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# Find the value of the following trigonometric equation.${\text{cos 2}}{{\text{0}}^{\text{0}}}{\text{ + cos 10}}{{\text{0}}^{\text{0}}}{\text{ + cos 14}}{{\text{0}}^{\text{0}}}$

Last updated date: 22nd Jul 2024
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Hint: Let us apply the trigonometric identity for cos A + cos B. So, that above equation can be reduced into a product of trigonometric functions.
Now to find the value of the given trigonometric equation first we will apply a trigonometric formula to find the sum of cosines of two angles.

As we know that if A and B are some angles, then cos A + cos B = 2cos$\left( {\dfrac{{{\text{A + B}}}}{{\text{2}}}} \right)$ cos$\left( {\dfrac{{{\text{A - B}}}}{{\text{2}}}} \right)$.
So, if A = ${\text{10}}{{\text{0}}^{\text{0}}}$ and B = ${\text{2}}{{\text{0}}^{\text{0}}}$.
Then the given equation becomes,
$\Rightarrow$ $\left( {{\text{cos 10}}{{\text{0}}^{\text{0}}}{\text{ + cos 2}}{{\text{0}}^{\text{0}}}} \right){\text{ + cos 14}}{{\text{0}}^{\text{0}}}{\text{ = }}\left( {{\text{2cos}}\left( {\dfrac{{{\text{10}}{{\text{0}}^0}{\text{ + 2}}{{\text{0}}^0}}}{{\text{2}}}} \right){\text{ cos}}\left( {\dfrac{{{\text{10}}{{\text{0}}^0}{\text{ - 2}}{{\text{0}}^0}}}{{\text{2}}}} \right)} \right){\text{ + cos 14}}{{\text{0}}^{\text{0}}}$ ___________(1)
On solving equation 1. We get,
$\Rightarrow$ $\left( {{\text{cos 10}}{{\text{0}}^{\text{0}}}{\text{ + cos 2}}{{\text{0}}^{\text{0}}}} \right){\text{ + cos 14}}{{\text{0}}^{\text{0}}}{\text{ = 2}}\left( {{\text{cos 6}}{{\text{0}}^{\text{0}}}} \right)\left( {{\text{cos 4}}{{\text{0}}^{\text{0}}}} \right){\text{ + cos 14}}{{\text{0}}^{\text{0}}}$ _________________________ (2)
Now as we know that ${\text{cos 6}}{{\text{0}}^{\text{0}}}$ = $\dfrac{{\text{1}}}{{\text{2}}}$.
So, equation 2 becomes,
$\Rightarrow$ $\left( {{\text{cos 10}}{{\text{0}}^{\text{0}}}{\text{ + cos 2}}{{\text{0}}^{\text{0}}}} \right){\text{ + cos 14}}{{\text{0}}^{\text{0}}}{\text{ = 2}}\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right)\left( {{\text{cos 4}}{{\text{0}}^{\text{0}}}} \right){\text{ + cos 14}}{{\text{0}}^{\text{0}}}$
$\Rightarrow$ $\left( {{\text{cos 10}}{{\text{0}}^{\text{0}}}{\text{ + cos 2}}{{\text{0}}^{\text{0}}}} \right){\text{ + cos 14}}{{\text{0}}^{\text{0}}}{\text{ = cos 4}}{{\text{0}}^{\text{0}}}{\text{ + cos 14}}{{\text{0}}^{\text{0}}}$ ________________________ (3)
Now as we know that to find the value of given equation, we had to find the value of ${\text{cos 4}}{{\text{0}}^{\text{0}}}{\text{ + cos 14}}{{\text{0}}^{\text{0}}}$.
So, applying the trigonometric identity of cos A + cos B in ${\text{cos 14}}{{\text{0}}^{\text{0}}}{\text{ + cos 4}}{{\text{0}}^{\text{0}}}$.
So, let A = ${\text{14}}{{\text{0}}^{\text{0}}}$ and B = ${\text{4}}{{\text{0}}^{\text{0}}}$.
So, equation 3 becomes,
$\Rightarrow$ $\left( {{\text{cos 10}}{{\text{0}}^{\text{0}}}{\text{ + cos 2}}{{\text{0}}^{\text{0}}}} \right){\text{ + cos 14}}{{\text{0}}^{\text{0}}}{\text{ = }}\left( {{\text{2cos}}\left( {\dfrac{{{\text{14}}{{\text{0}}^{\text{0}}}{\text{ + 4}}{{\text{0}}^{\text{0}}}}}{{\text{2}}}} \right){\text{ cos}}\left( {\dfrac{{{\text{14}}{{\text{0}}^{\text{0}}}{\text{ - 4}}{{\text{0}}^{\text{0}}}}}{{\text{2}}}} \right)} \right)$
On solving the above equation. It becomes,
$\Rightarrow$ $\left( {{\text{cos 10}}{{\text{0}}^{\text{0}}}{\text{ + cos 2}}{{\text{0}}^{\text{0}}}} \right){\text{ + cos 14}}{{\text{0}}^{\text{0}}}{\text{ = 2}}\left( {{\text{cos 9}}{{\text{0}}^{\text{0}}}} \right)\left( {{\text{cos 5}}{{\text{0}}^{\text{0}}}} \right)$ __________________________ (4)
Now as we know that ${\text{cos 9}}{{\text{0}}^{\text{0}}}$ = 0.
So, equation 4 becomes,
$\Rightarrow$ ${\text{cos 2}}{{\text{0}}^{\text{0}}}{\text{ + cos 10}}{{\text{0}}^{\text{0}}}{\text{ + cos 14}}{{\text{0}}^{\text{0}}}$ = 0
Hence, ${\text{cos 2}}{{\text{0}}^{\text{0}}}{\text{ + cos 10}}{{\text{0}}^{\text{0}}}{\text{ + cos 14}}{{\text{0}}^{\text{0}}}$ = 0.

Note: Whenever we come up with this type of problem, where we are asked to find the sum of cosine of three different angles then first, we find the sum of cosine of any two angle from them using trigonometric identity which states that cos A + cos B = 2cos$\left( {\dfrac{{{\text{A + B}}}}{{\text{2}}}} \right)$ cos$\left( {\dfrac{{{\text{A - B}}}}{{\text{2}}}} \right)$. And then we are left to find the sum of the cosine of the third angle and the cosine of angles we got from the previous sum. So, we had to again apply the trigonometric identity of cos A + cos B to get the required value of the given equation.