Question

Express $\sin 5\theta $ in terms of $\sin \theta $ and hence find the value of $\sin {{36}^{\circ }}$.

Answer 1

Hint: Take $5\theta $ as $3\theta +2\theta $. Now use trigonometry identity of $\sin \left( A+B \right),\sin 3\theta ,\cos 3\theta ,\sin 2\theta ,\cos 2\theta $ to solve it further and get the value of $\sin 5\theta $ in terms of $\sin \theta.$ Now put the value of $\theta $ as ${{36}^{\circ }}$ to get $\sin {{36}^{\circ }}$. Solve the biquadratic equation formed.

Complete step-by-step answer:

We can break $5\theta $ to $3\theta $ and $2\theta $ for expressing $\sin 5\theta $ in terms of $\sin \theta $.

Hence, we can write $\sin 5\theta $ as

$\sin 5\theta =\sin \left( 3\theta +2\theta \right)..........\left( i \right)$

Now, using the trigonometry identity \[\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B\], we get

$\sin 5\theta =\sin 3\theta \cos 2\theta +\cos 3\theta \sin 2\theta .........\left( ii \right)$

Now, we can use trigonometric identities of $\sin 3\theta,cos2\theta ,cos3\theta $ and $\sin 2\theta $ as

$\begin{align}

  & \sin 2\theta =2\sin \theta .cos\theta \\

 & \cos 2\theta =1-2{{\sin }^{2}}\theta \\

 & \sin 3\theta =3\sin \theta -4{{\sin }^{3}}\theta \\

 & \cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta \\

\end{align}$

Hence, applying all the above identities with equation (ii) we get,

$\sin 5\theta =\left( 3\sin \theta -4{{\sin }^{3}}\theta \right)\left( 1-2{{\sin }^{2}}\theta \right)+\left( 4{{\cos }^{3}}\theta -3\cos \theta \right)\left( 2\sin \theta \cos \theta \right)$

Now, simplifying the above equation, we get

$\sin 5\theta =3\sin \theta -6{{\sin }^{3}}\theta -4{{\sin }^{3}}\theta +8{{\sin }^{5}}\theta +8\sin \theta {{\cos }^{4}}\theta -6\sin \theta {{\cos }^{2}}\theta $

Now, we can convert ${{\cos }^{2}}\theta $ to $1-{{\sin }^{2}}\theta $ by using trigonometric identity as ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.

Hence, $\sin 5\theta $ can be fully converted to $\sin \theta $ as

$\sin 5\theta =3\sin \theta -10{{\sin }^{3}}\theta +8{{\sin }^{5}}\theta +8\sin \theta {{\left( 1-{{\sin }^{2}}\theta \right)}^{2}}-6\sin \theta \left( 1-{{\sin }^{2}}\theta \right)$

Now, using ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-ab$ , we can expand ${{\left( 1-{{\sin }^{2}}\theta \right)}^{2}}$ so, we get

$\sin 5\theta =3\sin \theta -10{{\sin }^{3}}\theta +8{{\sin }^{5}}\theta +8\sin \theta \left( 1+{{\sin }^{4}}-2{{\sin }^{2}}\theta \right)-6\sin \theta +6{{\sin }^{3}}\theta $

Or

$\sin 5\theta =-3\sin \theta -4{{\sin }^{3}}\theta +8{{\sin }^{5}}\theta +8\sin \theta +8{{\sin }^{5}}\theta -16{{\sin }^{3}}\theta $

Hence, $\sin 5\theta $ can be given as

$\sin 5\theta =5\sin \theta -20{{\sin }^{3}}\theta +16{{\sin }^{5}}\theta ..........\left( iii \right)$
Now, we can evaluate $\sin {{36}^{\circ }}$ by putting $\theta ={{36}^{\circ }}$ in the calculated expression of $\sin 5\theta $ in terms of $\sin \theta $ as given in equation (ii).

