Sin 18

Trigonometry is one of the most interesting concepts of mathematics. In trigonometry, we come across various methods all incorporating three major or fundamental functions, Sine, Cosine and Tangent, which are generally used to determine the angles and length of the right-angled triangle. Before we start our discussion of Sin 18, let us first understand the Sine function.

Sine function represents a unique relation between the angle and perpendicular side and the hypotenuse side. In other words, we can say Sin is equivalent to the ratio of perpendicular and hypotenuse of a right-angled triangle.

Let us start our discussion of Sin 18 degrees and estimate the sin18 value with the help of trigonometric identities.

Sin 18 Degrees:

The word sine originated from the Latin word sinus which literally translates to 'bay' or 'fold'. This implies that it is either derived from a term that means bowstring. As we know that, the sin of angle theta is basically the ratio of the length of the opposite side, (perpendicular side) and hypotenuse of the given right-angled triangle.

The value of the sine function at an angle of 18 degrees is known as Sin 18 degrees and it can be written as $\sin18^\circ$ or Sin 18 degrees. In a circular system, Sin 18 degrees can be written as $\sin\frac{\pi}{10}$. The value of $\sin18^\circ$ is given by:

$\Rightarrow sin18^\circ = \frac{\sqrt{5-1}}{4} \\ \Rightarrow sin18^\circ = 0.30901$

How to Calculate the Sin18 Value?

These kinds of special angles are solved easily using double angles. The Double angle formulas are generally applied to express the trigonometric ratios of double angles $2\theta$ in terms of trigonometric ratios of single angle $\theta$ and similarly, the half-angle formulas are usually employed to express the trigonometric ratios of half angles $\frac{1}{2}\theta$ in terms of trigonometric ratios of single angle $\theta$. Thus, the exact value of $\sin18^\circ$ can be found using the sum and difference, double angle, or half-angle formulas and it is found to be $sin18^\circ = \frac{\sqrt{5-1}}{4}$

Let us have a look at how to find the value of sin 18 degrees:

Let us assume that $\theta = 18^\circ$ ….(1)

We have to analyse the problem as to how we can write standard angles such as 180 in terms of the 18. Here, we can write 18 in terms of 90 since sin 90 will be equal to 1. This step is followed in order to simplify the solution.

Now, we know that,

$ \Rightarrow 90^{\circ} = 36^{\circ} + 54^{\circ} \\ $

$\Rightarrow 90^{\circ} = (2\times 18^{\circ}) + (3\times 18^{\circ}) \\$

$\Rightarrow 90^{\circ} = 2\theta + 3\theta \\$

(We assumed 18 as θ in the beginning)

$3\theta = 90^{\circ} - 2\theta -------->(2)$

Now taking sine on both LHS and RHS of the equation (2),

$\Rightarrow\sin3\theta = \sin(90^{\circ} - 2\theta )$

Using the trigonometric identity $\sin 3A = 3\sin A -4\sin^{3}A$ and $\sin (90-\theta) =\cos\theta$ in the above equation we get:

$\Rightarrow3\sin\theta - 4sin^{3}\theta = \cos 2\theta -----> (3)$

We know that:

$\cos 2\theta =1 -2\sin^{2} \theta—------>(4)$

Substituting equation (4) in equation (3) and simplify:

$ \Rightarrow3\sin\theta - 4sin^{3}\theta = 1 -2\sin^{2}\theta \\$

$\Rightarrow3\sin\theta - 4sin^{3}\theta + 2\sin^{2} \theta -1 =0 \\$

$\Rightarrow 4sin^{3}\theta -2\sin^{2} \theta -3sin\theta +1 =0 -------->(5)$

Let us assume that $\sin\theta = x$ , substitute $\sin\theta = x $ in the equation (5) we will get an algebraic equation of order 3.

$\Rightarrow 4x^{3}-2x^{2}-3x+1 = 0$

Simplifying the above expression using factor method, we get:

$\Rightarrow (x-1)(4x^{2}+2x-1) = 0 \\ \Rightarrow (x-1)=0 , (4x^{2}+2x-1) = 0 $

Therefore, the roots of the above equation x=1can not be accepted as we know that the maximum value of the sine function is 1 and the Sin18 Value must be less than 1. Because the roots of the sine will always lie between 1 and -1, we know that the range of sine function is $ -1\leqslant \sin \leqslant 1 $. -1 ≤ Sin ≤ 1.

Thus, let us verify the root $4x^{2}+2x-1 = 0$

$\Rightarrow 4x^{2}+2x-1 = 0$

Using quadratic equation formula, we write:

$\Rightarrow x={\frac{-b \pm\sqrt{b^{2}-4ac}}{2a} }$

Here,

a = 4, b = 2, c = -1

$ x={\frac{-2 \pm\sqrt{2^{2}-4(4)(-1)}}{2\times4} } \\$

$x={\frac{-2 \pm\sqrt{20}}{8} }\\ $

$ x={\frac{2(-1 \pm\sqrt{5})}{8} }\\$

$x={\frac{(-1 \pm\sqrt{5})}{4} }\\ $

$x={\frac{-1 +\sqrt{5}}{4} }~~~~~ {\frac{\sqrt{5}-1}{4} }$

Therefore, the Sin18 value is given by the above equation. While finding Sin 18 degrees value can not consider $\frac{-1-\sqrt{5}}{4}$root, since we know that sin18 value must be less than 1 and should lie between -1 and 1. And sin 18 degrees in fraction is equal to 0.30901.

