Question

The value of cos15 – cos105 is:
$
  A.\dfrac{1}{2} \\
  B.\dfrac{1}{{\sqrt 2 }} \\
  C.\dfrac{{\sqrt 3 }}{2} \\
  D.\sqrt {\dfrac{3}{2}} \\
$

Answer 1

Hint: In order to solve this problem we need to use the formula of cos(a-b) and cos(a+b) and then put the value of a = 60 and that of b = 45 then you will get the equation which is given in problem solving that equation by using the formula we will get the right answer.

Complete step-by-step answer:
After writing the term 15 in a – b and 105 in a + b we get the values as,
So, 15 = 60 – 45 and 105 = 60 + 45
Therefore on putting those above values in the angles we have we get the values as: cos15 = cos(60 - 45) and cos105 = cos(60 + 45)
And we know the formula that cos(A+B)= cosAcosB – sinAsinB and cos(A–B) = cosAcosB + sinAsinB
On putting a = 60 and b = 45 we will be getting the equation we have in the question.
Therefore, we can do cos15 = cos(60–45) = cos 60 cos 45 + sin 60 sin 45
$\cos 15 = \dfrac{1}{2} \times \dfrac{1}{{\sqrt 2 }} + \dfrac{{\sqrt 3 }}{2} \times \dfrac{1}{{\sqrt 2 }} = \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}$………….(1)
And cos 105 = cos(60+45) = cos 60 cos 45 – sin 60 sin 45
$\cos 105 = \dfrac{1}{2} \times \dfrac{1}{{\sqrt 2 }} - \dfrac{{\sqrt 3 }}{2} \times \dfrac{1}{{\sqrt 2 }} = \dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }}$…………….(2)
We need to calculate the value of cos15 – cos105.
From (1) and (2) we get,
So, the value of cos 15 – cos105 = $\dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}$ - $\dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }}$ = $\dfrac{{2\sqrt 3 }}{{\sqrt 2 }} = \sqrt {\dfrac{3}{2}} $.
So, the correct option is option D.

Note: Whenever you face such types of problems you need to use the general formula of angles in trigonometry since doing that will be the easiest way you can find the answer. Generally we try to put the values if it can be calculated then solve it. We should not need to do that. We have to convert the terms into simpler form and then use their respective values. Doing this will solve such types of problems and will give you the right answer.