Question

Solve the following :
$\dfrac{{\sin \left( {{{10}^ \circ }} \right) + \sin \left( {{{20}^ \circ }} \right) + \sin \left( {{{40}^ \circ }} \right) + \sin \left( {{{50}^ \circ }} \right)}}{{\cos \left( {{{10}^ \circ }} \right) + \cos \left( {{{20}^ \circ }} \right) + \cos \left( {{{40}^ \circ }} \right) + \cos \left( {{{50}^ \circ }} \right)}} =$ ?

Answer 1

In this question, first we need to identify how we can use our trigonometric identities. Here we can see that we can club angle of ten degrees with fifty degrees and angle of twenty degrees with forty degrees. Here we can use the identity of $\sin C + \sin D$ and $\cos C + \cos D$.
Formula used: $\left( 1 \right)\,\,\sin C + \sin D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$
$\left( 2 \right)\,\,\cos C + \cos D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$

Complete step by step answer:
In the above question, we have
$= \dfrac{{\sin \left( {{{10}^ \circ }} \right) + \sin \left( {{{20}^ \circ }} \right) + \sin \left( {{{40}^ \circ }} \right) + \sin \left( {{{50}^ \circ }} \right)}}{{\cos \left( {{{10}^ \circ }} \right) + \cos \left( {{{20}^ \circ }} \right) + \cos \left( {{{40}^ \circ }} \right) + \cos \left( {{{50}^ \circ }} \right)}}$
We can also write it as
$= \dfrac{{\sin \left( {{{10}^ \circ }} \right) + \sin \left( {{{50}^ \circ }} \right) + \sin \left( {{{40}^ \circ }} \right) + \sin \left( {{{20}^ \circ }} \right)}}{{\cos \left( {{{10}^ \circ }} \right) + \cos \left( {{{50}^ \circ }} \right) + \cos \left( {{{40}^ \circ }} \right) + \cos \left( {{{20}^ \circ }} \right)}}$
Now use the identities $\sin C + \sin D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$ and $\cos C + \cos D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$.
$= \dfrac{{2\sin \left( {\dfrac{{50 + 10}}{2}} \right)\cos \left( {\dfrac{{50 - 10}}{2}} \right) + 2\sin \left( {\dfrac{{40 + 20}}{2}} \right)\cos \left( {\dfrac{{40 - 20}}{2}} \right)}}{{2\cos \left( {\dfrac{{50 + 10}}{2}} \right)\cos \left( {\dfrac{{50 - 10}}{2}} \right) + 2\cos \left( {\dfrac{{40 + 20}}{2}} \right)\cos \left( {\dfrac{{40 - 20}}{2}} \right)}}$
$= \dfrac{{2\sin \left( {30} \right)\cos \left( {20} \right) + 2\sin \left( {30} \right)\cos \left( {10} \right)}}{{2\cos \left( {30} \right)\cos \left( {20} \right) + 2\cos \left( {30} \right)\cos \left( {10} \right)}}$
Now we know that $\sin \left( {30} \right) = \dfrac{1}{2}\,\,and\,\,\cos \left( {30} \right) = \dfrac{{\sqrt 3 }}{2}$.
$= \dfrac{{2 \times \dfrac{1}{2}\cos \left( {20} \right) + 2 \times \dfrac{1}{2}\cos \left( {10} \right)}}{{2 \times \dfrac{{\sqrt 3 }}{2}\cos \left( {20} \right) + 2 \times \dfrac{{\sqrt 3 }}{2}\cos \left( {10} \right)}}$
On simplification in numerator and denominator, we get
$= \dfrac{{\cos \left( {20} \right) + \cos \left( {10} \right)}}{{\sqrt 3 \cos \left( {20} \right) + \sqrt 3 \cos \left( {10} \right)}}$
Now taking the common $\sqrt 3$ from the denominator.
$= \dfrac{{\cos \left( {20} \right) + \cos \left( {10} \right)}}{{\sqrt 3 \left( {\cos \left( {20} \right) + \cos \left( {10} \right)} \right)}}$
$= \dfrac{1}{{\sqrt 3 }}$
Therefore, the value of the given trigonometric expression is $\dfrac{1}{{\sqrt 3 }}$.

Note:
We can also do this question by simply finding the value of each quantity with the help of a calculator and then putting them in numerator and denominator and then we can solve it. But whenever we see a question from trigonometry, the first thing which would come in our mind is the use of identities.