Sin 18 Value in Radians

We got the value of sin 18 in both fraction and decimal form. Now here let us find the value of sin 18 in terms of radian. We can find the value by converting the radians measure value to degree measure.

We know that,

$1 rad = \frac{\pi}{180^{\circ}}$

$\frac{\pi }{180^0}$ degrees

Then here

1 rad = 57.2958 degrees

18 rad = 1031.32 degrees

Thus, sin 18 rad = sin 1031.32°

.Hence, sin 18 rad = 0.769214…

We know that 1031.32° lies in the fourth quadrant, where sin is negative. Therefore, sin 18 rad = -0.769214…

Solved Examples

1. Calculate Cos 18 value.

Solution:

We are asked to calculate Cos 18. The Cosine 18 degrees can be easily found using one of the trigonometric identities given by:

$\Rightarrow \sin^{2}A + \cos^{2}A =1$

Here, consider A=18°

Then, we get:

$\Rightarrow[{\frac{-1 +\sqrt{5}}{4} }]^{2} + \cos ^{2} 18^{\circ} =1 \\$

$ \Rightarrow {\frac{(-1 +\sqrt{5})^{2}}{16} } + \cos ^{2} 18^{\circ} =1 \\$

$ \Rightarrow \cos ^{2} 18^{\circ} = 1-\frac{6-2\sqrt{5}}{16} = \frac{16-6+2\sqrt{5}}{16} \\ $

$\Rightarrow \cos ^{2} 18^{\circ} = \frac{16-6+2\sqrt{5}}{16} \\ $

$ \Rightarrow \cos ^{2} 18^{\circ} = \frac{10+2\sqrt{5}}{16} \\$

$ \Rightarrow \cos 18^{\circ} = \sqrt{\frac{10+2\sqrt{5}}{16}} \\ $

$\Rightarrow \cos 18^{\circ} = \frac{\sqrt{10+2\sqrt{5}}}{4}$

Therefore, Cos 18 value is found using Sin 18 degrees and is equal to $\frac{\sqrt{10+2\sqrt{5}}}{4}$.

2. Calculate $\frac{\sin 18^{\circ}}{\cos 72^{\circ}}$ using Sin 18 degrees.

Solution:

In this example, we are asked to determine the ratio of $\frac{\sin 18^{\circ}}{\cos 72^{\circ}} $. We know the sin18 value, in order to find the ratio first we have to calculate the Cos 72.

We will start finding the value of Cos 72 by rewriting the cosine value in the simplified form. We will write cos 72 as:

$\Rightarrow {\cos 72^{\circ}} = cos (90^{\circ} - 18^{\circ}) \\ $

$\Rightarrow {\cos 72^{\circ}} = \sin 18^{\circ}$

(from trigonometric allied angles we know that $ cos (90^{\circ} - x) = \sin x$).

$\Rightarrow \cos 72^{\circ} = \frac{\sqrt{5}-1}{4}$

Now, the ratio of sin 18 degrees to the cos 72 will be:

$\frac{\sin 18^{\circ}}{\cos 72^{\circ}} = \frac{\frac{\sqrt{5}-1}{4}}{\frac{\sqrt{5}-1}{4}} =1$

Therefore, the value of $ \frac{\sin 18^{\circ}}{\cos 72^{\circ}}$ using the Sin 18 degree value is 1.

3. Find the value of $\sin 18 ^{\circ}$ if $\csc 18 ^{\circ}$ is 3.2360.

Solution:

We know that $\sin =\frac{1}{\csc}$

$\Rightarrow \sin 18 ^{\circ} =\frac{1}{\csc 18 ^{\circ}} \\ $

$\Rightarrow \sin 18 ^{\circ} =\frac{1}{3.2360} \\ \Rightarrow \sin 18 ^{\circ} = 0.309$

Practise Questions

1.If $\tan \theta$ = 34 then the value of $\sin\theta$ is _________

a) $\dfrac{3}{5}$

b) $\dfrac{6}{5}$

c) $\dfrac{3}{4}$

d) $\dfrac{-3}{5}$

2. Find the value of $\frac{2\cos 18^{\circ}}{3\sin 72^{\circ}}$.

a) $\dfrac{3}{5}$

b) $\dfrac{1}{5}$

c) $\dfrac{2}{3}$

d) $\dfrac{4}{5}$

3. Using the value of $\sin18^{\circ}$, solve: $1-\cos^{2} (18^{\circ})$

a) 0.0955

b) 0.737

c) 0.9988

d) 0.1437

Answer

1. a

2. c

3. a

Conclusion

Here we have discussed sin 18 value and we also found the sin 18 value in different terms like degrees, fraction, decimals and radians. To get the values of sin for other angle measures, you can visit our website. Hope this article was helpful.