Question

If \[x = \sin \dfrac{{2\pi }}{7} + \sin \dfrac{{4\pi }}{7} + \sin \dfrac{{8\pi }}{7}\] and \[y = \cos \dfrac{{2\pi }}{7} + \cos \dfrac{{4\pi }}{7} + \cos \dfrac{{8\pi }}{7}\] then \[{x^2} + {y^2}\] equals
A. 1
B. 2
C. 3
D. 4

Answer 1

Hint: We have been given the value of \[x\] and \[y\] in the question.
We will first find the numerical values of these two variables and use them to find \[{x^2} + {y^2}\].
We will solve for \[y\] by taking \[a = \dfrac{{2\pi }}{7}\] and simplify to find the values of both the variables.
on using these two values to find \[{x^2} + {y^2}\] we get the required answer.

Formula used: \[2\cos A\sin B = \sin (A + B) - \sin (A - B)\]
\[2\sin A\cos B = \sin (A + B) + \sin (A - B)\]
\[\cos (A - B) = \sin A\sin B + \cos A\cos B\]
\[{(a + b + c)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca\]
Trigonometric identity, \[{\sin ^2}x + {\cos ^2}x = 1\]

Complete step-by-step answer:
We first take \[y = \cos \dfrac{{2\pi }}{7} + \cos \dfrac{{4\pi }}{7} + \cos \dfrac{{8\pi }}{7} \to (1)\]
Let us consider\[a = \dfrac{{2\pi }}{7}\].
\[ \Rightarrow 7a = 2\pi \].
Substituting \[7a = 2\pi \] in \[(1)\]and we get,
\[ \Rightarrow y = \cos a + \cos 2a + \cos 4a\]
Now we multiply and divide \[2\sin \dfrac{a}{2}\] on the right-hand side.
\[ \Rightarrow \dfrac{1}{{2\sin \dfrac{a}{2}}}\left( {2\sin \dfrac{a}{2}\cos a + 2\sin \dfrac{a}{2}\cos 2a + 2\sin \dfrac{a}{2}\cos 4a} \right)\]
Since \[2\cos A\sin B = \sin (A + B) - \sin (A - B)\]we can write the above as
\[ \Rightarrow \dfrac{1}{{2\sin \dfrac{a}{2}}}\left( {\sin \left( {a + \dfrac{a}{2}} \right) - \sin \left( {a - \dfrac{a}{2}} \right) + \sin \left( {2a + \dfrac{a}{2}} \right) - \sin \left( {2a - \dfrac{a}{2}} \right) + \sin \left( {4a + \dfrac{a}{2}} \right) - \sin \left( {4a - \dfrac{a}{2}} \right)} \right)\]
On simplification we get,
\[ \Rightarrow \dfrac{1}{{2\sin \dfrac{a}{2}}}\left( {\sin \left( {\dfrac{{3a}}{2}} \right) - \sin \left( {\dfrac{a}{2}} \right) + \sin \left( {\dfrac{{5a}}{2}} \right) - \sin \left( {\dfrac{{3a}}{2}} \right) + \sin \left( {\dfrac{{9a}}{2}} \right) - \sin \left( {\dfrac{{7a}}{2}} \right)} \right)\]
Cancelling the like terms with opposite signs and substituting \[7a = 2\pi \] in the last term.
\[ \Rightarrow \dfrac{1}{{2\sin \dfrac{a}{2}}}\left( { - \sin \left( {\dfrac{a}{2}} \right) + \sin \left( {\dfrac{{5a}}{2}} \right) + \sin \left( {\dfrac{{9a}}{2}} \right) - \sin \left( {\dfrac{{2\pi }}{2}} \right)} \right)\]
We know that \[{{sin n\pi = 0 for}}\,{\text{all}}\,{\text{n}}\], hence \[\sin \left( {\dfrac{{2\pi }}{2}} \right) = 0\]
\[ \Rightarrow \dfrac{1}{{2\sin \dfrac{a}{2}}}\left( { - \sin \left( {\dfrac{a}{2}} \right) + \sin \left( {\dfrac{{5a}}{2}} \right) + \sin \left( {\dfrac{{9a}}{2}} \right)} \right) \to (2)\]
