Question

Prove the trigonometric expression: $\cos {{20}^{o}}+\cos {{100}^{o}}+\cos {{140}^{o}}=0$

Answer 1

Hint: Use the trigonometric identity of $\cos C+\cos D$ which is given as
$\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$. And hence simplify it further to prove the given relation.

Complete step-by-step answer:

To prove
$\cos {{20}^{o}}+\cos {{100}^{o}}+\cos {{140}^{o}}=0$ …………. (i)
Let us solve the LHS of equation (i) to prove the given relation.
So, we have LHS of equation (i) as
$LHS=\cos {{20}^{o}}+\cos {{100}^{o}}+\cos {{140}^{o}}$……….. (ii)
Now, let us convert the above angle trigonometry functions to acute angles by using the relation
$\cos \left( {{180}^{o}}-\theta \right)=-\cos \theta $……………… (iii)
Hence, $\cos {{100}^{o}}$ can be written as $\cos \left( {{180}^{o}}-{{80}^{o}} \right)$from equation iii, so, we get
$\cos {{100}^{o}}=\cos \left( {{180}^{o}}-{{80}^{o}} \right)=-\cos {{80}^{o}}$ ……………… (iv)
Similarly, we can convert $\cos {{140}^{o}}$ to $\cos \left( {{180}^{o}}-{{40}^{o}} \right)$and hence, we get
$cos{{140}^{o}}=\cos \left( {{180}^{o}}-{{40}^{o}} \right)=-\cos {{40}^{o}}$ ………….. (v)
Now, we can replace $\cos {{100}^{o}}$ and $\cos {{140}^{o}}$ to $-\cos {{80}^{o}}$ and $-\cos {{40}^{o}}$respectively in equation (ii); hence, we get
$\begin{align}
  & LHS=\cos {{20}^{o}}-\cos {{80}^{o}}-\cos {{40}^{o}} \\
 & \Rightarrow LHS=\cos {{20}^{o}}-\left( \cos {{80}^{o}}+\cos {{40}^{o}} \right)............(vi) \\
\end{align}$
Now, we can use trigonometric identity of $\operatorname{cosC}+cosD$which can be given as\[\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)\]……………..(vii)
Now, we can simplify the equation (vi), hence, we get
$\begin{align}
  & LHS=\cos {{20}^{o}}-\left( 2\cos \left( \dfrac{80+40}{2} \right)\cos \left( \dfrac{80-40}{2} \right) \right) \\
 & LHS=\cos {{20}^{o}}-2\cos {{60}^{o}}\cos {{20}^{o}} \\
\end{align}$
Now, we know the value of $\cos {{60}^{o}}$ is $'\dfrac{1}{2}'$. So, above relation can be given as
$\begin{align}
  & LHS=\cos {{20}^{o}}-2\times \dfrac{1}{2}\cos {{20}^{o}} \\
 & LHS=\cos {{20}^{o}}-\cos {{20}^{o}} \\
 & LHS=0 \\
\end{align}$
So, we can observe the value of LHS and RHS that both are ‘0’.
Hence, LHS = RHS = 0.
Hence, proved the given relation that $\cos {{20}^{o}}+\cos {{100}^{0}}+\cos {{140}^{0}}=0$.

Note: We can convert $\cos 100$ and $\cos 140$to $-\sin 10$ and $-\sin 50$by applying identity $\cos \left( 90+\theta \right)=-\sin \theta $. Hence, apply the formula $\sin C+\sin D=2\sin \dfrac{C+D}{2}.\cos \dfrac{C-D}{2}$ to get the answer.
One can get confused with the formula of $\cos C+\cos D$. He/she may apply formula as $2\sin \dfrac{C+D}{2}\sin \dfrac{C-D}{2}$ Or $2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$ which are wrong; so be clear with the trigonometric identities to get the accurate solution.
One can try to calculate exact values of $\cos {{20}^{o}}$, $\cos {{100}^{o}}$ and $\cos {{140}^{o}}$ which is the waste of time and wrong approach as well.
So, try to use trigonometric identities to simplify the given expression. Getting values of given trigonometric functions is a complex task, so go by concept not waste your time.