Euler formula in Trigonometry

With these identities, we can connect trigonometry and Euler''s formula

(1) \(cos(\alpha+\beta) = cos(\alpha)*cos(\beta) - sin(\alpha)*sin(\beta)\)

(2) \(sin(\alpha+\beta) = sin(\alpha)*cos(\beta) + cos(\alpha)*sin(\beta)\)

(3) \(e^{i\Theta} = cos\Theta + i*sin\Theta\)

We can imagine the equation (3) as a circle in a complex plane with \(cos\Theta\) as \(\hat{x}\) and \(sin\Theta\) as \(\hat{y}\). It''s much clearer than looking at the formula :)

In complex number, we can define a complex conjugate by changing the sign of every imaginary part that can be notated into something like this.

(4) \(z^* = (a + ib)^* \equiv (a - ib)\) ¹

If we add and subtract equation (3) with its complex conjugate, we can get another term to define \(cos\Theta\) and \(sin\Theta\) ².

(3) \(e^{i\Theta} = cos\Theta + i*sin\Theta\)

(5) \(e^{-i\Theta} = cos\Theta - i*sin\Theta\)

Addition $$e^{i\Theta} + e^{-i\Theta} = 2 * cos\Theta$$

(6) \(cos\Theta = (e^{i\Theta} + e^{-i\Theta})/2\)

Subtraction $$e^{i\Theta} - e^{-i\Theta} = 2i * sin\Theta$$

(7) \(sin\Theta = (e^{i\Theta} - e^{-i\Theta})/2i\)

Then we can continue to relate equations (6) and (7) to hyperbolic trigonometry by multiply \(\Theta\) with i.

(8) \(cos(i\Theta) = (e^{-\Theta} + e^{\Theta})/2 = cosh\Theta\)

(9) \(sin(i\Theta) = (e^{-\Theta} - e^{\Theta})/2i = sinh\Theta\)

To circle back to equations (1) and (2), we can use Euler''s function and its derivative to proof them.

(1) \(cos(\alpha+\beta) = cos(\alpha)*cos(\beta) - sin(\alpha)*sin(\beta)\)

(6) \(cos\Theta = (e^{i\Theta} + e^{-i\Theta})/2\)

$$cos(\alpha + \beta) = (e^{i(\alpha + \beta)} + e^{-i(\alpha + \beta)})/2$$

$$cos(\alpha + \beta) = (e^{i\alpha}*e^{i\beta} + e^{-i\alpha} * e^{-i\beta})/2$$

substitute \(e^{ix}\) with the equation (3), we get

$$cos(\alpha + \beta) = \frac{[(cos(\alpha) + i * sin(\alpha)) * (cos(\beta) + i * sin(\beta)) + ((cos(-\alpha) + i * sin(-\alpha)) * (cos(-\beta) + i * sin(-\beta)))]}{2}$$

Remember the properties of cos and sin functions that:

• Even function: \(cos(-x) = cos(x)\)
• Odd function: \(sin(-x) = -sin(x)\)

Hence,

$$cos(\alpha + \beta) = \frac{[(cos(\alpha) + i * sin(\alpha)) * ((cos(\beta) + i * sin(\beta)) + ((cos(\alpha) - i * sin(\alpha)) * (cos(\beta) - i * sin(\beta)))]}{2}$$

$$ cos(\alpha + \beta) = \frac{[cos(\alpha) * cos(\beta) + cos(\alpha) * i * sin(\beta) + cos(\beta) * i * sin(\alpha) - sin(\alpha) * sin(\beta) + cos(\alpha) * cos(\beta) - cos(\alpha) * i * sin(\beta) - cos(\beta) * i * sin(\alpha) - sin(\alpha) * sin(\beta)]}{2} $$

$$ cos(\alpha + \beta) = \frac{[cos(\alpha) * cos(\beta) - sin(\alpha) * sin(\beta) + cos(\alpha) * cos(\beta) - sin(\alpha) * sin(\beta)]}{2}$$

(1) \(cos(\alpha + \beta) = cos(\alpha)*cos(\beta) - sin(\alpha)*sin(\beta)\)

Similar process for equation (2)

(2) \(sin(\alpha+\beta) = sin(\alpha)*cos(\beta) + cos(\alpha)*sin(\beta)\) (7) \(sin\Theta = (e^{i\Theta} - e^{-i\Theta})/2i\)

$$ sin(\alpha + \beta) = (e^{i(\alpha + \beta)} - e^{-i(\alpha + \beta)})/2i $$

$$ sin(\alpha + \beta) = (e^{i\alpha}*e^{i\beta} - e^{-i\alpha}*e^{-i\beta})/2i $$

$$ sin(\alpha + \beta) = \frac{[(cos(\alpha) + i * sin(\alpha)) * ((cos(\beta) + i * sin(\beta)) - ((cos(\alpha) - i * sin(\alpha)) * ((cos(\beta) - i * sin(\beta))]}{2i}$$

$$ sin(\alpha + \beta) = \frac{[cos(\alpha) * cos(\beta) + cos(\alpha) * i * sin(\beta) + cos(\beta) * i * sin(\alpha) - sin(\alpha) * sin(\beta) - cos(\alpha) * cos(\beta) + cos(\alpha) * i * sin(\beta) + cos(\beta) * i * sin(\alpha) + sin(\alpha) * sin(\beta)]}{2i} $$

$$ sin(\alpha + \beta) = \frac{[cos(\alpha) * i * sin(\beta) + cos(\beta) * i * sin(\alpha) + cos(\alpha) * i * sin(\beta) + cos(\beta) * i * sin(\alpha)]}{2i} $$

(2) \(sin(\alpha+\beta) = sin(\alpha)*cos(\beta) + cos(\alpha)*sin(\beta)\)

References

Tags: math