Answer
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Hint: Here we have to solve the trigonometric expression. Here we will use the trigonometric identities to find the value of the above trigonometric expression. We will simplify the terms one by one using trigonometric identities i.e. we will use sum to product trigonometric identities and product trigonometric identities to simplify the above expression.
Formula used:
We will use the following formulas:
1. \[\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)\]
2. \[\cos A \cdot \cos B = \dfrac{{\cos \left( {A + B} \right) + \cos \left( {A - B} \right)}}{2}\]
Complete Complete Step by Step Solution:
A. \[\cos 20^\circ + \cos 80^\circ - \sqrt 3 \cos 50^\circ \]
Therefore, we will use the trigonometric identity, \[\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)\], for finding the sum \[\cos 20^\circ + \cos 80^\circ \].
Therefore,
\[ \Rightarrow \cos 20^\circ + \cos 80^\circ - \sqrt 3 \cos 50^\circ = 2\cos \left( {\dfrac{{80^\circ + 20^\circ }}{2}} \right)\cos \left( {\dfrac{{80^\circ - 20^\circ }}{2}} \right) - \sqrt 3 \cos {50^ \circ }\]
Simplifying the terms inside the bracket, we get
\[ \Rightarrow \cos 20^\circ + \cos 80^\circ - \sqrt 3 \cos 50^\circ = 2 \times \cos {50^ \circ } \times \cos {30^ \circ } - \sqrt 3 \cos {50^ \circ }\]
We know the value of \[\cos {30^ \circ }\] is \[\dfrac{{\sqrt 3 }}{2}\].
Therefore, putting the value of \[\cos {30^ \circ }\]here, we get
\[ \Rightarrow \cos 20^\circ + \cos 80^\circ - \sqrt 3 \cos 50^\circ = 2 \times \cos {50^ \circ } \times \dfrac{{\sqrt 3 }}{2} - \sqrt 3 \cos {50^ \circ }\]
On simplifying the terms, we get
\[ \Rightarrow \cos 20^\circ + \cos 80^\circ - \sqrt 3 \cos 50^\circ = \sqrt 3 \cos {50^ \circ } - \sqrt 3 \cos {50^ \circ } = 0\]
Hence, option (S) is the correct option for (A).
B. \[\cos 0 + \cos \dfrac{\pi }{7} + \cos \dfrac{{2\pi }}{7} + \cos \dfrac{{3\pi }}{7} + \cos \dfrac{{4\pi }}{7} + \cos \dfrac{{5\pi }}{7} + \cos \dfrac{{6\pi }}{7}\]
Let \[\cos 0 + \cos \dfrac{\pi }{7} + \cos \dfrac{{2\pi }}{7} + \cos \dfrac{{3\pi }}{7} + \cos \dfrac{{4\pi }}{7} + \cos \dfrac{{5\pi }}{7} + \cos \dfrac{{6\pi }}{7} = S\]
We can write \[\cos \dfrac{{4\pi }}{7}\] as \[\cos \left( {\pi - \dfrac{{3\pi }}{7}} \right)\]. Similarly, we can write \[\cos \dfrac{{5\pi }}{7}\] as \[\cos \left( {\pi - \dfrac{{2\pi }}{7}} \right)\] and we can write \[\cos \dfrac{{6\pi }}{7}\] as \[\cos \left( {\pi - \dfrac{\pi }{7}} \right)\].
Substituting these values, we get the expression as
\[S = \cos 0 + \cos \dfrac{\pi }{7} + \cos \dfrac{{2\pi }}{7} + \cos \dfrac{{3\pi }}{7} + \cos \left( {\pi - \dfrac{{3\pi }}{7}} \right) + \cos \left( {\pi - \dfrac{{2\pi }}{7}} \right) + \cos \left( {\pi - \dfrac{\pi }{7}} \right)\]
We know that \[\cos \left( {\pi - A} \right) = - \cos A\]
Therefore, using this identity in the given expression, we get
\[\begin{array}{l} \Rightarrow S = \cos 0 + \cos \dfrac{\pi }{7} + \cos \dfrac{{2\pi }}{7} + \cos \dfrac{{3\pi }}{7} - \cos \dfrac{\pi }{7} - \cos \dfrac{{2\pi }}{7} - \cos \dfrac{{3\pi }}{7}\\ \Rightarrow S = \cos 0\\ \Rightarrow S = 1\end{array}\]
So, \[\cos 0 + \cos \dfrac{\pi }{7} + \cos \dfrac{{2\pi }}{7} + \cos \dfrac{{3\pi }}{7} + \cos \dfrac{{4\pi }}{7} + \cos \dfrac{{5\pi }}{7} + \cos \dfrac{{6\pi }}{7} = 1\]
Hence, option (R) is the correct option for (B).
