Find the value of the following trigonometric equation.
\[{\text{cos 2}}{{\text{0}}^{\text{0}}}{\text{ + cos 10}}{{\text{0}}^{\text{0}}}{\text{ + cos 14}}{{\text{0}}^{\text{0}}}\]
Answer
363k+ views
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.
Complete step-by-step answer:
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.
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.
Complete step-by-step answer:
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.
Last updated date: 26th Sep 2023
•
Total views: 363k
•
Views today: 4.63k
Recently Updated Pages
What do you mean by public facilities

Paragraph on Friendship

Slogan on Noise Pollution

Disadvantages of Advertising

Prepare a Pocket Guide on First Aid for your School

10 Slogans on Save the Tiger

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

One cusec is equal to how many liters class 8 maths CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

What is the color of ferrous sulphate crystals? How does this color change after heating? Name the products formed on strongly heating ferrous sulphate crystals. What type of chemical reaction occurs in this type of change.

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE
