## Trigonometric Ratios of Multiple Angles

If A is an angle, then 2A, 3A, 4A, 5A, etc are called multiple angles of A. In this section we will learn about the trigonometric ratios of multiple angles 2A and 3A in terms of A.

### Geometrical Proof of 2A Formulae:

Let O be the centre of the circle ABC and AB be a diameter. Let CAB = A. Then COB = 2A. [Because, central angle is double than inscribed angle on same arc.]

Here, ACB = 90°. Let CM is perpendicular to AB. Then ACM = 90° - A and hence, BCM = A.

Now, sin2A = CM/OC  = 2CM/2OC  = 2CM/AB

= 2 . CM/AC . AC/AB

= 2sinA cosA

cos2A = OM/OC  = 2OM/2OC  = 2OM/AB

= [(AO + OM) - (AO - OM)]/AB  = [(AO + OM) - (BO - OM)]/AB

= (AM - BM)/AB  = AM/AB - BM/AB

= AM/AC . AC/AB - BM/BC . BC/AB

= cosA cosB – sinA sinB

= cos2A – sin2A

tan2A = CM/OM  = 2CM/2OM  = 2CM/[(AO + OM) – (AO – OM)]

= 2CM/[(AO + OM) – (BO – OM)]  = 2CM/(AM – BM)

= [2CM/AM]/[(AM – BM)/AM]  = [2CM/AM]/[AM/AM – BM/AM]

= [2CM/AM]/[1 – BM/CM . CM/AM]

= 2tanA/(1 – tanA tanA)

= 2tanA/(1 – tan2A)

### Worked Out Examples

Example 1: If sinA = 5/13, find the value of sin2A, cos2A and tan2A.

Solution: Here,

sinA = 5/13

cosA = √(1 – sin2A) = √(1 – 25/169) = √(144/169) = 12/13

Now, sin2A = 2sinA cosA

= 2 × 5/13 × 12/13

= 120/169

cos2A = cos2A – sin2A

= 144/169 – 25/169

= 119/169

tan2A = sin2A/cos2A

= [120/169]/[119/169]

= 120/119

Example 3: Express cos5A in terms of cosA.

Solution:

cos5A = cos(2A + 3A)

= cos2A cos3A – sin2A sin3A

= (2cos2A – 1)(4cos3A – 3cosA) – 2sinA cosA(3sinA – 4sin3A)

= 8cos5A – 6cos3A – 4cos3A + 3cosA – 2cosA sin2A(3 – 4sin2A)

= 8cos5A – 6cos3A – 4cos3A + 3cosA – 2cosA(1 – cos2A)(3 – 4 + 4cos2A)

= 8cos5A – 6cos3A – 4cos3A + 3cosA – (2cosA – 2cos3A)(4cos2A - 1)

= 8cos5A – 6cos3A – 4cos3A + 3cosA – 8cos3A + 2cosA + 8cos5A – 2cos3A

= 16cos5A – 20cos3A + 5cosA

You can comment your questions or problems regarding the trigonometric ratios of multiple angles.