Hence putting $\theta ={{36}^{\circ }}$ in equation (ii) we get,

$\sin 36\times 5=5\sin {{36}^{\circ }}-20{{\sin }^{3}}{{36}^{\circ }}+16{{\sin }^{5}}{{36}^{\circ }}$

$\sin {{180}^{\circ }}=5\sin {{36}^{\circ }}-20{{\sin }^{3}}{{36}^{\circ }}+16{{\sin }^{5}}{{36}^{\circ }}$

As, we know value of $\sin {{180}^{\circ }}$ is zero, hence we can write above equation as
$16{{\sin }^{5}}{{36}^{\circ }}-20{{\sin }^{3}}{{36}^{\circ }}+5\sin {{36}^{\circ }}=0$

Let $\sin {{36}^{\circ }}=x$ , hence we can write above equation in form of ’x’ as

$16{{x}^{5}}-20{{x}^{3}}+5x=0...........\left( iv \right)$

Taking ‘x’ as common from equation (iv), we get

\[x\left( 16{{x}^{4}}-20{{x}^{2}}+5 \right)=0\]

Now, x = 0 or $16{{x}^{4}}-20{{x}^{2}}+5x=0$.

As $x=\sin {{36}^{\circ }}$, so $\sin {{36}^{\circ }}$ can never be zero.

So $x\ne 0.$

Hence,

$16{{x}^{4}}-20{{x}^{2}}+5=0$

Take ${{x}^{2}}=t$ in the above equation, we get

$16{{t}^{2}}-20t+5=0............\left( v \right)$

Now, we know that roots of any quadratic $A{{x}^{2}}+Bx+C$ is given by quadratic formula as

$x=\dfrac{-B\pm \sqrt{{{B}^{2}}-4AC}}{2A}$

So, value of ‘t’ from equation (v) can be given as

$t=\dfrac{20\pm \sqrt{{{\left( 20 \right)}^{2}}-4\times 16\times 5}}{2\times 16}$

$t=\dfrac{20\pm \sqrt{400-320}}{32}$

$t=\dfrac{20\pm \sqrt{80}}{32}$

Now $\sqrt{80}$ can be written as $\sqrt{16\times }5$ i.e., $4\sqrt{5}$.

Hence, ‘t’ will be given as

$t=\dfrac{20\pm 4\sqrt{5}}{32}$ or

$t=\dfrac{5\pm \sqrt{5}}{8}$

Now, we can put $t={{x}^{2}}$ as supposed earlier. Hence we get,

${{x}^{2}}=\dfrac{5\pm \sqrt{5}}{8}$

Hence, we have two values of ${{x}^{2}}$ i.e.,

${{x}^{2}}=\dfrac{5+\sqrt{5}}{8}$ or \[{{x}^{2}}=\dfrac{5-\sqrt{5}}{8}\]

Now, we know that $\sin {{36}^{\circ }}$ will lie in (0,1) and $\sin {{36}^{\circ }}$ will be less than $\sin {{45}^{\circ }}$ as well because curve of $\sin \theta $ is increasing in $\left( 0,\dfrac{\pi }{2} \right)$ as shown in diagram.

And we know very well that if any number is lying in (0,1) then if we square it, it will become more less.

Now, let us come to both values of ${{x}^{2}}$ i.e. ${{\sin }^{2}}{{36}^{\circ }}.$

As $\sin 45=\dfrac{1}{\sqrt{2}}$ , hence ${{\sin }^{2}}45=\dfrac{1}{2}=0.5$

Now putting value of $\sqrt{5}=2.236$ we get,

${{x}^{2}}=\dfrac{5+\sqrt{5}}{8}=\dfrac{5+2.236}{8}=0.9045$

Where ${{x}^{2}}$ i.e. ${{\sin }^{2}}{{36}^{\circ }}$ is higher than ${{\sin }^{2}}45$ or 0.5 which cannot be possible if we look by curve of sin x.

Hence,

${{x}^{2}}=\dfrac{5-\sqrt{5}}{8}$

Now taking root to both sides we get,

$x=\pm \sqrt{\dfrac{5-\sqrt{5}}{8}}$

As $\dfrac{5-\sqrt{5}}{8}$ is a positive number and if we put a negative sign in front of it then it will be a negative number. And we know very well that $\sin {{36}^{\circ }}$ should be a positive value.

Hence,
$\sin {{36}^{\circ }}=\sqrt{\dfrac{5-\sqrt{5}}{8}}$

Note: One can get confused with lots of values of ‘x’. So, take care while calculating the value of x i.e. $\sin {{36}^{\circ }}$ . Eliminate all those values which are not possible for $\sin {{36}^{\circ }}$ .
Another approach for expanding $\sin 5\theta $ would be that we can break it into $4\theta +\theta $ and apply identities of $\cos 2\theta $ and $\sin 2\theta $ whenever required.