Again we use the formula, \[2\sin A\cos B = \sin (A + B) + \sin (A - B)\] then we can take \[A = \dfrac{{7a}}{2}\] and \[B = a\]
That is we can write, \[\sin \left( {\dfrac{{5a}}{2}} \right) + \sin \left( {\dfrac{{9a}}{2}} \right) = \sin \left( {\dfrac{{7a}}{2} - a} \right) + \sin \left( {\dfrac{{7a}}{2} + a} \right) = 2\sin \left( {\dfrac{{7a}}{2}} \right)\cos (a)\]
Substituting this in equation (2)
\[ \Rightarrow \dfrac{1}{{2\sin \dfrac{a}{2}}}\left( { - \sin \left( {\dfrac{a}{2}} \right) + 2\sin \left( {\dfrac{{7a}}{2}} \right)\cos (a)} \right)\]
Again, we can substitute\[7a = 2\pi \]
\[ \Rightarrow \dfrac{1}{{2\sin \dfrac{a}{2}}}\left( { - \sin \left( {\dfrac{a}{2}} \right) + \sin 2\pi \cos a} \right)\]
\[{{sin n\pi = 0 for}}\,{\text{all}}\,{\text{n}}\], \[\sin 2\pi \cos a = 0 \times \cos a = 0\]
\[ \Rightarrow \dfrac{1}{{2\sin \dfrac{a}{2}}} \times - \sin \left( {\dfrac{a}{2}} \right)\]
Cancelling \[\dfrac{1}{{\sin \dfrac{a}{2}}} \times - \sin \left( {\dfrac{a}{2}} \right)\] we can get the value of y.
\[ \Rightarrow y = - \dfrac{1}{2}\]
Now to find \[x\] we need to find \[\sin \dfrac{{2\pi }}{7} + \sin \dfrac{{4\pi }}{7} + \sin \dfrac{{8\pi }}{7} \to (3)\].
By taking \[7a = 2\pi \] and solving for \[\left( 1 \right)\] and \[\left( 3 \right)\] as follows,
\[{(\sin a + \sin 2a + \sin 4a)^2} + {(\cos a + \cos 2a + \cos 4a)^2} \to (4)\] will give the value of \[x\].
By using \[{(a + b + c)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca\] we can write (4) as,
\[\begin{gathered}
   \Rightarrow {\sin ^2}a + {\sin ^2}2a + {\sin ^2}4a + {\cos ^2}a + {\cos ^2}2a + {\cos ^2}4a + 2(\sin a\sin 2a + \sin 2a\sin 4a + \sin 4a\sin a + \cos a\cos 2a \\
   + \cos 2a\cos 4a + \cos 4a\cos a) \\
\end{gathered} \]
By using the trigonometric identity \[{\sin ^2}x + {\cos ^2}x = 1\]
\[({\sin ^2}a + {\cos ^2}a) + ({\sin ^2}2a + {\cos ^2}2a) + ({\sin ^2}4a + {\cos ^2}4a) \Rightarrow 1 + 1 + 1 \to (5)\]
Now, by using (5)
\[ \Rightarrow 1 + 1 + 1 + 2(\sin a\sin 2a + \sin 2a\sin 4a + \sin 4a\sin a + \cos a\cos 2a + \cos 2a\cos 4a + \cos 4a\cos a)\]
On adding we get,
\[ \Rightarrow 3 + 2(\sin a\sin 2a + \sin 2a\sin 4a + \sin 4a\sin a + \cos a\cos 2a + \cos 2a\cos 4a + \cos 4a\cos a)\]
By using \[\cos (A - B) = \sin A\sin B + \cos A\cos B\] we can further do as follows,
\[ \Rightarrow 3 + 2((\sin a\sin 2a + \cos a\cos 2a) + (\sin 2a\sin 4a + \cos 2a\cos 4a) + (\sin 4a\sin a + \cos 4a\cos a))\] Using the formula and we get,
\[ \Rightarrow 3 + 2(\cos (2a - a) + \cos (4a - 2a) + \cos (4a - a))\]
On subtracting the bracket terms and we get,
\[ \Rightarrow 3 + 2(\cos (a) + \cos (2a) + \cos (3a))\]
Since \[{{cos(2\pi - \theta ) = cos(\theta )cos(2\pi ) + sin(\theta )sin(2\pi )}}\] and \[\cos 2\pi = 1,\sin 2\pi = 0\]