C. \[\cos {20^ \circ } + \cos {40^ \circ } + \cos {60^ \circ } - 4\cos {10^ \circ }\cos {20^ \circ }\cos {30^ \circ }\]
Let \[\cos {20^ \circ } + \cos {40^ \circ } + \cos {60^ \circ } - 4\cos {10^ \circ }\cos {20^ \circ }\cos {30^ \circ } = T\]
We know from trigonometric identities;
\[\begin{array}{l}\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)\\\cos A.\cos B = \dfrac{{\cos \left( {A + B} \right) + \cos \left( {A - B} \right)}}{2}\end{array}\]
Using this same identity for the sum \[\cos {20^\circ } + \cos {40^\circ }\] and for the product \[\cos {10^\circ }\cos {20^\circ }\] in the expression, we get
\[ \Rightarrow T = 2\cos \left( {\dfrac{{{{20}^ \circ } + {{40}^ \circ }}}{2}} \right)\cos \left( {\dfrac{{{{40}^ \circ } - {{20}^ \circ }}}{2}} \right) + \cos {60^ \circ } - 4\dfrac{{\cos \left( {{{20}^ \circ } + {{10}^ \circ }} \right) + \cos \left( {{{20}^ \circ } - {{10}^ \circ }} \right)}}{2}\cos {30^ \circ }\]
Simplifying the terms inside the bracket, we get
\[ \Rightarrow T = 2\cos {30^ \circ }\cos {10^ \circ } + \cos {60^ \circ } - 2\left( {\cos {{30}^ \circ } + \cos {{10}^ \circ }} \right)\cos {30^ \circ }\]
We know that \[\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}\] and \[\cos {60^ \circ } = \dfrac{1}{2}\].
Substituting these values in the expression, we get
\[ \Rightarrow T = 2\dfrac{{\sqrt 3 }}{2}\cos {10^ \circ } + \dfrac{1}{2} - 2\left( {\dfrac{{\sqrt 3 }}{2} + \cos {{10}^ \circ }} \right)\dfrac{{\sqrt 3 }}{2}\]
On further simplification, we get
\[ \Rightarrow T = \sqrt 3 \cos {10^ \circ } + \dfrac{1}{2} - \dfrac{3}{2} - \sqrt 3 \cos {10^ \circ }\]
Simplifying the likes terms, we get
\[ \Rightarrow T = - 1\]
So, \[\cos {20^ \circ } + \cos {40^ \circ } + \cos {60^ \circ } - 4\cos {10^ \circ }\cos {20^ \circ }\cos {30^ \circ } = - 1\]
Hence, option (P) is the correct option for (C).
D. \[\cos {20^ \circ }\cos {100^ \circ } + \cos {100^ \circ }\cos {140^ \circ } - \cos {140^ \circ }\cos {200^ \circ }\]
Let \[\cos {20^ \circ }\cos {100^ \circ } + \cos {100^ \circ }\cos {140^ \circ } - \cos {140^ \circ }\cos {200^ \circ } = U\]
We know from trigonometric identities;
\[\cos A.\cos B = \dfrac{{\cos \left( {A + B} \right) + \cos \left( {A - B} \right)}}{2}\]
Using this same identity for the entire product in the expression, we get
\[ \Rightarrow U = \dfrac{{\cos {{120}^ \circ } + \cos {{80}^ \circ }}}{2} + \dfrac{{\cos {{240}^ \circ } + \cos {{40}^ \circ }}}{2} - \dfrac{{\cos {{340}^ \circ } + \cos {{60}^ \circ }}}{2}\]
Simplifying the terms further, we get
\[ \Rightarrow U = \dfrac{1}{2}\left( {\cos {{120}^ \circ } + \cos {{80}^ \circ } + \cos {{240}^ \circ } + \cos {{40}^ \circ } - \cos {{340}^ \circ } - \cos {{60}^ \circ }} \right)\]
We know that
\[\begin{array}{l}\cos {120^ \circ } = - \dfrac{1}{2}\\\cos {60^ \circ } = \dfrac{1}{2}\end{array}\]
We will put these values here.
