Question

What is the value of \[\sin {{18}^{0}}\cos {{36}^{0}}\] equal to
A. 4
B. 2
C. 1
D. \[\dfrac{1}{4}\]

Answer 1

Hint: We have basic trigonometric identities as \[{{\cos }^{2}}A=1-{{\sin }^{2}}A\] and \[\cos 2A=1-2{{\sin }^{2}}A\] etc we use these identities to solve the problem and find the value of sin and cos separately.

Complete step-by-step answer:
We will first find the value of \[\sin {{18}^{0}}\].
Now we will first put A = \[{{18}^{0}}\]
Then, as (18)(5) = 90, we get 5A=90.
Now, we can write the above as,
\[\begin{align}
  & {{90}^{0}}=5A \\
 & \Rightarrow {{90}^{0}}=3A+2A \\
 & \Rightarrow 2A={{90}^{0}}-3A \\
\end{align}\]
Now because 2A is equal to 900-3A, then applying sin on both sides of the above equation we get,
 \[\sin (2A)=\sin ({{90}^{0}}-3A)\]
We know that \[\sin ({{90}^{0}}-\theta )=\cos \theta \], applying this to above obtained expression,
\[\Rightarrow \sin (2A)=\cos (3A)\]
Now using the trigonometric identities as \[\sin (2A)=2\sin A\cos A\] and \[\cos (3A)=4co{{s}^{3}}A-3\cos A\] in the above expression on both sides, we get,
\[\begin{align}
  & 2\sin A\cos A=4co{{s}^{3}}A-3\cos A \\
 & \Rightarrow 2\sin A\cos A-4co{{s}^{3}}A+3\cos A=0 \\
\end{align}\]
Taking cos common we get,
\[\Rightarrow \cos A(2\sin A-4{{\cos }^{2}}A+3)=0\]
Now because the above obtained expression is equal to 0 then one of the terms is 0. Now cosA can’t be 0 as A=180, which implies that \[2\sin A-4{{\cos }^{2}}A+3=0\].
We have \[2\sin A-4{{\cos }^{2}}A+3=0\]
Putting \[{{\cos }^{2}}A=1-{{\sin }^{2}}A\],
\[\begin{align}
  & \Rightarrow 2\sin A-4(1-{{\sin }^{2}}A)+3=0 \\
 & \Rightarrow 2\sin A-4+4{{\sin }^{2}}A+3=0 \\
 & \Rightarrow 4{{\sin }^{2}}A+2\sin A-1=0 \\
\end{align}\]
Let sinA = x in above we get,
\[4{{x}^{2}}+2x-1=0\]
Applying the formula as, \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] , where b = 2 a= 4 and c= -1 in the formula we get,
\[\begin{align}
  & x=\dfrac{-2\pm \sqrt{{{4}^{2}}+4}}{8} \\
 & \Rightarrow x=\dfrac{-2\pm \sqrt{20}}{8} \\
 & \Rightarrow x=\dfrac{-2\pm 2\sqrt{5}}{8} \\
 & \Rightarrow x=\dfrac{-1\pm \sqrt{5}}{4} \\
\end{align}\]
Then, \[\sin A=\dfrac{-1\pm \sqrt{5}}{4}\].
Now taking the negative in \[-\sqrt{5}\] we get a value of sinA less than -1, which is not possible.
Hence, we get the value of \[\sin A=\dfrac{-1+\sqrt{5}}{4}\].
\[\Rightarrow \sin {{18}^{^{0}}}=\dfrac{-1+\sqrt{5}}{4}..........(i)\]
And applying the property \[\cos 2A=1-2{{\sin }^{2}}A\], we get,
\[\begin{align}
  & cos{{36}^{0}}=\cos 2A=1-2{{\sin }^{2}}A \\
 & \Rightarrow cos{{36}^{0}}=1-2{{\sin }^{2}}{{18}^{0}} \\
 & \Rightarrow cos{{36}^{0}}=1-2{{\left( \dfrac{-1+\sqrt{5}}{4} \right)}^{2}} \\
 & \Rightarrow cos{{36}^{0}}=\dfrac{1+\sqrt{5}}{4}............(ii) \\
\end{align}\]
Now we calculate the value of \[\sin {{18}^{0}}\cos {{36}^{0}}\] using equation(i) and equation(ii) we have,
\[\begin{align}
  & \sin {{18}^{0}}\cos {{36}^{0}}=\left( \dfrac{-1+\sqrt{5}}{4} \right)\left( \dfrac{1+\sqrt{5}}{4} \right) \\
 & \Rightarrow \sin {{18}^{0}}\cos {{36}^{0}}=\dfrac{(-1+\sqrt{5})(1+\sqrt{5})}{16} \\
 & \Rightarrow \sin {{18}^{0}}\cos {{36}^{0}}=\dfrac{4}{16} \\
 & \Rightarrow \sin {{18}^{0}}\cos {{36}^{0}}=\dfrac{1}{4} \\
\end{align}\]
Hence, we obtain the value of \[\sin {{18}^{0}}\cos {{36}^{0}}=\dfrac{1}{4}\], which is option (d).

Note: We always show or calculate the values of the trigonometric functions using trigonometric identities. Always assume one of the functions or angles as x and then try to apply the identity to get the value of the problem.