Trigonometric values| Trigonometric ratio table

Trigonometric values  of different ratios are very important for the study  of relationships between lengths and angles of triangles.  The trigonometric values of \large 0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ}, 90^{\circ} etc.  are  commonly used to solve many problems related to triangles.

Trigonometric ratios of some specific angles

\mathbf{\angle A} \mathbf{0^{\circ}} \mathbf{30^{\circ}} \mathbf{45^{\circ}} \mathbf{60^{\circ}} \mathbf{90^{\circ}}
\mathbf{{\color{Red} sinA}} \mathbf{}\mathbf{0} \mathbf{\frac{1}{2}} \mathbf{\frac{1}{\sqrt{2}}} \mathbf{\frac{\sqrt{3}}{2}} \boldsymbol{1}
\mathbf{{\color{Red} cosA}} \mathbf{1} \mathbf{\frac{\sqrt{3}}{2}} \mathbf{\frac{1}{\sqrt{2}}} \mathbf{\frac{1}{2}} \mathbf{0}
\mathbf{{\color{Red} tanA}} \mathbf{0} \mathbf{\frac{1}{\sqrt{3}}} \mathbf{1} \mathbf{\sqrt{3}} not defined
\mathbf{{\color{Red} cosecA}} not defined 2 \mathbf{\sqrt{2}} \mathbf{\frac{2}{\sqrt{3}}} \mathbf{1}
\mathbf{{\color{Red} secA}} \mathbf{1} \mathbf{\frac{2}{\sqrt{3}}} \mathbf{\sqrt{2}} 2 not defined
\mathbf{{\color{Red} cotA}} not defined \mathbf{\sqrt{3}} \mathbf{1} \mathbf{\frac{1}{\sqrt{3}}} 0

 

Value  of all trigonometric  ratios at \large \mathbf{18^{\circ}, \, 36^{\circ}, \, 54^{\circ} \, \, and \, \, 72^{\circ}}

Let \large x=18^{\circ}. Then \large 2x+3x=90^{\circ}.Therefore  \large sin 2x=sin(90^{\circ}-3x)=cos 3x.

\large 2sinxcosx=4cos^{3}x-3cosx

\large \Rightarrow cosx(2sinx-4cos^{2}x+3)=0

Since \large cosx\neq 0 \, \, (\because x=18^{\circ})

\large \Rightarrow 2sinx-4cos^{2}x+3=0

\large \Rightarrow 2sinx-4(1-sin^{2}x)+3=0

\large 4sin^{2}x+2sinx-1=0,  Solving this we get

\large sinx=\frac{-2\pm \sqrt{4+16}}{8}=\frac{\pm \sqrt{5}-1}{4}

Since \large sin18^{\circ}> 0, we get that         \bg_red \large \boldsymbol{sin18^{\circ}=\frac{\sqrt{5}-1}{4}}

Now \large cos18^{\circ}=\sqrt{1-sin^{2}18^{\circ}}=\sqrt{1-\left(\frac{\sqrt{5}-1}{4}\right)^{2}}

=\large \sqrt{\frac{16-(5+1-2\sqrt{5})}{16}}=\frac{\sqrt{10+2\sqrt{5}}}{4}

\bg_red \large cos18^{\circ}=\frac{\sqrt{10+2\sqrt{5}}}{4}

Now            \bg_red \large tan18^{\circ}=\frac{sin18^{\circ}}{cos18^{\circ}}=\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}

Similarly we can find \large cosec18^{\circ}=\frac{1}{sin18^{\circ}}=\frac{4}{\sqrt{5}-1},

\large sec18^{\circ}=\frac{1}{cos18^{\circ}}=\frac{4}{\sqrt{10+2\sqrt{5}}}      and  \large cot 18^{\circ}=\frac{cos18^{\circ}}{sin18^{\circ}}=\frac{\sqrt{10+2\sqrt{5}}}{\sqrt{5}-1}.

Now \mathbf{cos 36^{\circ}=2cos^{2}18^{\circ}-1=2\left [ \frac{10+2\sqrt{5}}{16} \right ]-1=\frac{2+2\sqrt{5}}{8}=\frac{\sqrt{5}+1}{4}}

\small \mathbf{sin36^{\circ}}=\sqrt{1-cos^{2}36^{\circ}}=\sqrt{1-\left ( \frac{\sqrt{5}+1}{4} \right )^{2}}=\frac{\sqrt{16-(6+2\sqrt{5})}}{4}=\frac{\sqrt{10-2\sqrt{5}}}{4}

By using these , we get    \large cos72^{\circ}=cos(90^{\circ}-18^{\circ})=sin18^{\circ}=\frac{\sqrt{5}-1}{4}

\large sin54^{\circ}=sin(90^{\circ}-36^{\circ})=cos36^{\circ}=\frac{\sqrt{5}+1}{4}

\large sin72^{\circ}=cos18^{\circ}=\frac{\sqrt{10+2\sqrt{5}}}{4}

\large cos54^{\circ}=sin36^{\circ}=\frac{\sqrt{10-2\sqrt{5}}}{4} etc.

https://youtu.be/JdJcD7C4LMQ

Leave a Reply

Your email address will not be published. Required fields are marked *

NVS Recruitment 2022 For 1925 Group A, B & C Posts SOME SPECIAL NUMBERS : गणित के कुछ जादुई अंक Best SSC Exam 2022 Preparation Books and Preparation Tips SSC CGL 2021-22 Recruitment Notification and Important dates JEE 2022 : Registration date and Exam date