Question

Prove that \[\cos {{35}^{\circ }}+\cos {{85}^{\circ }}+\cos {{155}^{\circ }}=0\].

Answer 1

Hint: In order to find solution to this problem, we will use compound angle formula that is $\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cdot \cos \left( \dfrac{A-B}{2} \right)$ in our trigonometric equation to prove the right-hand side. First, we will use our compound angle formula in our two terms and then after solving it we will get our answer.

Complete step-by-step solution:
We have our trigonometric equation as:
\[\Rightarrow \cos {{35}^{\circ }}+\cos {{85}^{\circ }}+\cos {{155}^{\circ }}\]
On applying compound angle formula in our above trigonometric equation and taking $\cos A=\cos {{35}^{\circ }}$ and $\cos B=\cos {{85}^{\circ }}$, we get our equation as:
\[\Rightarrow 2\cos \left( \dfrac{{{35}^{\circ }}+{{85}^{\circ }}}{2} \right)\cdot \cos \left( \dfrac{{{35}^{\circ }}-{{85}^{\circ }}}{2} \right)+\cos {{155}^{\circ }}\]
On simplification, we get our equation as:
\[\Rightarrow 2\cos \left( \dfrac{{{120}^{\circ }}}{2} \right)\cdot \cos \left( \dfrac{-{{50}^{\circ }}}{2} \right)+\cos {{155}^{\circ }}\]
On simplifying, we get our equation as:
\[\Rightarrow 2\cos \left( {{60}^{\circ }} \right)\cdot \cos \left( -{{25}^{\circ }} \right)+\cos {{155}^{\circ }}\]
As we know that $\cos {{60}^{\circ }}=\dfrac{1}{2}$ from trigonometric functions.
Therefore, on applying in our equation, we get:
\[\Rightarrow 2\times \dfrac{1}{2}\cdot \cos \left( -{{25}^{\circ }} \right)+\cos {{155}^{\circ }}\]
On simplifying, we get:
\[\Rightarrow \cos \left( -{{25}^{\circ }} \right)+\cos {{155}^{\circ }}\]
As we know the rule, $\cos \left( -A \right)=\cos A$.
Therefore, on applying in our trigonometric equation, we get:
\[\Rightarrow \cos \left( {{25}^{\circ }} \right)+\cos {{155}^{\circ }}\]
Now, as we can see that we can again apply compound angle formula, therefore, we will use compound angle formula.
$\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cdot \cos \left( \dfrac{A-B}{2} \right)$
On applying, we get:
$\Rightarrow 2\cos \left( \dfrac{{{25}^{\circ }}+{{155}^{\circ }}}{2} \right)\cdot \cos \left( \dfrac{{{25}^{\circ }}-{{155}^{\circ }}}{2} \right)$
Now, on simplifying our equation, we get:
$\Rightarrow 2\cos \left( \dfrac{{{180}^{\circ }}}{2} \right)\cdot \cos \left( \dfrac{-{{130}^{\circ }}}{2} \right)$
On simplifying, we get:
$\Rightarrow 2\cos \left( {{90}^{\circ }} \right)\cdot \cos \left( -{{65}^{\circ }} \right)$
Now, as we know that $\cos {{90}^{\circ }}=0$ from trigonometric functions.
Therefore, on applying in our equation, we get:
$\Rightarrow 2\left( 0 \right)\cdot \cos \left( -{{65}^{\circ }} \right)$
On simplifying, we get:
$\Rightarrow 0\cdot \cos \left( -{{65}^{\circ }} \right)$
On further simplification, we get our solution as:
$\Rightarrow 0$, which is equal to the right-hand side, hence proved.
Therefore, we can say that \[\cos {{35}^{\circ }}+\cos {{85}^{\circ }}+\cos {{155}^{\circ }}=0\].

Note: There are three basic trigonometry functions are Sine, Cosine and Tangent.
The angles of sine, cosine, and tangent are the primary classification of functions of trigonometry. And the three functions which are cotangent, secant and cosecant can be derived from the primary functions.
In order to solve trigonometric function problems, we have to be familiar with the concepts of trigonometric functions.