
Calculate sin18 and cos18 without using trigonometric tables.
Answer
597.9k+ views
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,
\[\begin{align}
& 5A={{90}^{\circ }} \\
& 2A+3A={{90}^{\circ }} \\
& A={{90}^{\circ }}-3A \\
\end{align}\]
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
$\begin{align}
& 4{{\cos }^{2}}A-2\sin A-3=0 \\
& 4\left( 1-{{\sin }^{2}}A \right)-2\sin A-3=0 \\
& 4-4{{\sin }^{2}}A-2\sin A-3=0 \\
& 1=4{{\sin }^{2}}A+2\sin A \\
& 4{{\sin }^{2}}A+2\sin A-1=0 \\
\end{align}$
Now we will use quadratic formula to find sinA
$\begin{align}
& \sin A=\dfrac{-2\pm \sqrt{{{2}^{2}}-4\times \left( 4 \right)\left( -1 \right)}}{2\times 4} \\
& =\dfrac{-2\pm \sqrt{4+16}}{8} \\
& =\dfrac{-2\pm \sqrt{20}}{8} \\
& =\dfrac{-1\pm \sqrt{5}}{4}
\end{align}$
As we know sin18 > 0
So $\sin \left( 18 \right)=\dfrac{-1+\sqrt{5}}{4}$
Now to find cos18 we know that
$\begin{align}
& {{\sin }^{2}}18+{{\cos }^{2}}18=1 \\
& \Rightarrow {{\cos }^{2}}18=1-{{\sin }^{2}}18 \\
& \Rightarrow {{\cos }^{2}}18=1-{{\left( \dfrac{-1+\sqrt{5}}{4} \right)}^{2}} \\
& \Rightarrow {{\cos }^{2}}18=1-\left( \dfrac{6-2\sqrt{5}}{16} \right) \\
& \Rightarrow {{\cos }^{2}}18=\dfrac{10+2\sqrt{5}}{16} \\
& \Rightarrow \cos 18=\dfrac{\sqrt{10+2\sqrt{5}}}{4} \\
\end{align}$
We will ignore negative rules as cos18 > 0.
Note:To solve these types of question one should remember important trigonometric identities like
$\begin{align}
& {{\sin }^{2}}A+{{\cos }^{2}}A=1 \\
& \sin 2A=2\sin A\cos A \\
& \cos 3A=4{{\cos }^{3}}A-3\cos A \\
\end{align}$
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,
\[\begin{align}
& 5A={{90}^{\circ }} \\
& 2A+3A={{90}^{\circ }} \\
& A={{90}^{\circ }}-3A \\
\end{align}\]
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
$\begin{align}
& 4{{\cos }^{2}}A-2\sin A-3=0 \\
& 4\left( 1-{{\sin }^{2}}A \right)-2\sin A-3=0 \\
& 4-4{{\sin }^{2}}A-2\sin A-3=0 \\
& 1=4{{\sin }^{2}}A+2\sin A \\
& 4{{\sin }^{2}}A+2\sin A-1=0 \\
\end{align}$
Now we will use quadratic formula to find sinA
$\begin{align}
& \sin A=\dfrac{-2\pm \sqrt{{{2}^{2}}-4\times \left( 4 \right)\left( -1 \right)}}{2\times 4} \\
& =\dfrac{-2\pm \sqrt{4+16}}{8} \\
& =\dfrac{-2\pm \sqrt{20}}{8} \\
& =\dfrac{-1\pm \sqrt{5}}{4}
\end{align}$
As we know sin18 > 0
So $\sin \left( 18 \right)=\dfrac{-1+\sqrt{5}}{4}$
Now to find cos18 we know that
$\begin{align}
& {{\sin }^{2}}18+{{\cos }^{2}}18=1 \\
& \Rightarrow {{\cos }^{2}}18=1-{{\sin }^{2}}18 \\
& \Rightarrow {{\cos }^{2}}18=1-{{\left( \dfrac{-1+\sqrt{5}}{4} \right)}^{2}} \\
& \Rightarrow {{\cos }^{2}}18=1-\left( \dfrac{6-2\sqrt{5}}{16} \right) \\
& \Rightarrow {{\cos }^{2}}18=\dfrac{10+2\sqrt{5}}{16} \\
& \Rightarrow \cos 18=\dfrac{\sqrt{10+2\sqrt{5}}}{4} \\
\end{align}$
We will ignore negative rules as cos18 > 0.
Note:To solve these types of question one should remember important trigonometric identities like
$\begin{align}
& {{\sin }^{2}}A+{{\cos }^{2}}A=1 \\
& \sin 2A=2\sin A\cos A \\
& \cos 3A=4{{\cos }^{3}}A-3\cos A \\
\end{align}$
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