Question

Find the value of \[\dfrac{{{m}^{2}}-{{n}^{2}}}{{{n}^{2}}}\]. If \[{{m}^{2}}\cos .\dfrac{2\pi }{15}\cos .\dfrac{4\pi }{15}\cos .\dfrac{8\pi }{15}\cos .\dfrac{14\pi }{15}={{n}^{2}}\].

Answer 1

HINT: - In such questions, where cos or sin functions are multiplied with each other and the angles on which the cos or sin function is applied are two times the previous one, then we try to fit in the half angle formula which is as follows
 \[\sin 2x=2\cdot \sin x\cdot \cos x\]
Also, another important formula that will be used in this question is as follows
\[2\sin A\cos B=\sin (A+B)+\sin (A-B)\]
Firstly, we will simply evaluate the trigonometric part of the given expression and then later on we will try to find the value of the asked expression that is \[\dfrac{{{m}^{2}}-{{n}^{2}}}{{{n}^{2}}}\] .

Complete step-by-step answer:
As mentioned in the question, we are asked to find the value of \[\dfrac{{{m}^{2}}-{{n}^{2}}}{{{n}^{2}}}\].
Now, as mentioned in the hint, we will first evaluate the trigonometric part of the expression by first taking the non trigonometric part of the L.H S. to the R.H.S. as follows
\[\begin{align}
  & \Rightarrow \cos .\dfrac{2\pi }{15}\cos .\dfrac{4\pi }{15}\cos .\dfrac{8\pi }{15}\cos .\dfrac{14\pi }{15}=\dfrac{{{n}^{2}}}{{{m}^{2}}} \\
 & \Rightarrow \cos .\dfrac{2\pi }{15}\cos .\dfrac{4\pi }{15}\cos .\dfrac{8\pi }{15}\cos .\dfrac{14\pi }{15}=\dfrac{{{n}^{2}}}{{{m}^{2}}} \\
\end{align}\]

Now, we will first evaluate the L.H.S and so, we will get the value of \[\dfrac{{{n}^{2}}}{{{m}^{2}}}\] .
Now, on multiplying and dividing \[2\sin \left( \dfrac{2\pi }{15} \right)\] in the above expression, we will get the following expression
 \[\Rightarrow \dfrac{2\sin \left( \dfrac{2\pi }{15} \right)\cos .\dfrac{2\pi }{15}\cos .\dfrac{4\pi }{15}\cos .\dfrac{8\pi }{15}\cos .\dfrac{14\pi }{15}}{2\sin \left( \dfrac{2\pi }{15} \right)}\]
Now, using the half angle formula that is given in the hint, we can write as follows
\[\begin{align}
  & \Rightarrow \dfrac{2\sin \left( \dfrac{2\pi }{15} \right)\cdot \cos \dfrac{2\pi }{15}\cdot \cos \dfrac{4\pi }{15}\cdot \cos \dfrac{8\pi }{15}\cdot \cos \dfrac{14\pi }{15}}{2\sin \left( \dfrac{2\pi }{15} \right)} \\
 & \Rightarrow \dfrac{\sin \cdot \dfrac{4\pi }{15}\cos \dfrac{4\pi }{15}\cdot \cos \dfrac{8\pi }{15}\cdot \cos \dfrac{14\pi }{15}}{2\sin \left( \dfrac{2\pi }{15} \right)} \\
\end{align}\]
Now, multiplying and dividing the expression with 2, we get
\[\begin{align}
  & \Rightarrow \dfrac{2\sin \dfrac{4\pi }{15}\cdot \cos \dfrac{4\pi }{15}\cdot \cos \dfrac{8\pi }{15}\cdot \cos \dfrac{14\pi }{15}}{{{(2)}^{2}}\sin \left( \dfrac{2\pi }{15} \right)} \\
 & \Rightarrow \dfrac{\sin \dfrac{8\pi }{15}\cdot \cos \dfrac{8\pi }{15}\cdot \cos \dfrac{14\pi }{15}}{{{(2)}^{2}}\sin \left( \dfrac{2\pi }{15} \right)} \\
\end{align}\]
Now, multiplying and dividing the expression with 2, we get
\[\Rightarrow \dfrac{2\sin \dfrac{8\pi }{15}\cdot \cos \dfrac{8\pi }{15}\cdot \cos \dfrac{14\pi }{15}}{{{(2)}^{3}}\sin \left( \dfrac{2\pi }{15} \right)}\]
Now, using the half angle formula that is given in the hint, we can write as follows
\[\Rightarrow \dfrac{\sin \dfrac{16\pi }{15}\cdot \cos \dfrac{14\pi }{15}}{{{(2)}^{3}}\sin \left( \dfrac{2\pi }{15} \right)}\]
Now, on multiplying and dividing with 2 and then using the other formula that is given in the hint which is \[2\sin A\cos B=\sin (A+B)+\sin (A-B)\] , we get
\[\begin{align}
  & =\dfrac{2\sin \dfrac{16\pi }{15}\cdot \cos \dfrac{14\pi }{15}}{{{(2)}^{4}}\sin \left( \dfrac{2\pi }{15} \right)} \\
 & =\dfrac{\sin \left( \dfrac{16\pi }{15}+\dfrac{14\pi }{15} \right)+\sin \left( \dfrac{16\pi }{15}-\dfrac{14\pi }{15} \right)}{{{(2)}^{4}}\sin \left( \dfrac{2\pi }{15} \right)} \\
 & =\dfrac{\sin \left( \dfrac{30\pi }{15} \right)+\sin \left( \dfrac{2\pi }{15} \right)}{{{(2)}^{4}}\sin \left( \dfrac{2\pi }{15} \right)}=\dfrac{\sin 2\pi +\sin \left( \dfrac{2\pi }{15} \right)}{{{(2)}^{4}}\sin \left( \dfrac{2\pi }{15} \right)}=\dfrac{\sin \left( \dfrac{2\pi }{15} \right)}{{{(2)}^{4}}\sin \left( \dfrac{2\pi }{15} \right)}=\dfrac{1}{16} \\
\end{align}\]
(As \[\sin 2\pi \] is 0 )
Hence, we can write as follows
\[\begin{align}
  & \Rightarrow \dfrac{{{n}^{2}}}{{{m}^{2}}}=\dfrac{1}{16} \\
 & Taking\ the\ reciprocal, \\
 & \Rightarrow \dfrac{{{m}^{2}}}{{{n}^{2}}}=16 \\
\end{align}\]
 (Because the value that we get on evaluating the trigonometric expression, we get the value of \[\dfrac{{{n}^{2}}}{{{m}^{2}}}\] as mentioned before)
Now, on subtracting 1 from both sides, we get
\[\begin{align}
  & \dfrac{{{m}^{2}}}{{{n}^{2}}}-1=16-1 \\
 & \dfrac{{{m}^{2}}-{{n}^{2}}}{{{n}^{2}}}=15 \\
\end{align}\]
 Hence, this is the required value.

NOTE: - The students can make an error if they don’t know about the relations between the different trigonometric functions because without knowing them; one could never get to the correct solution.
Also, another important trick that was used in the solution and must be known to the students is that
If we are given a ratio \[\dfrac{a}{b}\] , then to find the value of \[\dfrac{a+b}{b}\], we simply add 1 and to find the value of \[\dfrac{a-b}{b}\] , we simply subtract 1.
Also, in such questions, calculation mistakes are very common, so, to get to the right answer, one should try to be extra careful while solving these questions.