   Question Answers

# Solve the equation ${\text{sin3}}{{\text{6}}^{\text{0}}}$${\text{sin7}}{{\text{2}}^0}$${\text{sin10}}{{\text{8}}^0}$${\text{sin14}}{{\text{4}}^0}$ =​$\dfrac{{\text{k}}}{{{\text{16}}}}$ and find the value of k.  Verified
148.5k+ views
Hint – In order to solve this problem use the concept of trigonometric functions that sin(180-a)=sin(a). Using this formula and using the value of some angles of sine and cosine will take you to the value of k.

Complete step-by-step solution -
The given equation is:
${\text{sin3}}{{\text{6}}^{\text{0}}}$${\text{sin7}}{{\text{2}}^0}$${\text{sin10}}{{\text{8}}^0}$${\text{sin14}}{{\text{4}}^0}$ =​$\dfrac{{\text{k}}}{{{\text{16}}}}$ ……(1)
We have to find the value of k.
We will solve only LHS to get the value of k.
As we know, sin (180-a)=sin(a)
Therefore, sin108=sin(180-72)=sin72 and sin144=sin(180-36)=sin36
Putting these values in the equation (1) the equation becomes,
${\text{sin3}}{{\text{6}}^{\text{0}}}$${\text{sin3}}{{\text{6}}^0}$${\text{sin7}}{{\text{2}}^0}$${\text{sin7}}{{\text{2}}^0}$ =​$\dfrac{{\text{k}}}{{{\text{16}}}}$
On multiplying 4 with the numerator and denominator n the LGS of equation 4 we
$\dfrac{4}{4}$(${\text{sin3}}{{\text{6}}^{\text{0}}}$$\sin {72^0}$${\text{sin7}}{{\text{2}}^0}$$\sin {36^0}$) =​$\dfrac{{\text{k}}}{{{\text{16}}}}$
$\dfrac{1}{4}$(2${\text{sin3}}{{\text{6}}^{\text{0}}}$$\sin {72^0}$)$^2$ =​$\dfrac{{\text{k}}}{{{\text{16}}}}$
We can further do,
$\dfrac{1}{4}$(2${\text{sin3}}{{\text{6}}^{\text{0}}}$$\sin {72^0}$)(2$\sin {36^0}$${\text{sin7}}{{\text{2}}^0}$) =​$\dfrac{{\text{k}}}{{{\text{16}}}}$

As we know the formula: 2sinasinb=cos(a-b)-cos(a+b) applying the same to the above equation we get,
$\dfrac{1}{4}$($\cos (36 - 72) - \cos (36 + 72)$)($\cos (36 - 72) - \cos (36 + 72)$) =​$\dfrac{{\text{k}}}{{{\text{16}}}}$
cos(-a) = cos(a) and cos(180-a) = -cos(a)
$\dfrac{1}{4}$${(\cos (36) - \cos (108))^2}=​\dfrac{{\text{k}}}{{{\text{16}}}} ……(2) cos(-a) = cos(a) and cos(90+a) = -sin(a) \dfrac{1}{4}$${(\cos (36) - \cos (90 + 18))^2}$=​$\dfrac{{\text{k}}}{{{\text{16}}}}$
$\dfrac{1}{4}$${(\cos (36) + \sin (18))^2}$=​$\dfrac{{\text{k}}}{{{\text{16}}}}$
As we know the value of cos36 =$\dfrac{{\sqrt 5 + 1}}{4}$, sin18 =$\dfrac{{\sqrt 5 - 1}}{4}$
On putting the values of cos90, cos18 and cos54 in equation number (2) we get,
$\dfrac{1}{4}$(${\left( {\dfrac{{\sqrt 5 + 1}}{4} + \dfrac{{\sqrt 5 - 1}}{4}} \right)^2}$=$\dfrac{{\text{k}}}{{{\text{16}}}}$
$\dfrac{1}{4}$(${\left( {\dfrac{{\sqrt 5 }}{2}} \right)^2}$=$\dfrac{{\text{k}}}{{{\text{16}}}}$
Further solving the equation we get the equation as:
$\dfrac{5}{{16}}$=​$\dfrac{{\text{k}}}{{{\text{16}}}}$
So, k = 5.
Hence, the value of k is 5.

Note – Whenever you face such types of problems you have to use the general formulas and value of trigonometric functions. Doing this will solve your half of the question. After that solving the equation algebraically we will reach the correct solution of the question.