August 8, 2024

# 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 $\large&space;0^{\circ},&space;30^{\circ},&space;45^{\circ},&space;60^{\circ},&space;90^{\circ}&space;etc.$  are  commonly used to solve many problems related to triangles.

Trigonometry formulas:

Trigonometric ratios of some specific angles

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

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

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

$\large&space;2sinxcosx=4cos^{3}x-3cosx$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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#### Bina singh

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