Cos 90 Degrees: Value of cos 90 with Proof, Examples and FAQ

 

Cos 90 Degrees

In a right-angled triangle, the cosine function of an angle is the ratio of the length of the adjacent side and the hypotenuse side (of angle θ).

In this article, we will discuss the cosine of angle 90 degrees value, which is equal to zero. We will also derive this value using the quadrants of a unit circle.

\cos \left(90^{\circ}\right)=\cos \pi / 2=0

Proof of  Cos 90 Degrees

Now let us calculate the value of cos 90 using a unit circle that has its centre at the origin of the coordinate axes ‘x’ and ‘y’. 

Let P (a, b) be a point on the circle’s circumference that forms an angle POA = x radian. It means that the length of the arc PA is equal to x. From this, we define the value of sin x = PM/OP = b and cos x = OM/OP = a.

Consider a right-angled triangle OMP in the given unit circle.

Using the Pythagoras theorem,

\mathrm{OM}^{2}+\mathrm{MP}^{2}=\mathrm{OP}^{2}
\Rightarrow \mathrm{a}^{2}+\mathrm{b}^{2}=1

It tells that every point on the unit circle is defined as,

a^{2}+b^{2}=1
\Rightarrow \cos ^{2} x+\sin ^{2} x=1(\text { Since } \sin x=b \text { and } \cos x=a)

Remember that one complete revolution subtends an angle of 2π radian at the centre of the circle, and from the unit circle – 

\angle \mathrm{AOB}=\pi / 2,
\angle \mathrm{AOC}=\pi,
\angle \mathrm{AOD}=3 \pi / 2 .

The coordinates of the points A, B, C and D will be (1, 0), (0, 1), (–1, 0) and (0, –1), respectively. 

Now for the ∠AOB = π/2, we will take the coordinates of point B, which is (0, 1).

This means, for the ∠AOB = π/2, a = 0 and b = 1

Since cos x = a, 

So, cos π/2 = cos 90° = 0

Therefore, the value of the cosine of angle 90 degrees is 0.

Trigonometric Method

We can derive the value of cosine of 90 degrees using sin 90 value using the below formula:

\cos ^{2} x=1-\sin ^{2} x

Putting x = 90°

\cos ^{2} 90^{\circ}=1-\sin ^{2} 90^{\circ}

The value of sin 90 is 1.

\text { So, } \cos ^{2} 90^{\circ}=1-(1)^{2}

\cos ^{2} 90^{\circ}=1-1
\cos ^{2} 90^{\circ}=0
\cos \left(90^{\circ}\right)=0

 

Examples

1. Evaluate: cos 90° + sin 0°

We know that cos (90°) = sin (0°) = 0
So, cos (90°) + sin (0°)
= 0 + 0
= 0

 

2. Evaluate: 2cos 90° + 3sin 90°

We know that cos (90°) = sin (0°) = 0
and sin (90°) = cos (0°) = 1
So, 2cos (90°) + 3sin (90°)
= (2 x 0) + (3 x 1)
= 0 + 3
= 3