In the previous section, we learnt the following formulae:
- Sin(A + B) = SinA*CosB + CosA*SinB
- Sin(A-B) = SinA*CosB – CosA*SinB
- Cos(A+B) = CosACosB – SinA*SinB
- Cos(A-B) = CosACosB + SinA*SinB
In this section, we will learn the formula for Sin2A, Cos2A, Sin(A/2), Cos(A/2) also called multiple and submultiple angles.
Multiple Angles
Deriving Sin2A
we know that
Sin(A+B) = SinA*CosB + CosA*SinB
If in this formula, we have A = B, Sin(A+B) = sin(A + A) = Sin2A
Sin(A+A) = SinA*CosA + CosA*SinA
or,
Sin2A = 2sinA*CosA
Deriving Cos2A
we know that
Cos(A+B) = CosA*CosB – SinA*SinB
If in this formula, we have A = B, Cos(A+B) = Cos(A + A) = Cos2A
Cos(A+A) = CosA*CosA + SinA*SinA
or,
Cos2A = Cos2A – Sin2A
We also know that
sin2A + cos2A = 1
or Cos2A = 1- Sin2A
so, we also have
Cos2A = 2cos2A – 1 = 1 – 2sin2A
Now, let us check formula for 3A
Sin3A can be written as sin(2A + A)
Sin(2A + A) = sin2A*CosA + Cos2A*SinA
using the formula for Cos2A and Sin2A, we get
Sin(2A + A) = 2SinACosA*CosA + (Cos2A – Sin2A)SinA
or, Sin3A = 2SinA*Cos2A + Cos2A*SinA – Sin3A
or, Sin3A = 3SinA*Cos2A – Sin3A
or, Sin3A = 3SinA(1-sin2A) – Sin3A
or, Sin3A = 3SinA – 3sin3A) – Sin3A
or,
Sin3A = 3SinA – 4sin3A
Use the same approach to find
Cos3A = 4cos3A – 3cosA
Sine and Cosine of Submultiple angles
We know:
cos2A = 2cos2A -1 or 1-2sin2A
If instead of A, we take x/2, 2A will become x and we will have
Cosx = cos(2(x/2)) = 2cos2(x/2) – 1
or, 2cos2(x/2) = Cosx + 1
or, cos2(x/2) = Cosx + 1⁄2
or, cos(x⁄2) = √(Cosx+1)⁄2
Replacing x by A, we have
Cos(A⁄2) = √(CosA+1)⁄√2
We also have
CosA = 1-2Sin2A⁄2
2Sin2A⁄2 = 1-CosA
Sin2(A⁄2) = (1 – Cosx)⁄2
Sin(A⁄2) = √ (1 – Cosx)⁄√2
So, for Submultiple, we have the formulae as:
Cos(A⁄2) = √(CosA+1)⁄√2
Sin(A⁄2) = √ (1 – Cosx)⁄√2