We can write \[cos(2\pi - \theta ) = \cos \theta \]
Using this in \[\cos 3a = \cos (2\pi - 4a) = \cos 4a\]
\[ \Rightarrow 3 + 2(\cos (a) + \cos (2a) + \cos (4a))\]
Since we already have \[y = \cos (a) + \cos (2a) + \cos (4a) = - \dfrac{1}{2}\]
\[ \Rightarrow 3 + 2 \times \left( { - \dfrac{1}{2}} \right)\]
On cancelling the term and we get.
\[ \Rightarrow 3 - 1\]
Let us subtracting we get,
\[ \Rightarrow 2\]
Now equation \[\left( 4 \right)\] can be written as,
\[ \Rightarrow {(\sin a + \sin 2a + \sin 4a)^2} + {(\cos a + \cos 2a + \cos 4a)^2} = 2\]
We can again substitute\[y = \cos (a) + \cos (2a) + \cos (4a) = - \dfrac{1}{2}\] in the above and we arrive at,
\[ \Rightarrow {(\sin a + \sin 2a + \sin 4a)^2} + {\left( { - \dfrac{1}{2}} \right)^2} = 2\]
On squaring the term and we get,
\[ \Rightarrow {(\sin a + \sin 2a + \sin 4a)^2} + \left( {\dfrac{1}{4}} \right) = 2\]
Taking the fraction term as RHS and change into the negative sign we get
\[ \Rightarrow {(\sin a + \sin 2a + \sin 4a)^2} = 2 - \left( {\dfrac{1}{4}} \right)\]
Taking LCM as RHS and we get,
\[ \Rightarrow {(\sin a + \sin 2a + \sin 4a)^2} = \dfrac{{8 - 1}}{4}\]
Let us subtract the numerator term on RHS we get,
\[ \Rightarrow {(\sin a + \sin 2a + \sin 4a)^2} = \dfrac{7}{4}\]
By taking square root on both sides we get,
\[ \Rightarrow \sin a + \sin 2a + \sin 4a = \dfrac{{\sqrt 7 }}{2}\]
This is because; the square of \[2\] is \[4\].
From the above calculations, we have obtained the values as
\[x = \sin \dfrac{{2\pi }}{7} + \sin \dfrac{{4\pi }}{7} + \sin \dfrac{{8\pi }}{7} = \dfrac{{\sqrt 7 }}{2}\]
\[y = \cos \dfrac{{2\pi }}{7} + \cos \dfrac{{4\pi }}{7} + \cos \dfrac{{8\pi }}{7} = - \dfrac{1}{2}\]
To find \[{x^2} + {y^2}\] we now use the above values,
\[ \Rightarrow {\left( {\dfrac{{\sqrt 7 }}{2}} \right)^2} + {\left( { - \dfrac{1}{2}} \right)^2}\]
On squaring the term and we get,
\[ \Rightarrow \left( {\dfrac{7}{4}} \right) + \left( {\dfrac{1}{4}} \right)\]
On adding the term we get,
\[ \Rightarrow \dfrac{8}{4}\]
Let us divide the term and we get,
\[ \Rightarrow 2\]
On simplifying this we arrive at the answer as \[2\].

The correct option for this question is B.

Note: We need to be aware of the trigonometric conversions and formulas to solve this problem.
It is important for us to know that \[{{sin n\pi = 0 for}}\,{\text{all}}\,{\text{n}}\] whereas \[{{cos n\pi = 1when}}\,{\text{n}}\,{\text{is}}\,{\text{even}}\] and \[{{cos n\pi = - 1when}}\,{\text{n}}\,{\text{is}}\,{\text{odd}}\].
When using formulas such as
\[2\cos A\sin B = \sin (A + B) - \sin (A - B)\]
\[2\sin A\cos B = \sin (A + B) + \sin (A - B)\]
\[\cos (A - B) = \sin A\sin B + \cos A\cos B\]
We need to be careful in choosing the values of A and B accordingly to get the required answer.