What is $\cos {{24}^{\circ }}+\cos {{5}^{\circ }}+\cos {{175}^{\circ }}+\cos {{204}^{\circ }}+\cos {{300}^{\circ }}=$
a) $\dfrac{1}{2}$
b) $\dfrac{-1}{2}$
c) $\dfrac{\sqrt{3}}{2}$
d) 1
Answer
683.4k+ views
Hint: Change the cosine terms with the higher angles $\left( >{{90}^{\circ }} \right)$ to the acute values. Use the following identities for the conversion:
$\begin{align}
& \cos \left( 180-\theta \right)=-\cos \theta \\
& \cos \left( 180+\theta \right)=-\cos \theta \\
& \cos \left( 300-\theta \right)=\cos \theta \\
\end{align}$
And, put the value of $\cos {{60}^{\circ }}\Rightarrow \dfrac{1}{2}$ wherever required.
Complete step-by-step answer:
Expression given in the question is
$\cos {{24}^{\circ }}+\cos {{5}^{\circ }}+\cos {{175}^{\circ }}+\cos {{204}^{\circ }}+\cos {{300}^{\circ }}=?$
Let the value of the given expression in the problem be ‘M’. So, we have
$M=\cos {{24}^{\circ }}+\cos {{5}^{\circ }}+\cos {{175}^{\circ }}+\cos {{204}^{\circ }}+\cos {{300}^{\circ }}.........\left( i \right)$
As we can observe that we do not know the exact values of the cosine terms given in the above equation. So, we need to observe the relation and have to apply some trigonometric identity. So, let us try to convert all the angles involved in the expression to acute angles with the help of identities expressed below
\[\begin{align}
& \cos \left( 180-\theta \right)=-\cos \theta .............\left( ii \right) \\
& \cos \left( 180+\theta \right)=-\cos \theta .............\left( iii \right) \\
& \cos \left( 300-\theta \right)=\cos \theta ...............\left( iv \right) \\
\end{align}\]
As we can observe that $\cos {{24}^{\circ }},\cos {{5}^{\circ }}$ has involved only acute angles $\left( {{0}^{\circ }}\to {{90}^{\circ }} \right)$. So, we do not need to convert them. We need to use the identities mentioned above with the terms \[\cos {{175}^{\circ }},\cos {{204}^{\circ }},\cos {{300}^{\circ }}\] . So, let us convert them one by one as we know we can write \[\cos {{175}^{\circ }}\Rightarrow \cos \left( {{180}^{\circ }}-{{5}^{\circ }} \right)\] and use the identity in the equation (ii) as
\[\cos {{175}^{\circ }}=\cos \left( 180-{{5}^{\circ }} \right)=-\cos {{5}^{\circ }}\] So, we get
\[\cos {{175}^{\circ }}=-\cos {{5}^{\circ }}...............\left( v \right)\]
For \[\cos {{204}^{\circ }}\] , we can rewrite it as $\cos \left( {{180}^{\circ }}+{{24}^{\circ }} \right)$ and can apply identity of the equation (iii) so, we get
\[\begin{align}
& \cos {{204}^{\circ }}=\cos \left( {{180}^{\circ }}+{{24}^{\circ }} \right)=-\cos {{24}^{\circ }}, \\
& \cos {{204}^{\circ }}=-\cos {{24}^{\circ }}................\left( vi \right) \\
\end{align}\]
Now, similarly \[\cos {{300}^{\circ }}\] can be written as $\cos \left( {{360}^{\circ }}-{{60}^{\circ }} \right)$ and we can use identity (iv) to simplify it as
$\cos {{300}^{\circ }}=\cos \left( {{360}^{\circ }}-{{60}^{\circ }} \right)=\cos {{60}^{\circ }}.................\left( vii \right)$
Now, we can replace the terms
$\begin{align}
& \cos {{204}^{\circ }},\cos {{300}^{\circ }} \\
& \Rightarrow -\cos {{5}^{\circ }},-\cos {{24}^{\circ }},\cos {{60}^{\circ }} \\
\end{align}$
Respectively from the equation (v), (vi) and (vi) with the equation (i). Hence, we get
$\begin{align}
& M=\cos {{24}^{\circ }}+\cos {{5}^{\circ }}-\cos {{5}^{\circ }}-\cos {{24}^{\circ }}+\cos {{60}^{\circ }} \\
& \Rightarrow M=\cos {{24}^{\circ }}-\cos {{24}^{\circ }}+\cos {{60}^{\circ }} \\
& \Rightarrow M=\cos {{60}^{\circ }} \\
\end{align}$
Now, we know the value of$\cos {{60}^{\circ }}\Rightarrow \dfrac{1}{2}$ , hence we get value of M is $\dfrac{1}{2}$ . So, value of$\cos {{24}^{\circ }}+\cos {{5}^{\circ }}+\cos {{175}^{\circ }}+\cos {{204}^{\circ }}+\cos {{300}^{\circ }}=\dfrac{1}{2}$ .
So, option (a) is correct.
Note: One may go wrong with the identities expressed in the solution. We don’t need to remember these identities as these can be calculated by the quadrant rule for the trigonometric functions. So, be clear with these identities to solve these kinds of questions and always try to convert higher angles in the given expression of these kinds to acute angles with the help of identities of trigonometric.
