Find the value of cos(11(pi)/6)

Trigonometry is a branch of standardized mathematics that deals with the relationship between lengths, heights, and angles. Trigonometry is the branch of mathematics that deals with the ratios and the relations among the sides and angles in a triangle. Using Trigonometry cab be calculated for various measurements connected to a triangle. Some standard ratios are defined for the ease of calculation of some common problems related to the length and angles of the sides of a right-angled triangle.

Trigonometric Ratios

A trigonometric ratio is the proportion of sides with either of the acute angles in the right-angled triangle. A simple trigonometric ratio can be defined in terms of sides of a right-angled triangle i.e. the hypotenuse, base side, and the perpendicular side. There are three simple trigonometric ratios wiz. sine, cosine, and tangent.

Sine is the function that takes in the parameter an angle Î¸, which is either of the acute-angles in the right-angled triangles and is defined as the ratio of the length of the opposite side to the hypotenuse of the right-angled triangle. In technical terms, it can be written as,

sin(Î¸) = opposite side / hypotenuse

Cosine is the function that takes in the parameter an angle Î¸, which is either of the acute-angles in the right-angled triangles and is defined as the ratio of the length of the adjacent side to the hypotenuse of the right-angled triangle. In technical terms, it can be written as,Â

cos(Î¸) = adjacent side / hypotenuse

Tangent is the function that takes in the parameter an angle Î¸, which is either of the acute-angles in the right-angled triangles and is defined as the ratio of the length of the opposite side to the adjacent side of the right-angled triangle. In technical terms, it can be written as,Â

tan(Î¸) = opposite side / adjacent side

These trigonometric ratios are related to one another using some trigonometric identities and formulas,

tan(Î¸) = sin(Î¸)/cos(Î¸)

sin2(Î¸) + cos2(Î¸) = 1

Each of the trigonometric ratios has other three derived trigonometric ratios which are deduced by taking the inverse of the respective ratios. The Other three Trigonometric ratios are Cosecant, Secant, and Cotangent used mathematically as cosec, sec, and cot.

These are related to the Primary trigonometric ratios as follows,

cosec(Î¸) = 1 / sin(Î¸)

sec(Î¸) = 1 / cos(Î¸)

cot(Î¸) = 1 / tan(Î¸) = cos(Î¸) / sin(Î¸)

Below are some of the identities related to the standard trigonometric ratios and the derived trigonometric ratios,

tan2(Î¸) + 1 = sec2(Î¸)

cot2(Î¸) + 1 =cosec2(Î¸)

Trigonometric Table

The below is the table for some common angles and the basic trigonometric ratios. The value of each angle in the trigonometry is fixed and known but the ones mentioned are more common and mostly used,

Ratio\Angle(Î¸)

030456090
sin(Î¸)01/21/âˆš2âˆš3/21
cos(Î¸)1âˆš3/21/âˆš21/20
tan(Î¸)01/âˆš31âˆš3âˆž
cosec(Î¸)âˆž2âˆš22/âˆš31
sec(Î¸)12/âˆš3âˆš22âˆž
cot(Î¸)âˆžâˆš311/âˆš30

There are also some other trigonometric ratios to apply beyond the right-angled triangles:

sin(-Î¸) = – sin(Î¸)

cos(-Î¸) = cos(Î¸)

tan(-Î¸) = – tan(Î¸)

There are special trigonometric formula for tangent function,

cos (A + B) = [cos(A) .cos(B)] – [sin(A).sin(B)]

cos (A – B) = [cos(A) .cos(B)] + [sin(A).sin(B)]

Find the value of cos(11(pi)/6).

Method 1

Make use of simple trigonometric identity to calculate the value of cos(11Ï€/6). Here we make use of the following identity or formula,

cos(2Ï€ – Î¸) = cos(Î¸)

Solution:

cos(11Ï€/6) Â

(11Ï€/6) as (2Ï€ – Ï€/6)

So, cos(11Ï€/6) Â = cos(2Ï€ – Ï€/6)

cos(2Ï€ – Î¸) = cos(Î¸)

Here, Î¸ = Ï€/6 = 30Â°

cos(11Ï€/6) = cos(Ï€/6)

= âˆš3/2

Therefore, cos(11Ï€/6) Â = âˆš3/2Â

= 0.866025.

Method 2

Make use of simple trigonometric identity to calculate the value of cos(11(pi)/6). Here we make use of the following identity,

cos(3Ï€/2 + Î¸) = sin(Î¸)

Solution:

cos(11Ï€/6)Â

(11Ï€/6) as (3Ï€/2 + Ï€/3),

So, cos(11Ï€/6) Â = cos(3Ï€/2 – Ï€/3)

cos(3Ï€/2 + Î¸) = sin(Î¸)

Here, Î¸ = Ï€/3 = 60Â°

cos(11Ï€/6) = sin(Ï€/3)

= âˆš3/2

Therefore, cos(11Ï€/6) = âˆš3/2

= 0.866025.

