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,

& {{90}^{0}}=5A \\\ & \Rightarrow {{90}^{0}}=3A+2A \\\ & \Rightarrow 2A={{90}^{0}}-3A \\\ \end{aligned}$$ Now because 2A is equal to 900-3A, then applying sin on both sides of the above equation we get, $$\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{aligned} & 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{aligned}$$ Taking cos common we get, $$\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{aligned} & \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{aligned}$$ Let sinA = x in above we get, $$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{aligned} & 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{aligned}$$ Then, $$\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{aligned} & 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{aligned}$$ Now we calculate the value of $$\sin {{18}^{0}}\cos {{36}^{0}}$$ using equation(i) and equation(ii) we have, $$\begin{aligned} & \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{aligned}$$ Hence, we obtain the value of $$\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.