trigonometric multiple angles formula and their uses | relation for sin2A, cos2A etc. | Knowledge of Physics

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If A, B and C are single angles, then the angles of the form 2A, 2B, 2C, 3A, 3B, 3C etc., which are formed by multiplying the individual angle by integers like 2, 3, 4, ....., are called multiple angles.
Relation of trigonometric ratios of multiple angles with those of single angles have major importance in various field of study such as physics, mathematics, engineering etc. 

This post helps readers to learn about multiple angle formulae, derive these relations as well as apply these relations to solve few trigonometric problems.

Some relations of multiple angles are given as below:
  1. sin2A =  2sinA. cosA = `\frac{2tanA}{1+tan^2A} =  \frac{2cotA}{cot^2A+1}`
  2. cos2A = `cos^2A - sin^2A = 2cos^2A-1 = 1-2sin^2A`
  3. cos2A = `\frac{1-tan^2A}{1+tan^2A} = \frac{cot^2A-1}{cot^2A+1}` 
  4. tan2A = `\frac{2tanA}{1-tan^2A}` 
  5. cot2A = `\frac{cot^2A-1}{2 cotA}` 
  6. sin3A = `3sinA - 4sin^3A` etc.
We try to prove these relations plus few other formulae as well on this post.

1. Proving the formula: sin2A = 2sinA cosA

We have from compound angle formula, 
sin(A+B) = sinA cosB + sinB cosA

For B =A, 
sin(A+A) = sinA cosA + sinA cosA

or, sin2A = 2sinA cosA

`\therefore` sin2A = 2sinA cosA -------(1)
This proves the relation.

Also, multiplying (1) by `\frac{cosA}{cosA}`, we get

sin2A = `\frac{2sinA cos^2A}{cosA}`

or, sin2A = 2`\frac{sinA}{cosA} \frac{1}{sec^2A}`

Since, `\frac{sinA}{cosA} = tanA` and `sec^2A = 1+tan^2A`, then

sin2A = 2 tanA `\frac{1}{1+tan^2A}`

`\therefore` `sin2A = \frac{2 tanA}{1+tan^2A}` -------(2)
This is another relation for sin2A.

Using `\frac{1}{cotA} = tanA` in (2), we get

`\therefore` `sin2A = \frac{2 cotA}{cot^2A+1}` -------(3)
This is sin2A in terms of cotA.

2. Proving the formula: cos2A = `cos^2A - sin^2A`

We have, cos(A+B) = cosA cosB - sinA sinB, from compound angle formula given in the link.

Put B =A, then

cos(A+A) = cosA cosA - sinA sinA

`therefore` cos2A = `cos^2A - sin^2A` ----(4)
This proves the formula.

Again, using the identity  `sin^2A+cos^2A = 1` in (4),

`cos2A = 1 - sin^2A - sin^2A`

`\therefore` `cos2A = 1 - 2sin^2A` ------(5)
This gives cos2A on terms of sinA.

Also,
`cos2A = cos^2A - (1-cos^2A) `

`cos2A = cos^2A -1+cos^2A`

`\therefore` `cos2A = 2cos^2A -1`------(6)
This gives cos2A in terms of cosA.

Let's further simplify (6), then

`cos2A = 2cos^2A-1`

or, `cos2A = 2\frac{1}{sec^2A}-1`

or, `cos2A = 2 \frac{1}{1+tan^2A} -1`

or, `cos2A =  \frac{2 - (1+tan^2A)}{1+tan^2A}`

or, `cos2A = \frac{1-tan^2A}{1+tan^2A}`

`therefore` `cos2A = \frac{1-tan^2A}{1+tan^2A}` ------(7)
This gives cos2A in terms of tanA.

Again, from (5),
`cos2A = 1 - 2 sin^2A`

Using the reciprocal relation, `sinA = \frac{1}{cosecA}`,  we get

`cos2A = 1 - 2 \frac{1}{cosec^2A}`

Again using the identity, `cosec^A-cot^2A = 1`, we have

`cos2A = 1 -2 \frac{1}{1+cot^2A}`

or, `cos2A = \frac{1 + cot^2A - 2}{1+cot^2A}`

`\therefore` `cos2A = \frac{cot^2A-1}{cot^2A+1}` ----(8)
This relation gives cos2A in terms of cotA.

3. Proving the formula: `tan2A = \frac{2tanA}{1-tan^2A}`

We have,
`tan(A+B) = \frac{tanA+tanB}{1-tanA tanB}`, then for B=A we have

`tan(A+A) = \frac{tanA+tanA}{1-tanA tanA}`

`\therefore` `tan2A = \frac{2 tanA}{1-tan^2A}` -----(9)
This gives tan2A in terms of tanA.

