you will get the final answer.
Therefore, substituting these values in the formula, we get,
$ \Rightarrow \cos 105^{\circ} + \cos 15^{\circ} = \cos \left(90 + 15 \right) ^{\circ} + \cos \left(90 - 75 \right) ^{\circ} $
$ \Rightarrow \cos 105^{\circ} + \cos 15^{\circ} = -\sin 15^{\circ} + \sin 75^{\circ} $
$ \Rightarrow \cos 105^{\circ} + \cos 15^{\circ} = \sin 75^{\circ} - \sin 15^{\circ} $
From the above equation, we have proved that the right hand side is equal to the left hand side. Therefore, this is the proof that ${\sin75^{\circ}} – {\sin 15^{\circ}} = {\cos105^{\circ}} + {\cos 15^{\circ}}$.

Note: In order to solve these questions, you must certainly know the trigonometric identities which are very important for not only these questions, but also the other questions. Also, the above question can also be solved by converting the left hand side sine to cosine, to prove that the left hand side is equal to the right hand side.