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.

Trigonometry formulas:

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

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