Also, using `tanA = \frac{1}{cotA}` in (9), we have

`tan2A = 2\frac{\frac{1}{cotA}}{1-\frac{1}{cot^2A}}`

`\therefore` `tan2A =  \frac{2 cotA}{cot^2A-1}` -----(10)
This relation gives tan2A in terms of cotA.

4. Proving the formula: `cot2A = \frac{cot^2A-1}{2 cotA}`
We have,

`cot(A+B) = \frac{cotA cotB -1 }{cotB+cotA}`

Then using B=A, we get

`cot(A+A) = \frac{cotA cotA-1}{cotA+cotA}`

`cot2A = \frac{cot^2A-1}{2 cotA}`

`\therefore` `cot2A = \frac{cot^2A-1}{2 cotA}` ------(11)
This formula gives cot2A in terms of cotA.

Again.
Using `cotA = \frac{1}{tanA}` in (11), we get

`cot2A = \frac{\frac{1}{tan^2A}-1}{2 \frac{1}{tanA}}`

`cot2A = \frac{1-tan^2A}{2tanA}` 

`\therefore` `cot2A = \frac{1-tan^2A}{2 tanA}` ------(12)
This formula gives cot2A in terms of tanA.


Formula of multiple Angle 3A:

1. Proving the formula: `sin3A =3sinA - 4sin^3A`

This formula can also be proved by using both the compound angle formula and multiple angle formula of type 2A. 
We have, 
sin(3A) = sin(2A+A)

or, sin3A = sin2A  cosA + sinA cos2A

or, `sin3A = 2 sinA cosA . cosA + sinA (cos^2A - sin^2A)` 

or, `sin3A = 2sinA cos^2A + sinA (1-sin^2A-sin^2A)`

or, `sin3A = 2sinA (1-sin^2A) + sinA(1-2sin^2A)`

or, `sin3A = 2sinA - 2sin^3A + sinA - 2sin^3A`

or, `sin3A = 3sinA - 4sin^3A`

 `\therefore` `sin3A = 3sinA-4sin^3A` ----(1)
This formula gives sin3A in terms of sinA.

2. Proving the formula: `cos3A = 4cos^3A - 3cosA `
We have,
cos3A = cos(2A+A)

or, cos3A = cos2A cosA - sin2A sinA

or, `cos3A = (2cos^2A-1)cosA - 2sinA cosA sinA` 

or, `cos3A = 2cos^3A - cosA - 2sin^2A cosA`

or, `cos3A = 2cos^3A - cosA - 2 (1-cos^2A)cosA`

or, `cos3A = 2cos^3A - cosA - 2cosA + 2cos^3A`

`\therefore` `cos3A = 4cos^3A - 3cosA`-----(2)
This relation gives cos3A in terms of cosA.

3. Proving the formula: `tan3A =\frac{3tanA-tan^3A}{1-3tan^2A}`

We have,

`tan3A = tan(2A+A)`

or, `tan3A = \frac{tan2A+tanA}{1-tan2A tanA}`

or, `tan3A = \frac{\frac{2 tanA}{1-tan^2A}+tanA}{1- \frac{2tanA}{1-tan^2A}tanA}`

or, `tan3A = \frac{2tanA + tanA - tan^3A}{1-tan^2A-2tan^2A}`

`\therefore` `tan3A = \frac{3tanA-tan^3A}{1-3tan^2A}` -----(3)
This relation gives tan3A in terms of tanA.

4. Proving the formula: `cot3A = \frac{cot^3A-3cotA}{3cot^2A-1}`

We have,

cot3A = cot(2A+A)

or, `cot3A = \frac{cot2A cotA-1}{cot2A+cotA}`

or, `cot3A = \frac{\frac{cot^2A-1}{2cotA}cotA-1}{\frac{cot^2A-1}{2cotA}+cotA}`

or, `cot3A = \frac{cot^3A-cosA-2cotA}{cot^2A-1+2cot^2A}`

or, `cot3A = \frac{cot^3A-3cotA}{3cot^2A-1}`

`\therefore` `cot3A = \frac{cot^3A-3cotA}{3cot^2A-1}` ------(4)
This relation gives cot3A in terms of cotA.

Also, equation (4) can obtained from (3) as follows,

`tan3A = \frac{3tanA-tan^3A}{1-3tan^2A}`
Taking reciprocal,

`\frac{1}{cot3A} = \frac{3tanA-tan^3A}{1-3tan^2A}`

`cot3A = \frac{1-3tan^2A}{3tanA-tan^3A}`

or, `cot3A = \frac{1-3\frac{1}{cot^2A}}{3 \frac{1}{cotA} - \frac{1}{cot^3A}}`

or, `cot3A = \frac{cot^2A-3}{cot^2A} \frac{cot^3A}{3cot^2A-1}`

or, `cot3A = \frac{cot^2A-3}{3cot^2A-1}cotA`

or, `cot3A = \frac{cot^3A-3cotA}{3cot^2A-1}`

`\therefore` `cot3A = \frac{cot^3A-3cotA}{3cot^2A-1}` ------(5)
This relation gives cot3A in terms of cotA.


