## Integral Calculus, Double Angle Formula

### Double Angle Formula

Set u = v in the angle addition formulas to get
the double angle formulas.
sin(2u) = 2sin(u)cos(u)

cos(2u) = cos2(u) - sin2(u)

The latter is sometimes written:

cos(2u) = 2×cos2(u) - 1

cos(2u) = 1 - 2×sin2(u)

There is of course a triple angle formula.
Expand sin(2θ+θ) using the angle addition formula,
then expand cos(2θ) and sin(2θ) using the double angle formulas.
Do this again to get the quadruple angle formula, the quintuple angle formula, and so on.
sin(3θ) = 3sin(θ)cos2(θ) - sin3(θ)

cos(3θ) = cos3(θ) - 3cos(θ)sin2(θ)

sin(4θ) = 4sin(θ)cos3(θ) - 4sin3(θ)cos(θ)

cos(4θ) = cos4(θ)
- 6cos2(θ)sin2(θ)
+ sin4(θ)

If you are familiar with
the binomial theorem,
you will recognize a pattern.
Let c and s represent cosine and sine respectively,
and expand (c+s)n.
Take every other term starting with cn;
this is the cosine of nθ.
Well almost; you have to negate every other term in this series.
To find the sine of nθ,
Take every other term in the expansion,
starting with ncn-1s,
and again, negate every other term in this series.

Look at our last example, sine and cosine of 4θ.
Expand (c+s)4 and put the minus signs where they belong.
The cosine is every other term starting with c4,
and the sine is every other term starting with 4c3s.

c4 + 4c3s - 6c2s2 - 4cs3 + s4

We can prove the formula, in general, by induction on n.
Apply the angle addition formula to θ + nθ.
Imagine cosine and sine of nθ written together, as above.
It looks like the nth row of Pascal's triangle,
except some of the entries have been negated.
Now move on to the next level.

The sine is s times the previous formula for cosine
plus c times the previous formula for sine.
This adds adjacent terms from the previous row,
giving the next row of Pascal's triangle.
Similarly, the cosine is
c times the previous formula for cosine
minus s times the previous formula for sine.
This fills in the rest of the row, and all the minus signs are where they belong.
By induction, the formula holds for all n.

There is another proof that is more intuitive,
if you are familiar with complex exponentiation.

Eiθ has, as its real and imaginary components,
cos(θ) and sin(θ) respectively.
Now Eniθ is the same as Eiθ
raised to the nth power.
Expand (c+si)n and see what you get.

The real component is the cosine of nθ,
and that happens to be every other term,
starting with cn, with alternate terms negated.
Similarly, sin(nθ) is the imaginary component,
found by adding up the imaginary terms.
We need every other term, starting with ncns,
and alternate terms are negated.
This proves the formula for all values of n in one go.