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

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

Explanation

Solution

Here we will first find the value of sin⁡15∘\sin {15^ \circ } using the formula:
sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B
Then we will find the value of cos⁡105∘\cos {105^ \circ } using the formula:
cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\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∘+cos⁡105∘\sin {15^ \circ } + \cos {105^ \circ }
We will first find the value of sin⁡15∘\sin {15^ \circ }
Now we know that 15∘=60∘−45∘{15^ \circ } = {60^ \circ } - {45^ \circ }
Hence replacing this value we get:-
sin⁡15∘=sin⁡(60∘−45∘)\sin {15^ \circ } = \sin \left( {{{60}^ \circ } - {{45}^ \circ }} \right)
Now applying the following formula:
sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B
We get:-
sin⁡(60∘−45∘)=sin⁡60∘cos⁡45∘−cos⁡60∘sin⁡45∘\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∘=32\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2}
cos⁡45∘=sin⁡45∘=12\cos {45^ \circ } = \sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}
cos⁡60∘=12\cos {60^ \circ } = \dfrac{1}{2}
Therefore, putting the respective values we get:-
sin⁡(15∘)=32×12−12×12\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⁡(15∘)=3−122\sin \left( {{{15}^ \circ }} \right) = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}……………………….(1)
Now we will calculate the value of cos⁡105∘\cos {105^ \circ }
We know that:-
105∘=60∘+45∘{105^ \circ } = {60^ \circ } + {45^ \circ }
Hence replacing this value we get:-
cos⁡105∘=cos⁡(60∘+45∘)\cos {105^ \circ } = \cos \left( {{{60}^ \circ } + {{45}^ \circ }} \right)
Now applying the following formula:
cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B
We get:-
cos⁡(60∘+45∘)=cos⁡60∘cos⁡45∘−sin⁡60∘sin⁡45∘\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∘=32\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2}
cos⁡45∘=sin⁡45∘=12\cos {45^ \circ } = \sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}
cos⁡60∘=12\cos {60^ \circ } = \dfrac{1}{2}
Therefore, putting the respective values we get:-
cos⁡105∘=12×12−32×12\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∘=1−322\cos {105^ \circ } = \dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }}……………………….(2)
Adding (1) and (2) we get:-
sin⁡15∘+cos⁡105∘=3−122+1−322\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∘+cos⁡105∘=3−1+1−322\sin {15^ \circ } + \cos {105^ \circ } = \dfrac{{\sqrt 3 - 1 + 1 - \sqrt 3 }}{{2\sqrt 2 }}
⇒sin⁡15∘+cos⁡105∘=0\Rightarrow \sin {15^ \circ } + \cos {105^ \circ } = 0

Hence the answer is 0.

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