Some problems and their Solutions by Using Multiple Angle Formula:

1. If `cos2A = \frac{11}{25}` then prove that `cosA = \frac{6}{5\sqrt{2}}`.
Solution:

Here, given relation is,

`cos2A = \frac{11}{25}` 

By using multiple angle formula, `cos2A = 2cos^2A-1` on left hand side, we have

`2cos^2A -1  = \frac{11}{25}`

or,   `2cos^2A = \frac{11}{25}+1`

or, `2cos^2A = \frac{11+25}{25}`

or, `cos^2A = \frac{1}{2} \frac{36}{25}`

or, `cosA = \sqrt{\frac{36}{2\times25}}`

or, `cosA = \sqrt{\frac{6^2}{5^2}} \frac{1}{\sqrt{2}}`

or, `cosA = \frac{6}{5} \frac{1}{\sqrt{2}}`

`\therefore` `cosA = \frac{6}{5\sqrt{2}}`. Hence proved.


2. If `sin\theta = \frac{4}{5}`, then find the value of `sin2\theta, cos2\theta` and `tan2\theta`.
Solution:
Given, `sin\theta = \frac{4}{5}` ----(i)

So, `cos\theta = \sqrt{1-sin^2\theta}`

or, `cos\theta = \sqrt{1-(\frac{4}{5})^2}`

or, `cos\theta = \sqrt{1-\frac{16}{25}}`

or, `cos\theta = \sqrt{\frac{25-16}{25}}`

or, `cos\theta = \sqrt{\frac{9}{25}}`

or, `cos\theta = \frac{3}{5}`

`\therefore` `cos\theta = \frac{3}{5}` ------(ii)

Now, 
a.) `sin2\theta = 2sin\theta cos\theta`

Using (i) and (ii) on right hand side,

or, `sin2\theta = 2 \frac{4}{5} \frac{3}{5}`

`\therefore` `sin2\theta = \frac{24}{25}`

b.) `cos2\theta = cos^2\theta - sin^2\theta`

or, `cos2\theta = (\frac{3}{5}^2 - (\\frac{4}{5})^2 `

or, `cos2\theta = \frac{9}{25}-\frac{16}{25}`

or, `cos2\theta = \frac{9-16}{25}`

or, `cos2\theta = -\frac{7}{25}` 

`\therefore` `cos2\theta = -\frac{7}{25}`.

c.) `tan2\theta = \frac{sin2\theta}{cos2\theta}`

or, `tna2\theta = \frac{\frac{24}{25}}{-\frac{7}{25}}`

or, `tan2\theta = -\frac{24}{7}`

`\therefore` `tan2\theta = -\frac{24}{7}`.



3. Prove that `\frac{cos2A}{1+sin2A} = \frac{1-tanA}{1+tanA} = \frac{cosA-sinA}{cosA+sinA}`
Proof:
Taking,
`\frac{cos2A}{1+sin2A}`

= `\frac{cos^2A-sin^2A}{sin^2A+cos^2A+2sinA cosA}`

= `\frac{(cosA+sinA)(cosA-sinA)}{sin^2A+sinA cosA + cos^2A+sinA cosA}`

=`\frac{(cosA+sinA)(cosA-sinA)}{sinA(sinA+cosA)+cosA(cosA+sinA)}`

=`\frac{(cosA+sinA)(cosA-sinA)}{(sinA+cosA)(sinA+cosA)}`

=`\frac{cosA-sinA}{sinA+cosA}`

`\therefore``\frac{cos2A}{1+sin2A} = \frac{cosA-sinA}{cosA+sinA}`----(a)
This proves the second part.

Again, dividing numerator and denominator of equation (a) by cosA on right side, we get

`\frac{cosA}{1+sin2A} = \frac{\frac{cosA-sinA}{cosA}}{\frac{cosA+sinA}{cosA}}`

or, `\frac{cos2A}{1+sin2A} = \frac{1+\frac{sinA}{cosA}}{1+\frac{sinA}{cosA}}`

or, `\frac{cos2A}{1+sin2A} = \frac{1-tanA}{1+tanA}`

`\therefore`  `\frac{cos2A}{1+sin2A} = \frac{1-tanA}{1+tanA}` -----(b)
This proves the first part

Therefore, `\frac{cos2A}{1+sin2A} = \frac{1-tanA}{1+tanA} = \frac{cosA-sinA}{cosA+sinA}`. Hence proved. 

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