\[ \Rightarrow U = \dfrac{1}{2}\left( { - \dfrac{1}{2} + \cos {{80}^ \circ } + \cos {{240}^ \circ } + \cos {{40}^ \circ } - \cos {{340}^ \circ } - \dfrac{1}{2}} \right)\]
We can know, \[\cos {240^ \circ } = - \dfrac{1}{2}\]
\[ \Rightarrow U = \dfrac{1}{2}\left( { - \dfrac{1}{2} + \cos {{80}^ \circ } - \dfrac{1}{2} + \cos {{40}^ \circ } - \cos {{340}^ \circ } - \dfrac{1}{2}} \right)\]
Applying trigonometric identities for the sum \[\cos {80^ \circ } + \cos {40^ \circ }\] , we get
\[ \Rightarrow U = \dfrac{1}{2}\left( { - \dfrac{1}{2} + \cos \left( {\dfrac{{{{80}^ \circ } + {{40}^ \circ }}}{2}} \right)\cos \left( {\dfrac{{{{80}^ \circ } + {{40}^ \circ }}}{2}} \right) - \dfrac{1}{2} - \cos {{340}^ \circ } - \dfrac{1}{2}} \right)\]
Simplifying the terms inside the bracket, we get
\[\begin{array}{l} \Rightarrow U = \dfrac{1}{2}\left( { - \dfrac{1}{2} + \cos {{20}^ \circ } + \dfrac{1}{2} - \dfrac{1}{2} - \cos \left( {{{180}^ \circ } \times 2 - {{20}^ \circ }} \right) - \dfrac{1}{2}} \right)\\ \Rightarrow U = \dfrac{1}{2}\left( { - \dfrac{1}{2} + \cos {{20}^ \circ } + \dfrac{1}{2} - \dfrac{1}{2} - \cos {{20}^ \circ } - \dfrac{1}{2}} \right)\end{array}\]
Adding and subtracting the terms, we get
\[ \Rightarrow U = - \dfrac{3}{4}\]
So, \[\cos {20^ \circ }\cos {100^ \circ } + \cos {100^ \circ }\cos {140^ \circ } - \cos {140^ \circ }\cos {200^ \circ } = - \dfrac{3}{4}\]
Hence, option (Q) is the correct option for (D).
Note:
We have used many trigonometric identities to solve this problem of trigonometry. Trigonometric identities are defined as equalities that include trigonometric functions, and for every value of variables, these are true. We break the complex terms of trigonometric functions using product and sum to product identities of trigonometry.
Formula used:
We will use the following formulas:
1. \[\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)\]
2. \[\cos A \cdot \cos B = \dfrac{{\cos \left( {A + B} \right) + \cos \left( {A - B} \right)}}{2}\]
Complete Complete Step by Step Solution:
A. \[\cos 20^\circ + \cos 80^\circ - \sqrt 3 \cos 50^\circ \]
Therefore, we will use the trigonometric identity, \[\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)\], for finding the sum \[\cos 20^\circ + \cos 80^\circ \].
Therefore,
\[ \Rightarrow \cos 20^\circ + \cos 80^\circ - \sqrt 3 \cos 50^\circ = 2\cos \left( {\dfrac{{80^\circ + 20^\circ }}{2}} \right)\cos \left( {\dfrac{{80^\circ - 20^\circ }}{2}} \right) - \sqrt 3 \cos {50^ \circ }\]
Simplifying the terms inside the bracket, we get
\[ \Rightarrow \cos 20^\circ + \cos 80^\circ - \sqrt 3 \cos 50^\circ = 2 \times \cos {50^ \circ } \times \cos {30^ \circ } - \sqrt 3 \cos {50^ \circ }\]
We know the value of \[\cos {30^ \circ }\] is \[\dfrac{{\sqrt 3 }}{2}\].
Therefore, putting the value of \[\cos {30^ \circ }\]here, we get
\[ \Rightarrow \cos 20^\circ + \cos 80^\circ - \sqrt 3 \cos 50^\circ = 2 \times \cos {50^ \circ } \times \dfrac{{\sqrt 3 }}{2} - \sqrt 3 \cos {50^ \circ }\]
On simplifying the terms, we get
\[ \Rightarrow \cos 20^\circ + \cos 80^\circ - \sqrt 3 \cos 50^\circ = \sqrt 3 \cos {50^ \circ } - \sqrt 3 \cos {50^ \circ } = 0\]
Hence, option (S) is the correct option for (A).
