Question

Find the value of \[\sin {15^ \circ } + \cos {105^ \circ } = \].

Answer 1

Hint: Here we will first find the value of \[\sin {15^ \circ }\] using the formula:
\[\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B\]
Then we will find the value of \[\cos {105^ \circ }\] using the formula:
\[\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B\] and then add the values so obtained to get the desired answer.

Complete step-by-step answer:
The given expression is:
\[\sin {15^ \circ } + \cos {105^ \circ }\]
We will first find the value of \[\sin {15^ \circ }\]
Now we know that \[{15^ \circ } = {60^ \circ } - {45^ \circ }\]
Hence replacing this value we get:-
\[\sin {15^ \circ } = \sin \left( {{{60}^ \circ } - {{45}^ \circ }} \right)\]
Now applying the following formula:
\[\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B\]
We get:-
\[\sin \left( {{{60}^ \circ } - {{45}^ \circ }} \right) = \sin {60^ \circ }\cos {45^ \circ } - \cos {60^ \circ }\sin {45^ \circ }\]
Now we know that:
\[\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2}\]
\[\cos {45^ \circ } = \sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}\]
\[\cos {60^ \circ } = \dfrac{1}{2}\]
Therefore, putting the respective values we get:-
\[\sin \left( {{{15}^ \circ }} \right) = \dfrac{{\sqrt 3 }}{2} \times \dfrac{1}{{\sqrt 2 }} - \dfrac{1}{2} \times \dfrac{1}{{\sqrt 2 }}\]
Simplifying it further we get:-
\[\sin \left( {{{15}^ \circ }} \right) = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}\]……………………….(1)
Now we will calculate the value of \[\cos {105^ \circ }\]
We know that:-
\[{105^ \circ } = {60^ \circ } + {45^ \circ }\]
Hence replacing this value we get:-
\[\cos {105^ \circ } = \cos \left( {{{60}^ \circ } + {{45}^ \circ }} \right)\]
Now applying the following formula:
\[\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B\]
We get:-
\[\cos \left( {{{60}^ \circ } + {{45}^ \circ }} \right) = \cos {60^ \circ }\cos {45^ \circ } - \sin {60^ \circ }\sin {45^ \circ }\]
Now we know that:
\[\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2}\]
\[\cos {45^ \circ } = \sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}\]
\[\cos {60^ \circ } = \dfrac{1}{2}\]
Therefore, putting the respective values we get:-
\[\cos {105^ \circ } = \dfrac{1}{2} \times \dfrac{1}{{\sqrt 2 }} - \dfrac{{\sqrt 3 }}{2} \times \dfrac{1}{{\sqrt 2 }}\]
Simplifying it further we get:-
\[\cos {105^ \circ } = \dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }}\]……………………….(2)
Adding (1) and (2) we get:-
\[\sin {15^ \circ } + \cos {105^ \circ } = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }} + \dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }}\]
Taking LCM and solving it further we get:-
\[\sin {15^ \circ } + \cos {105^ \circ } = \dfrac{{\sqrt 3 - 1 + 1 - \sqrt 3 }}{{2\sqrt 2 }}\]
\[ \Rightarrow \sin {15^ \circ } + \cos {105^ \circ } = 0\]

Hence the answer is 0.

Note: Students can also use the fact that \[\cos \left( {{{90}^ \circ } + \theta } \right) = - \sin \theta \] and then simplify the expression so obtained.
\[\cos {105^ \circ } = \cos \left( {{{90}^ \circ } + {{15}^ \circ }} \right)\]
Applying the above identity we get:-
\[\cos {105^ \circ } = - \sin {15^ \circ }\]
Now we have to find the value of \[\sin {15^ \circ } + \cos {105^ \circ }\]
Hence substituting the value we get:-
\[\sin {15^ \circ } + \cos {105^ \circ } = \sin {15^ \circ } - \sin {15^ \circ }\]
\[ \Rightarrow \sin {15^ \circ } + \cos {105^ \circ } = 0\]