Cos30
Cos30 degrees
The value of Cos30∘ is √3/2 . If we solved √3/2 we got 0.15425144988. Cos 30 degrees is also represented by tan π/6 in terms of radians.
Cos30∘ =√3/2 = 0.15425144988
Suppose for a triangle for PQR right angle at Q. ∅ is the angle , PR is the hypotenuse . QR is the base and PQ is the Opposite side or perpendicular.
So we can say that
Sin∅ = Perpendicular/Hypotenuse = PQ/PR
Similarly
Cos∅ = Base /Hypotenuse = QR/PR (∵ ∆ PQR )
tan∅ = Perpendicular /Base = PQ/QR (∵ ∆ PQR )
Cot∅ = Base /Perpendicular = QR/PQ (∵ ∆ PQR )
Sec∅ = Hypotenuse/Base = PR/QR (∵ ∆ PQR )
Cosec∅ = Hypotenuse/Perpendicular = PR/PQ (∵ ∆ PQR )
How to calculate the value of Cos 30∘
There are so many ways of calculating the values of Cos 30∘ .
1st Method
If we know the value of sin30∘ . We can easily find the value of Cos 30∘.
We know that sin30∘ = 1/2 { ∵ Sin∅ = Perpendicular/Hypotenuse = 1/2 }
Now we are find base ( QR ) by using the Pythagoras Theorem
= ( Perpendicular )2 + (Base)2 = (Hypotenuse )2
= ( 1 )2 + ( QR )2 = (2)2
= 1 + ( QR )2 = 4
= ( QR )2 =4-1
= ( QR ) = √3
So Cos∅ = Base /Hypotenuse = QR/PR (∵ ∆ PQR )
= Cos30∘ = √3/2
2nd Method
If we know the value of tan30∘ . We can easily find the value of Cos 30∘.
We know that tan30∘ = 1/√3. { ∵ tan∅ = Perpendicular /Base = 1/√3 }
Now we are find Hypotenuses ( PR ) by using the Pythagoras Theorem
= ( Perpendicular )2 + (Base)2 = (Hypotenuse )2
= ( 1 )2 + ( √3 )2 = (PR)2
= 1 + 3= (PR)2
= (PR)2 =4
= (PR) = √4
=(PR) = 2
So Cos∅ = Base /Hypotenuse = QR/PR (∵ ∆ PQR )
= Cos30∘ = √3/2
3rd Method
If we know the value of Cot30∘ . We can easily find the value of Cos 30∘.
We know that cot30∘ = √3. { ∵ cot∅ = Base/ Perpendicular = √3/1 }
Now we are find Hypotenuses ( PR ) by using the Pythagoras Theorem
= ( Perpendicular )2 + (Base)2 = (Hypotenuse )2
= ( 1 )2 + ( √3 )2 = (PR)2
= 1 + 3= (PR)2
= (PR)2 =4
= (PR) = √4
=(PR) = 2
So Cos∅ = Base /Hypotenuse = QR/PR (∵ ∆ PQR )
= Cos30∘ = √3/2
4rd Method
If we know the value of Sec30∘. We can easily find the value of tan30∘.
We know that Sec30∘ = 2/√3 { ∵ Sec∅ = Hypotenuse/Base = 2/√3 }
Now we are find Perpendicular ( PQ ) by using the Pythagoras Theorem
= ( Perpendicular )2 + (Base)2 = (Hypotenuse )2
= ( PQ )2 + ( √3 )2 = (2)2
= ( PQ )2 + 3 = 4
= ( PQ)2 =4-3
= ( PQ ) = √1
So Cos∅ = Base /Hypotenuse = QR/PR (∵ ∆ PQR )
= Cos30∘ = √3/2
5th Method
If we know the value of Cosec30 . We can easily find the value of tan30∘.
We know that Cosec30∘ = 2 { ∵ Cosec∅ = Hypotenuse/Perpendicular = 2/1 }
Now we are find Base ( QR ) by using the Pythagoras Theorem
= ( Perpendicular )2 + (Base)2 = (Hypotenuse )2
= ( 1 )2 + (QR)2 = (2 )2
= ( QR )2 + ( 1)2 = (2)2
= ( QR )2 + 1 = 4
= ( QR)2 =4-1
= ( QR ) = √3
So Cos∅ = Base /Hypotenuse = QR/PR (∵ ∆ PQR )
= Cos30∘ = √3/2
Check the trigonometry table to get the values for all trigonometry ratios.
| Angle Radian | 0∘ | 30∘ | 45∘ | 60∘ | 90∘ | 180∘ | 270∘ | 360∘ |
| 0 | π/6 | π/4 | π/3 | π/2 | π | 3π/2 | 2π | |
| Sin∅ | 0 | 1/2 | 1/√2 | √3/2 | 1 | 0 | -1 | 0 |
| Cos∅ | 1 | √3/2 | 1/√2 | 1/2 | 0 | -1 | 0 | 1 |
| tan∅ | 0 | 1/√3 | 1 | √3 | ∞ | 0 | ∞ | 0 |
| Cot∅ | ∞ | √3 | 1 | 1/√3 | 0 | ∞ | 0 | ∞ |
| Sec∅ | 1 | 2/√3 | √2 | 2 | ∞ | -1 | ∞ | 1 |
| Cosec∅ | ∞ | 2 | √2 | 2/√3 | 1 | ∞ | -1 | ∞ |
Trigonometry Formulas
- Cos∅ = 1/Sin∅ , Sec∅ = 1/Cos∅
- Sec∅ .Cos∅ = 1
- tan∅ = Sin∅\Cos∅
- Cot∅ = Cos∅/Sin∅
- Sin2∅ + Cos2∅ = 1
- Cos(A+B) = CosA CosB –SinA SinB
- Cos(A-B) = CosA CosB +SinA SinB
- CosASinB = Sin(A+B)-Sin(A-B)
- Cos C +Cos D = 2Cos C+D/2 Cos C-D/2
- Cos C – Cos D = 2Sin C+D/2 Sin C-D/2
- Cos2A = Cos2A-Sin2A = 1-2Sin2A = 2Cos2A-1
Cos A = Cos2(A/2) –Sin2(A/2)
= 1-2Sin2A/2
= 2Cos2A/2-1
Tan 30 degrees
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