Don’t go for calculating exact values of the given trigonometric functions. Try to observe the relation between them i.e. sum of 5 and 175 is ${{180}^{\circ }}$ and difference of 204, 24 is${{180}^{\circ }}$ and difference of ${{360}^{\circ }}$ and 300 is${{60}^{\circ }}$ . These are the key points of the given question.
$\begin{align}
& \cos \left( 180-\theta \right)=-\cos \theta \\
& \cos \left( 180+\theta \right)=-\cos \theta \\
& \cos \left( 300-\theta \right)=\cos \theta \\
\end{align}$
And, put the value of $\cos {{60}^{\circ }}\Rightarrow \dfrac{1}{2}$ wherever required.
Complete step-by-step answer:
Expression given in the question is
$\cos {{24}^{\circ }}+\cos {{5}^{\circ }}+\cos {{175}^{\circ }}+\cos {{204}^{\circ }}+\cos {{300}^{\circ }}=?$
Let the value of the given expression in the problem be ‘M’. So, we have
$M=\cos {{24}^{\circ }}+\cos {{5}^{\circ }}+\cos {{175}^{\circ }}+\cos {{204}^{\circ }}+\cos {{300}^{\circ }}.........\left( i \right)$
As we can observe that we do not know the exact values of the cosine terms given in the above equation. So, we need to observe the relation and have to apply some trigonometric identity. So, let us try to convert all the angles involved in the expression to acute angles with the help of identities expressed below
\[\begin{align}
& \cos \left( 180-\theta \right)=-\cos \theta .............\left( ii \right) \\
& \cos \left( 180+\theta \right)=-\cos \theta .............\left( iii \right) \\
& \cos \left( 300-\theta \right)=\cos \theta ...............\left( iv \right) \\
\end{align}\]
As we can observe that $\cos {{24}^{\circ }},\cos {{5}^{\circ }}$ has involved only acute angles $\left( {{0}^{\circ }}\to {{90}^{\circ }} \right)$. So, we do not need to convert them. We need to use the identities mentioned above with the terms \[\cos {{175}^{\circ }},\cos {{204}^{\circ }},\cos {{300}^{\circ }}\] . So, let us convert them one by one as we know we can write \[\cos {{175}^{\circ }}\Rightarrow \cos \left( {{180}^{\circ }}-{{5}^{\circ }} \right)\] and use the identity in the equation (ii) as
\[\cos {{175}^{\circ }}=\cos \left( 180-{{5}^{\circ }} \right)=-\cos {{5}^{\circ }}\] So, we get
\[\cos {{175}^{\circ }}=-\cos {{5}^{\circ }}...............\left( v \right)\]
For \[\cos {{204}^{\circ }}\] , we can rewrite it as $\cos \left( {{180}^{\circ }}+{{24}^{\circ }} \right)$ and can apply identity of the equation (iii) so, we get
\[\begin{align}
& \cos {{204}^{\circ }}=\cos \left( {{180}^{\circ }}+{{24}^{\circ }} \right)=-\cos {{24}^{\circ }}, \\
& \cos {{204}^{\circ }}=-\cos {{24}^{\circ }}................\left( vi \right) \\
\end{align}\]
Now, similarly \[\cos {{300}^{\circ }}\] can be written as $\cos \left( {{360}^{\circ }}-{{60}^{\circ }} \right)$ and we can use identity (iv) to simplify it as
$\cos {{300}^{\circ }}=\cos \left( {{360}^{\circ }}-{{60}^{\circ }} \right)=\cos {{60}^{\circ }}.................\left( vii \right)$
Now, we can replace the terms
$\begin{align}
& \cos {{204}^{\circ }},\cos {{300}^{\circ }} \\
& \Rightarrow -\cos {{5}^{\circ }},-\cos {{24}^{\circ }},\cos {{60}^{\circ }} \\
\end{align}$
Respectively from the equation (v), (vi) and (vi) with the equation (i). Hence, we get
$\begin{align}
& M=\cos {{24}^{\circ }}+\cos {{5}^{\circ }}-\cos {{5}^{\circ }}-\cos {{24}^{\circ }}+\cos {{60}^{\circ }} \\
& \Rightarrow M=\cos {{24}^{\circ }}-\cos {{24}^{\circ }}+\cos {{60}^{\circ }} \\
& \Rightarrow M=\cos {{60}^{\circ }} \\
\end{align}$
Now, we know the value of$\cos {{60}^{\circ }}\Rightarrow \dfrac{1}{2}$ , hence we get value of M is $\dfrac{1}{2}$ . So, value of$\cos {{24}^{\circ }}+\cos {{5}^{\circ }}+\cos {{175}^{\circ }}+\cos {{204}^{\circ }}+\cos {{300}^{\circ }}=\dfrac{1}{2}$ .
So, option (a) is correct.
Note: One may go wrong with the identities expressed in the solution. We don’t need to remember these identities as these can be calculated by the quadrant rule for the trigonometric functions. So, be clear with these identities to solve these kinds of questions and always try to convert higher angles in the given expression of these kinds to acute angles with the help of identities of trigonometric.
Don’t go for calculating exact values of the given trigonometric functions. Try to observe the relation between them i.e. sum of 5 and 175 is ${{180}^{\circ }}$ and difference of 204, 24 is${{180}^{\circ }}$ and difference of ${{360}^{\circ }}$ and 300 is${{60}^{\circ }}$ . These are the key points of the given question.
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