Question

Calculate sin18 and cos18 without using trigonometric tables.

Answer 1

Hint: We will be using the concepts of trigonometric functions to solve the question. We will be using
Sin2A = 2sinAcosA
$\cos 3A=4{{\cos }^{3}}A-3\cos A$

Complete step by step answer:
Now, we have to find the value of sin18.So let us assume $A={{18}^{\circ }}$ .
So,

& 5A={{90}^{\circ }} \\\ & 2A+3A={{90}^{\circ }} \\\ & A={{90}^{\circ }}-3A \\\ \end{aligned}$$ Taking sin on both sides, we get $$\sin \left( 2A \right)=\sin \left( 90-3A \right)$$ Now, we know that sin(90 – 3A) = cos3A Sin 2A = cos3A Now we need to convert them in single angle function of trigonometry We know that sin2A = 2sinAcosA $\cos 3A=4{{\cos }^{3}}A-3\cos A$ We will substitute these values to get $2\sin A\cos A=4{{\cos }^{3}}A-3\cos A$ Now, we will rearrange the terms to get, $4{{\cos }^{3}}A-3\cos A-2\sin A\cos A=0$ $\cos A\left( 4{{\cos }^{2}}A-2\sin A-3 \right)=0$ Now, we know that $\operatorname{cosA}\ne 0$, so $\begin{aligned} & 4{{\cos }^{2}}A-2\sin A-3=0 \\\ & 4\left( 1-{{\sin }^{2}}A \right)-2\sin A-3=0 \\\ & 4-4{{\sin }^{2}}A-2\sin A-3=0 \\\ & 1=4{{\sin }^{2}}A+2\sin A \\\ & 4{{\sin }^{2}}A+2\sin A-1=0 \\\ \end{aligned}$ Now we will use quadratic formula to find sinA $\begin{aligned} & \sin A=\dfrac{-2\pm \sqrt{{{2}^{2}}-4\times \left( 4 \right)\left( -1 \right)}}{2\times 4} \\\ & =\dfrac{-2\pm \sqrt{4+16}}{8} \\\ & =\dfrac{-2\pm \sqrt{20}}{8} \\\ & =\dfrac{-1\pm \sqrt{5}}{4} \end{aligned}$ As we know sin18 > 0 So $\sin \left( 18 \right)=\dfrac{-1+\sqrt{5}}{4}$ Now to find cos18 we know that $\begin{aligned} & {{\sin }^{2}}18+{{\cos }^{2}}18=1 \\\ & \Rightarrow {{\cos }^{2}}18=1-{{\sin }^{2}}18 \\\ & \Rightarrow {{\cos }^{2}}18=1-{{\left( \dfrac{-1+\sqrt{5}}{4} \right)}^{2}} \\\ & \Rightarrow {{\cos }^{2}}18=1-\left( \dfrac{6-2\sqrt{5}}{16} \right) \\\ & \Rightarrow {{\cos }^{2}}18=\dfrac{10+2\sqrt{5}}{16} \\\ & \Rightarrow \cos 18=\dfrac{\sqrt{10+2\sqrt{5}}}{4} \\\ \end{aligned}$ We will ignore negative rules as cos18 > 0. Note:To solve these types of question one should remember important trigonometric identities like $\begin{aligned} & {{\sin }^{2}}A+{{\cos }^{2}}A=1 \\\ & \sin 2A=2\sin A\cos A \\\ & \cos 3A=4{{\cos }^{3}}A-3\cos A \\\ \end{aligned}$