*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*is

^{i·x}*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

== 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)= 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)= 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)= 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*[

**/**d**x***]*

**e**^{i·x}*[*

**= i·e**

= i·^{i·x}== i·

*]*

**cos(x) + i·sin(x)**

**=**

= −sin(x) + i·cos(x)= −sin(x) + i·cos(x)

On the right side the result of differentiation is

*d*[

**/**d**x***]*

**cos(x)***[*

**+ i·**d**/**d**x***]*

**sin(x)**The results of differentiation of left and right sides of the Euler's formula must be equal to each other:

*d*[

**/**d**x***]*

**cos(x)***[*

**+ i·**d**/**d**x***]*

**sin(x)**

**=**

= −sin(x) + i·cos(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*[

**/**d**x***]*

**cos(x)**

**= −sin(x)***d*[

**/**d**x***]*

**sin(x)**

**= cos(x)**
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