# 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 .

### 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 )=4-1

= ( QR ) = √3

So Cos∅ = Base /Hypotenuse = QR/PR  (∵ ∆ PQR )

= Cos30∘ = √3/2

### 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

### 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

### 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)=4-3

= ( PQ ) = √1

So Cos∅ = Base /Hypotenuse = QR/PR (∵ ∆ PQR )

Cos30∘ =  √3/2

### 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)=4-1

= ( QR ) = √3

So Cos∅ = Base /Hypotenuse = QR/PR (∵ ∆ PQR )

Cos30∘ =  √3/2

### 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