Question

Show that $\cos {10^\circ} + \cos {110^\circ} + \cos {130^\circ} = 0$

Answer 1

Hint: According to the question given in the question we have to prove that $\cos {10^\circ} + \cos {110^\circ} + \cos {130^\circ} = 0$. So, first of all we have to solve the left hand side term of the given trigonometric expression which is $\cos {10^\circ} + \cos {110^\circ} + \cos {130^\circ}$ to obtain the right hand side of the trigonometric expression.
Now, to solve the left hand side of the given expression which is $\cos {10^\circ} + \cos {110^\circ} + \cos {130^\circ}$ we have to use the formula as given below:

Formula used: $ \Rightarrow \cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right).....................(a)$
Hence, with the help of the formula (a) above we will solve \[\cos {110^\circ} + \cos {10^\circ}\]to obtain the simplified form of \[\cos {110^\circ} + \cos {10^\circ}\]in term of $\cos $ and to solve the expression we have to place the value of $\cos {60^\circ}$which is given below:
\[ \Rightarrow \cos {60^\circ} = \dfrac{1}{2}....................(b)\]
Now we will again apply the formula (a) above to determine the value of the left hand side $\cos {10^\circ} + \cos {110^\circ} + \cos {130^\circ}$to obtain the right hand side that should be 0.
Now, to solve the obtained expression we have to place the value of $\cos {90^\circ}$which is given below:
\[ \Rightarrow \cos {90^\circ} = 0....................(c)\]

Complete step-by-step answer:
Step 1: First of all we have to solve the left hand side of the given trigonometric expression which is $\cos {10^\circ} + \cos {110^\circ} + \cos {130^\circ}$ with the help of the formula (a) as mentioned in the solution hint. But before that we have to rearrange the order of the trigonometric terms given in the expression,
$ = \cos {110^\circ} + \cos {10^\circ} + \cos {130^\circ}$
On applying formula (a) as mentioned in the solution hint,
$ = \cos \left( {\dfrac{{{{110}^\circ} + {{10}^\circ}}}{2}} \right)\cos \left( {\dfrac{{{{110}^\circ} - {{10}^\circ}}}{2}} \right) + \cos {130^\circ}$………………….(1)
Step 2: Now, in this step we have to solve the trigonometric expression (1) as obtained in the step (1) with the help of simply calculations hence,
$ = \cos \left( {\dfrac{{{{120}^\circ}}}{2}} \right)\cos \left( {\dfrac{{{{100}^\circ}}}{2}} \right) + \cos {130^\circ}$
$ = \cos {60^\circ}\cos {50^\circ} + \cos {130^\circ}$…………………….(2)
Step 3: Now, to solve to solve the expression (2) as obtained in the step 2 we have to substitute the value of $\cos {60^\circ}$ as mentioned in the solution hint as (b),
$ = 2 \times \dfrac{1}{2}\cos {50^\circ} + \cos {130^\circ}$
On solving the expression obtained just above,
$ = \cos {50^\circ} + \cos {130^\circ}..................(3)$
Step 4: Now, to solve the expression (3) as obtained in the step 3 we have to apply the formula (a) again as mentioned in the solution hint.
$ = 2\cos \left( {\dfrac{{{{130}^\circ} + {{50}^\circ}}}{2}} \right)\cos \left( {\dfrac{{{{130}^\circ} - {{50}^\circ}}}{2}} \right)$
On solving the expression obtained just above,
\[
   = 2\cos \left( {\dfrac{{{{180}^\circ}}}{2}} \right)\cos \left( {\dfrac{{{{80}^\circ}}}{2}} \right) \\
   = 2\cos {90^\circ}\cos {40^\circ}...........(4)
 \]
Step 5: Now, to solve the expression (4) as obtained in the solution step 4 we have to substitute the value of $\cos {90^\circ}$as mentioned in the solution hint as (c),
$
   = 2 \times 0 \times \cos {40^\circ} \\
   = 0
 $
Which is equal to the right hand side of the given trigonometric expression.

Hence, with the help of formula (a) we have prove that $\cos {10^\circ} + \cos {110^\circ} + \cos {130^\circ} = 0$

Note: To solve the trigonometric terms like $\cos A,\cos B$ we can use the identities to solve the given expressions in which their terms are to be added or subtracted by themselves.
If we obtain the trigonometric terms like $\cos {30^\circ},\sin {60^\circ}$…….e.t.c so we can simplify them by substituting the values of $\cos {30^\circ},\sin {60^\circ}$ to obtain the final solution or result.