B. \[\cos 0 + \cos \dfrac{\pi }{7} + \cos \dfrac{{2\pi }}{7} + \cos \dfrac{{3\pi }}{7} + \cos \dfrac{{4\pi }}{7} + \cos \dfrac{{5\pi }}{7} + \cos \dfrac{{6\pi }}{7}\]
Let \[\cos 0 + \cos \dfrac{\pi }{7} + \cos \dfrac{{2\pi }}{7} + \cos \dfrac{{3\pi }}{7} + \cos \dfrac{{4\pi }}{7} + \cos \dfrac{{5\pi }}{7} + \cos \dfrac{{6\pi }}{7} = S\]
We can write \[\cos \dfrac{{4\pi }}{7}\] as \[\cos \left( {\pi - \dfrac{{3\pi }}{7}} \right)\]. Similarly, we can write \[\cos \dfrac{{5\pi }}{7}\] as \[\cos \left( {\pi - \dfrac{{2\pi }}{7}} \right)\] and we can write \[\cos \dfrac{{6\pi }}{7}\] as \[\cos \left( {\pi - \dfrac{\pi }{7}} \right)\].
Substituting these values, we get the expression as
\[S = \cos 0 + \cos \dfrac{\pi }{7} + \cos \dfrac{{2\pi }}{7} + \cos \dfrac{{3\pi }}{7} + \cos \left( {\pi - \dfrac{{3\pi }}{7}} \right) + \cos \left( {\pi - \dfrac{{2\pi }}{7}} \right) + \cos \left( {\pi - \dfrac{\pi }{7}} \right)\]
We know that \[\cos \left( {\pi - A} \right) = - \cos A\]
Therefore, using this identity in the given expression, we get
\[\begin{array}{l} \Rightarrow S = \cos 0 + \cos \dfrac{\pi }{7} + \cos \dfrac{{2\pi }}{7} + \cos \dfrac{{3\pi }}{7} - \cos \dfrac{\pi }{7} - \cos \dfrac{{2\pi }}{7} - \cos \dfrac{{3\pi }}{7}\\ \Rightarrow S = \cos 0\\ \Rightarrow S = 1\end{array}\]
So, \[\cos 0 + \cos \dfrac{\pi }{7} + \cos \dfrac{{2\pi }}{7} + \cos \dfrac{{3\pi }}{7} + \cos \dfrac{{4\pi }}{7} + \cos \dfrac{{5\pi }}{7} + \cos \dfrac{{6\pi }}{7} = 1\]
Hence, option (R) is the correct option for (B).
C. \[\cos {20^ \circ } + \cos {40^ \circ } + \cos {60^ \circ } - 4\cos {10^ \circ }\cos {20^ \circ }\cos {30^ \circ }\]
Let \[\cos {20^ \circ } + \cos {40^ \circ } + \cos {60^ \circ } - 4\cos {10^ \circ }\cos {20^ \circ }\cos {30^ \circ } = T\]
We know from trigonometric identities;
\[\begin{array}{l}\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)\\\cos A.\cos B = \dfrac{{\cos \left( {A + B} \right) + \cos \left( {A - B} \right)}}{2}\end{array}\]
Using this same identity for the sum \[\cos {20^\circ } + \cos {40^\circ }\] and for the product \[\cos {10^\circ }\cos {20^\circ }\] in the expression, we get
\[ \Rightarrow T = 2\cos \left( {\dfrac{{{{20}^ \circ } + {{40}^ \circ }}}{2}} \right)\cos \left( {\dfrac{{{{40}^ \circ } - {{20}^ \circ }}}{2}} \right) + \cos {60^ \circ } - 4\dfrac{{\cos \left( {{{20}^ \circ } + {{10}^ \circ }} \right) + \cos \left( {{{20}^ \circ } - {{10}^ \circ }} \right)}}{2}\cos {30^ \circ }\]
Simplifying the terms inside the bracket, we get
\[ \Rightarrow T = 2\cos {30^ \circ }\cos {10^ \circ } + \cos {60^ \circ } - 2\left( {\cos {{30}^ \circ } + \cos {{10}^ \circ }} \right)\cos {30^ \circ }\]
We know that \[\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}\] and \[\cos {60^ \circ } = \dfrac{1}{2}\].
