
Calculate sin18 and cos18 without using trigonometric tables.
Answer
596.4k+ 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
$\begin{align}
& {{\sin }^{2}}A+{{\cos }^{2}}A=1 \\
& \sin 2A=2\sin A\cos A \\
& \cos 3A=4{{\cos }^{3}}A-3\cos A \\
\end{align}$
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

Which animal has three hearts class 11 biology CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

