Solveeit Logo

Question

Question: Find the value of \(\cos {10^ \circ } + \cos {110^ \circ } + \cos {130^ \circ }\) ....

Find the value of cos10+cos110+cos130\cos {10^ \circ } + \cos {110^ \circ } + \cos {130^ \circ } .

Explanation

Solution

Hint : The given question deals with basic simplification of trigonometric functions by using some of the simple trigonometric formulae. All the trigonometric formulae, rules and identities must be remembered by heart before solving such a complex question. Basic algebraic rules and simplification techniques are to be kept in mind while doing simplification in the given problem.

Complete step-by-step answer :
In the given problem, we have to find the value of the trigonometric expression cos10+cos110+cos130\cos {10^ \circ } + \cos {110^ \circ } + \cos {130^ \circ }.
We can simplify the trigonometric expression given to us by using the trigonometric formula involving the sum of two cosines of different angles. First, we have to group the cosine functions on which we will apply the identity. So, we get,
(cos10+cos110)+cos130\Rightarrow \left( {\cos {{10}^ \circ } + \cos {{110}^ \circ }} \right) + \cos {130^ \circ }
Now, using the trigonometric identity cosC+cosD=2cos(C+D2)cos(CD2)\cos C + \cos D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right) , we get,
2cos(10+1102)cos(101102)+cos130\Rightarrow 2\cos \left( {\dfrac{{{{10}^ \circ } + {{110}^ \circ }}}{2}} \right)\cos \left( {\dfrac{{{{10}^ \circ } - {{110}^ \circ }}}{2}} \right) + \cos {130^ \circ }
Simplifying the expression, we get,
2cos(1202)cos(1002)+cos130\Rightarrow 2\cos \left( {\dfrac{{{{120}^ \circ }}}{2}} \right)\cos \left( { - \dfrac{{{{100}^ \circ }}}{2}} \right) + \cos {130^ \circ }
2cos(60)cos(50)+cos130\Rightarrow 2\cos \left( {{{60}^ \circ }} \right)\cos \left( { - {{50}^ \circ }} \right) + \cos {130^ \circ }
Now, we know that cos(x)=cosx\cos \left( { - x} \right) = \cos x. So, using this in the expression, we get,
2cos60cos50+cos130\Rightarrow 2\cos {60^ \circ }\cos {50^ \circ } + \cos {130^ \circ }
We also know that the value of cos60\cos {60^ \circ } as (12)\left( {\dfrac{1}{2}} \right). So, we get,
2×12cos50+cos130\Rightarrow 2 \times \dfrac{1}{2}\cos {50^ \circ } + \cos {130^ \circ }
Canceling the common factors in numerator and denominator and using the trigonometric formula cos(180x)=cosx\cos \left( {{{180}^ \circ } - x} \right) = - \cos x, we get,
cos50+cos130\Rightarrow \cos {50^ \circ } + \cos {130^ \circ }
We can write 130{130^ \circ } as 18050{180^ \circ } - {50^ \circ } . So, we get,
cos50+cos(18050)\Rightarrow \cos {50^ \circ } + \cos \left( {{{180}^ \circ } - {{50}^ \circ }} \right)
cos50cos50\Rightarrow \cos {50^ \circ } - \cos {50^ \circ }
So, cancelling the like terms with opposite signs, we get,
0\Rightarrow 0
Hence, the value of cos10+cos110+cos130\cos {10^ \circ } + \cos {110^ \circ } + \cos {130^ \circ } is zero.

Note : Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such types of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations. However, questions involving this type of simplification of trigonometric ratios may also have multiple interconvertible answers.