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

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,

\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.