Here, we will discuss the value for cos 0 degrees and how the values are derived using the quadrants of a unit circle. The trigonometric functions are also known as an angle function that relates the angles of a triangle to the length of the triangle sides. Trigonometric functions are one of the most important topics which are used in the study of periodic phenomena like sound and light waves, the study of harmonic oscillators and finding the average temperature variations. The three familiar trigonometric ratios are sine function, cosine function, and tangent function. It is commonly defined for the angles less than a right angle, trigonometric functions are stated as the ratio of two sides of a right triangle containing the angle in which the values can be found in the length of various line segments around a unit circle. The angles of a triangle are calculated with respect to sin, cos and tan functions. Usually, the degrees are represented as 0Â°, 30Â°, 45Â°, 60Â°, 90Â°, 180Â°, 270Â°Â and 360Â°.
Cos 0 Degree Value
To define the cosine function of an acute angle, start with the rightangled triangle ABC with the angle of interest and the sides of a triangle. The three sides of the triangle are defined as follows:
 The opposite side is defined as the side opposite to the angle of interest.
 The hypotenuse side is the side opposite the right angle and it should be the longest side of a right triangle
 The adjacent side is the remaining side where it forms a side of both the angle of interest and the right angle
The cosine function of an angle is equal to the length of the adjacent side divided by the length of the hypotenuse side and the formula is given by
Cos Î¸ = Adjacent Side / Hypotenuse Side
Value of Cos 0 Using Unit Circle
Assume a unit circle with the center at the origin of the coordinate axes. Consider that P (a, b) be any point on the circle which forms an angle AOP = x radian. This means that the length of the arc AP is equal to x. So we define that cos x = a and sin x = b
Now consider a triangle OMP is a right triangle,
By using the Pythagorean theorem, we get
OM^{2}+ MP^{2}= OP^{2} (or) a^{2}+ b^{2}= 1
So for every point on the unit circle, we define it as
a^{2}+ b^{2} = 1 (or) cos^{2} x + sin^{2} x = 1
It is noted that the one complete revolution subtends an angle of 2Ï€ radian at the centre of the circle,
âˆ AOB=Ï€/2,
âˆ AOC = Ï€ and
âˆ AOD =3Ï€/2.
Since all angles of a triangle are the integral multiples of Ï€/2 and it is commonly called as quadrantal angles. Therefore, the coordinates of the points A, B, C and D are (1, 0), (0, 1), (â€“1, 0) and (0, â€“1) respectively. Therefore, from the quadrantal angles, we can get the cos 0 value
Cos 0Â°Â = 1
Now, when we take one complete revolution from the point P, again it comes back to the same point P. So, we also observe that the values of sine and cosine functions do not change, if x increases or decreases by an integral multiple of 2Ï€,
cos (2nÏ€ + x) = cos x, where n âˆˆ Z
Further, it is observed that
cos x = 0, when x = Â±Ï€/2, Â±3Ï€/2, Â±5Ï€/2, … It means that cos x vanishes when x is an odd multiple of Ï€/2.
So, cos x = 0 implies x = (2n + 1)Ï€/2 , where n takes the value of any integer.
For a triangle, ABC having the sides a, b, and c opposite the angles A, B, and C, the cosine law is defined.
Consider for an angle C, the law of cosines is stated as
c^{2 }= a^{2 }+ b^{2}– 2ab cos(C)
In the same way, we can derive other values of cos degrees like 30Â°, 45Â°, 60Â°, 90Â°, 180Â°, 270Â°and 360Â°. Also, it is easy to remember the special values like 0Â°, 30Â°, 45Â°, 60Â°, and 90Â° since all the values are present in the first quadrant. All the sine and cosine functions in the first quadrant take the form \(\frac{\sqrt{n}}{2}\) or Â \(\sqrt{\frac{n}{4}}\). Once we find the values of sine functions it is easy to find the cosine functions.
Sin 0Â° = \(\sqrt{\frac{0}{4}}\)
Sin 30Â° = \(\sqrt{\frac{1}{4}}\)
Sin 45Â° = \(\sqrt{\frac{2}{4}}\)
Sin 60Â° = \(\sqrt{\frac{3}{4}}\)
Sin 90Â° = \(\sqrt{\frac{4}{4}}\)
Now Simplify all the sine values obtained and put in the tabular form:
0Â° 
30Â°  45Â°  60Â° 
90Â° 

Sin 
0 
1/2  \(\frac{1}{\sqrt{2}}\)  \(\frac{\sqrt{3}}{2}\)  1 
From the values of sine, we can easily find the cosine function values. Now, to find the cos values, fill the opposite order the sine function values. It means that
Cos 0Â° = Sin 90Â°
Cos 30Â° = Sin 60Â°
Cos 45Â° = sin 45Â°
Cos 60Â° = sin 30Â°
Cos 90Â° = sin 0Â°
So the value of cos 0 degrees is equal to 1 since cos 0Â° = sin 90Â°
0Â° 
30Â°  45Â°  60Â° 
90Â° 

Sin 
0 
1/2  \(\frac{1}{\sqrt{2}}\)  \(\frac{\sqrt{3}}{2}\) 
1 
Cos 
1 
\(\frac{\sqrt{3}}{2}\)  \(\frac{1}{\sqrt{2}}\)  1/2 
0 
In a similar way, we can find the values of other degrees of trigonometric functions depends on the quadrant value. Below is the trigonometry table, which defines all the values of cosine function along with other trigonometric ratios.
Trigonometry Ratio Table  
Angles (In Degrees)  0  30  45  60  90  180  270  360 
Angles (In Radians)  0  Ï€/6  Ï€/4  Ï€/3  Ï€/2  Ï€  3Ï€/2  2Ï€ 
sin  0  1/2  1/âˆš2  âˆš3/2  1  0  âˆ’1  0 
cos  1  âˆš3/2  1/âˆš2  1/2  0  âˆ’1  0  1 
tan  0  1/âˆš3  1  âˆš3  Not Defined  0  Not Defined  0 
cot  Not Defined  âˆš3  1  1/âˆš3  0  Not Defined  0  Not Defined 
cosec  Not Defined  2  âˆš2  2/âˆš3  1  Not Defined  âˆ’1  Not Defined 
sec  1  2/âˆš3  âˆš2  2/âˆš3  Not Defined  âˆ’1  Not Defined  1 
Example
Question:
Find the value of cos 15Â°
Solution:
Cos 15Â°= cos(45Â°30Â°)
Now, take the values a = 45Â°Â and b = 30Â°
By using the formula, Cos (ab) = cos a cos b + sin a sin b
So, it becomes Cos 15Â°Â = cos 45Â°Â cos 30Â°Â +sin 45Â°Â sin 30Â°
\(\cos 15^{\circ}= \frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\frac{1}{2}\) \(\cos 15^{\circ}= \frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}\) \(\cos 15^{\circ}= \frac{\sqrt{3}+1}{2\sqrt{2}}\)For more information on cos 0 and other trigonometric functions, visit BYJUâ€™S and also watch the interactive videos to clarify the doubts.
Related Links  
Sin 0 Degree  Trigonometry Formulas 
Inverse Cosine  Cosine Rule 