Method 3

Make use of Compound Angle Formulae to calculate the value of cos(11(pi)/6). Here the use of the following identities and formulas is present,

cos(A + B) = (cos(A).cos(B)) – (sin(A).sin(B))

Solution:

cos(11Ï€/6)

(11Ï€/6) as (2Ï€/6 + 9Ï€/6),

So, cos(11Ï€/6) Â = cos(2Ï€/6 – 9Ï€/6)

The compound angle formulae,

cos (A + B) = (cos(A).cos(B)) – (sin(A).sin(B)),

here, A = 2Ï€/6 and B = 9Ï€/6

Therefore, cos(11Ï€/6) = cos( 2Ï€/6 + 9Ï€/6)Â

= [cos(2Ï€/6).cos(9Ï€/6)] – [sin(2Ï€/6).sin(9Ï€/6)]

= [cos(Ï€/3).cos(3Ï€/2)] – [sin(Ï€/3).sin(3Ï€/2)]

= [(1/2).(0)] – Â (âˆš3/2).(-1)]Â

= 0 – (-âˆš3/2)

= âˆš3/2

Therefore,cos(11Ï€/6) = âˆš3/2

= 0.866025.

So from the above methods, we were able to calculate the value of cos(11(pi)/6) which is cos(330Â°) to be âˆš3/2 or approximately 0.86602.

Similar Problems

Question 1: Find the value of cos(5Ï€/6)

Solution:

cos(5Ï€/6)Â

(5Ï€/6) as ( Ï€/2 + Ï€/3)

So, cos(5Ï€/6) = cos (Ï€/2 + Ï€/3)

cos(Ï€/2 + Î¸) = -sin(Î¸)

So, cos(5Ï€/6) = cos (Ï€/2 + Ï€/3)

= – sin(Ï€/3)

= – âˆš3/2

Therefore,Â

cos(5Ï€/6) = – âˆš3/2

Question 2: Find the value of cos(5Ï€/12)

Solution:

cos(5Ï€/12)

(5Ï€/12) as (Ï€/6 + Ï€/4)

So, cos(5Ï€/12) = cos (Ï€/6 + Ï€/4)

cos (A + B) = [cos(A) Ã— cos(B)] – [sin(A) Ã— sin(B)]

here, A = Ï€/6 and B= Ï€/4

So, cos(5Ï€/12) = cos (Ï€/6 + Ï€/4)

= [cos(Ï€/6) Ã— cos(Ï€/4)] – [sin(Ï€/6) Ã— sin(Ï€/4)]

= [(âˆš3/2) Ã— (1/âˆš2)] – [(1/2) Ã— (1/âˆš2)]

= (âˆš3/2âˆš2) – (1/2âˆš2)

= (âˆš(3)-1) / (2âˆš2)

= 0.258819.

Therefore,Â

cos(5Ï€/12) = (âˆš(3)-1)/(2âˆš2)Â

= 0.258819.

Question 3: If A = 3Ï€/4, find the values of all trigonometric ratios.

Solution:

A = 3Ï€/4,

Find the value of all trigonometric ratiosÂ

That is, sin(A), cos(A), tan(A), cosec(A), sec(A) and cot(A)

(3Ï€/4) as (Ï€ – Ï€/4) can be written,

So, sin(A) = sin(3Ï€/4)Â

= sin (Ï€ – Ï€/4)

sin(Ï€ – Î¸) = sin(Î¸)

Therefore, sin(A) = sin (Ï€ – Ï€/4)

= sin( Ï€/4)

= 1 / âˆš2 Â  Â  Â  Â  Â  Â  Â

Thus, sin(3Ï€/4) = Â 1 / âˆš2

Similarly, cos(A) = cos (3Ï€/4)

= cos (Ï€ -Ï€/4)

= – cos (Ï€/4)

= – 1 / âˆš2 Â  Â  Â

Thus, cos(3Ï€/4) = -1 / âˆš2

Now, tan(A) = sin(A) / cos(A)

Thus, tan(A) = (1 / âˆš2) / (- 1/âˆš2)

tan(A) = -1

Now, cosec(A) = 1/sin(A) = 1 / (1/âˆš2) = âˆš2

sec(A) = 1/ cos(A) = 1 / (-1 / âˆš2) = -âˆš2

cot(A) = 1/ tan(A) = 1/ (-1) = -1

Therefore,

sin(A) = 1/âˆš2, cosec(A) = âˆš2

cos(A) = – 1/âˆš2, Â sec(A) = – âˆš2

tan(A) = -1, cot(A) = -1

Previous
Next