Trigonometry+ 01: UNIZOR.COM - Math+ & Problems - Trigonometry

Notes to a video lecture on http://www.unizor.com

Trigonometry+ 01


Problem A

Using the Euler Formula
e^i·x = cos(x)+i·sin(x)
prove the formulas for sine and cosine of a sum of two angles.

Solution A

The Euler formula states that the real part of e^i·x is cos(x) and its imaginary part s i·sin(x)

From i² = −1,
e^i·(α+β) = cos(α+β)+i·sin(α+β),
and a^(p+q) = a^p·a^q
follows:
cos(α+β)+i·sin(α+β) =
= e^i·(α+β) = e^i·α·e^i·β =
= [cos(α)+i·sin(α)] ·
· [cos(β)+i·sin(β)] =
= cos(α)·cos(β)−sin(α)·sin(β) +
+ i·cos(α)·sin(β)+i·sin(α)·cos(β)

The real part of the last expression is
cos(α)·cos(β) − sin(α)·sin(β)
Its imaginary part is
i·[cos(α)·sin(β) + sin(α)·cos(β)]

Therefore,
cos(α+β)
(the real part of e^i·(α+β))
equals to
cos(α)·cos(β) − sin(α)·sin(β),
hence
cos(α+β) =
= cos(α)·cos(β) − sin(α)·sin(β)

Analogously,
i·sin(α+β)
(the imaginary part of e^i·(α+β))
equals to
i·[cos(α)·sin(β) + sin(α)·cos(β)],
hence
sin(α+β) =
= cos(α)·sin(β) + sin(α)·cos(β)


Problem B

Using the Euler Formula
e^i·x = cos(x)+i·sin(x)
prove the formulas for derivatives of sine and cosine functions.

Solution B

Let's differentiate the Euler's formula
e^i·x = cos(x)+i·sin(x)

On the left side the result is
d/dx[e^i·x] = i·e^i·x =
= i·[cos(x) + i·sin(x)] =
= −sin(x) + i·cos(x)

On the right side the result of differentiation is
d/dx[cos(x)] + i·d/dx[sin(x)]

The results of differentiation of left and right sides of the Euler's formula must be equal to each other:
d/dx[cos(x)] + i·d/dx[sin(x)] =
= −sin(x) + i·cos(x)

When two complex numbers are equal to each other, their real and imaginary parts must be equal.
Therefore,
d/dx[cos(x)] = −sin(x)
d/dx[sin(x)] = cos(x)