Question

Solve the following:
$\sin \left( {40} \right) + \cos \left( {50} \right) + \sin \left( {80} \right) - \cos \left( {190} \right) = ?$

Answer 1

Hint: To do this question you should have good knowledge of trigonometric formulas. Formulas are the base of trigonometry. In this question we should use the formula of $\sin C + \sin D = 2\sin \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}$, $\cos \left( {90 - \theta } \right) = \sin \theta $ and $\cos (270 - \theta ) = - \sin \left( \theta \right)$.

Complete step by step answer:
In the above question, we have
$ \Rightarrow \sin \left( {40} \right) + \cos \left( {50} \right) + \sin \left( {80} \right) - \cos \left( {190} \right)$
We can also write $\cos \left( {50} \right) = \cos \left( {90 - 40} \right)$ and $\cos \left( {190} \right) = \cos \left( {270 - 80} \right)$.
Now,
 $ \Rightarrow \sin \left( {40} \right) + \cos \left( {90 - 40} \right) + \sin \left( {80} \right) - \cos \left( {270 - 80} \right)$
We know that $\cos \left( {90 - \theta } \right) = \sin \theta $ and also $\cos (270 - \theta ) = - \sin \left( \theta \right)$.
$ \Rightarrow \sin \left( {40} \right) + \sin \left( {40} \right) + \sin \left( {80} \right) + \sin \left( {80} \right)$
Now we will add the terms having the same angle of sine function.
$ \Rightarrow 2\sin \left( {40} \right) + 2\sin \left( {80} \right)$
Taking two common out from the equation
$ \Rightarrow 2\left( {\sin \left( {40} \right) + \sin \left( {80} \right)} \right)$
Now we will use the identity $\sin C + \sin D = 2\sin \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}$ in the above equation.
$ \Rightarrow 2\left( {2\sin \dfrac{{80 + 40}}{2}\cos \dfrac{{80 - 40}}{2}} \right)$
$ \Rightarrow 2\left( {2\sin \dfrac{{120}}{2}\cos \dfrac{{40}}{2}} \right)$
$ \Rightarrow 4\sin 60\cos 20$
On simplification, we get
$ \Rightarrow 2\sqrt 3 \cos \left( {20} \right)$
Therefore, the value of $\sin \left( {40} \right) + \cos \left( {50} \right) + \sin \left( {80} \right) - \cos \left( {190} \right) = \,2\sqrt 3 \cos \left( {20} \right)$.

Note:
There is one more method to answer this question. in that method we will write $\cos \left( {190} \right) = \cos \left( {180 + 10} \right)$ and on further calculation we will write it as $\cos \left( {10} \right) = \cos \left( {90 - 80} \right)$ and it will convert into sine function and then we will get to the final result which will also in the form of sine function.