Substituting these values in the expression, we get
\[ \Rightarrow T = 2\dfrac{{\sqrt 3 }}{2}\cos {10^ \circ } + \dfrac{1}{2} - 2\left( {\dfrac{{\sqrt 3 }}{2} + \cos {{10}^ \circ }} \right)\dfrac{{\sqrt 3 }}{2}\]
On further simplification, we get
\[ \Rightarrow T = \sqrt 3 \cos {10^ \circ } + \dfrac{1}{2} - \dfrac{3}{2} - \sqrt 3 \cos {10^ \circ }\]
Simplifying the likes terms, we get
\[ \Rightarrow T = - 1\]
So, \[\cos {20^ \circ } + \cos {40^ \circ } + \cos {60^ \circ } - 4\cos {10^ \circ }\cos {20^ \circ }\cos {30^ \circ } = - 1\]
Hence, option (P) is the correct option for (C).
D. \[\cos {20^ \circ }\cos {100^ \circ } + \cos {100^ \circ }\cos {140^ \circ } - \cos {140^ \circ }\cos {200^ \circ }\]
Let \[\cos {20^ \circ }\cos {100^ \circ } + \cos {100^ \circ }\cos {140^ \circ } - \cos {140^ \circ }\cos {200^ \circ } = U\]
We know from trigonometric identities;
\[\cos A.\cos B = \dfrac{{\cos \left( {A + B} \right) + \cos \left( {A - B} \right)}}{2}\]
Using this same identity for the entire product in the expression, we get
\[ \Rightarrow U = \dfrac{{\cos {{120}^ \circ } + \cos {{80}^ \circ }}}{2} + \dfrac{{\cos {{240}^ \circ } + \cos {{40}^ \circ }}}{2} - \dfrac{{\cos {{340}^ \circ } + \cos {{60}^ \circ }}}{2}\]
Simplifying the terms further, we get
\[ \Rightarrow U = \dfrac{1}{2}\left( {\cos {{120}^ \circ } + \cos {{80}^ \circ } + \cos {{240}^ \circ } + \cos {{40}^ \circ } - \cos {{340}^ \circ } - \cos {{60}^ \circ }} \right)\]
We know that
\[\begin{array}{l}\cos {120^ \circ } = - \dfrac{1}{2}\\\cos {60^ \circ } = \dfrac{1}{2}\end{array}\]
We will put these values here.
\[ \Rightarrow U = \dfrac{1}{2}\left( { - \dfrac{1}{2} + \cos {{80}^ \circ } + \cos {{240}^ \circ } + \cos {{40}^ \circ } - \cos {{340}^ \circ } - \dfrac{1}{2}} \right)\]
We can know, \[\cos {240^ \circ } = - \dfrac{1}{2}\]
\[ \Rightarrow U = \dfrac{1}{2}\left( { - \dfrac{1}{2} + \cos {{80}^ \circ } - \dfrac{1}{2} + \cos {{40}^ \circ } - \cos {{340}^ \circ } - \dfrac{1}{2}} \right)\]
Applying trigonometric identities for the sum \[\cos {80^ \circ } + \cos {40^ \circ }\] , we get
\[ \Rightarrow U = \dfrac{1}{2}\left( { - \dfrac{1}{2} + \cos \left( {\dfrac{{{{80}^ \circ } + {{40}^ \circ }}}{2}} \right)\cos \left( {\dfrac{{{{80}^ \circ } + {{40}^ \circ }}}{2}} \right) - \dfrac{1}{2} - \cos {{340}^ \circ } - \dfrac{1}{2}} \right)\]
Simplifying the terms inside the bracket, we get
\[\begin{array}{l} \Rightarrow U = \dfrac{1}{2}\left( { - \dfrac{1}{2} + \cos {{20}^ \circ } + \dfrac{1}{2} - \dfrac{1}{2} - \cos \left( {{{180}^ \circ } \times 2 - {{20}^ \circ }} \right) - \dfrac{1}{2}} \right)\\ \Rightarrow U = \dfrac{1}{2}\left( { - \dfrac{1}{2} + \cos {{20}^ \circ } + \dfrac{1}{2} - \dfrac{1}{2} - \cos {{20}^ \circ } - \dfrac{1}{2}} \right)\end{array}\]
Adding and subtracting the terms, we get
\[ \Rightarrow U = - \dfrac{3}{4}\]
So, \[\cos {20^ \circ }\cos {100^ \circ } + \cos {100^ \circ }\cos {140^ \circ } - \cos {140^ \circ }\cos {200^ \circ } = - \dfrac{3}{4}\]
Hence, option (Q) is the correct option for (D).
Note:
We have used many trigonometric identities to solve this problem of trigonometry. Trigonometric identities are defined as equalities that include trigonometric functions, and for every value of variables, these are true. We break the complex terms of trigonometric functions using product and sum to product identities of trigonometry.
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