Parametric formulas in trigonometry

In trigonometry, parametric formulas allow us to compute the sine and cosine of an angle α based on the tangent of half the angle, α/2.

Parametric formula for sine: $$ \sin \alpha = \frac{2 \tan \frac{\alpha}{2} }{1+ \tan^2 \frac{\alpha}{2} } $$
Parametric formula for cosine: $$ \cos \alpha = \frac{1 - \tan^2 \frac{\alpha}{2} }{1+ \tan^2 \frac{\alpha}{2} } $$

If we introduce the parameter

$$ t = \tan \frac{ \alpha }{ 2 } $$

the formulas can be rewritten in a simpler parametric form:

$$ \sin \alpha = \frac{2 \cdot t }{1+ t^2 } $$ $$ \cos \alpha = \frac{1 - t^2 }{1+ t^2 } $$

The meaning remains unchanged.

Why use parametric formulas?

If you already know the value of the tangent of α/2, these formulas make it easy to find the sine and cosine of the angle α.

Practical example

Let's take the tangent of 45° (π/4 radians):

$$ \tan 45° = 1 $$

Using α/2=45°, we can calculate the sine of α=90° with the parametric sine formula:

$$ \sin \alpha = \frac{2 \cdot \tan \frac{\alpha}{2} }{1+ \tan^2 \frac{\alpha}{2} } $$

$$ \sin 90° = \frac{2 \cdot \tan 45° }{1+ \tan^2 45° } $$

Since tan 45° = 1

$$ \sin 90° = \frac{2 \cdot 1 }{1+ 1 } $$

$$ \sin 90° = \frac{2}{2} $$

$$ \sin 90° = 1 $$

This gives us the sine of 90° starting from the tangent of 45°.

Example 2

Starting again with the tangent of 45° (π/4 radians):

$$ \tan 45° = 1 $$

We can now calculate the cosine of α=90° using the parametric cosine formula with α/2=45°:

$$ \cos \alpha = \frac{1 - \tan^2 \frac{\alpha}{2} }{1+ \tan^2 \frac{\alpha}{2} } $$

$$ \cos 90° = \frac{1 - \tan^2 45° }{1+ \tan^2 45° } $$

Since tan 45° = 1

$$ \cos 90° = \frac{1 - 1 }{1+ 1 } $$

$$ \cos 90° = \frac{0}{2} $$

$$ \cos 90° = 0 $$

We’ve now calculated the cosine of 90° starting from the tangent of 45°.

The proof

Proof of the parametric formula for sine

The sine of an angle a

$$ \sin a $$

can be rewritten in terms of a/2 using the sine doubling formula:

$$ \sin a = 2 \cdot \sin \frac{a}{2} \cos \frac{a}{2} $$

The denominator is 1, which is usually omitted, but here it helps clarify the next steps in the proof.

$$ \sin a = \frac{ 2 \cdot \sin \frac{a}{2} \cos \frac{a}{2} }{1} $$

According to the fundamental trigonometric identity, the sum of the squares of sine and cosine equals 1, i.e., sin2(θ) + cos2(θ) = 1.

We can use this identity with the angle a/2 to replace the denominator:

$$ \sin a = \frac{ 2 \cdot \sin \frac{a}{2} \cos \frac{a}{2} }{ \sin^2 \frac{a}{2} + \cos^2 \frac{a}{2} } $$

This expression is equivalent to the previous one.

Now, applying the invariant property of fractions, we divide both the numerator and denominator by cos2(a/2):

$$ \sin a = \frac{ \frac{ 2 \cdot \sin \frac{a}{2} \cos \frac{a}{2} }{ \cos^2 \frac{a}{2} } }{ \frac { \sin^2 \frac{a}{2} + \cos^2 \frac{a}{2} }{ \cos^2 \frac{a}{2} } } $$

Simplifying the cosine in the numerator gives us:

$$ \sin a = \frac{ \frac{ 2 \cdot \sin \frac{a}{2} }{ \cos \frac{a}{2} } }{ \frac { \sin^2 \frac{a}{2} }{ \cos^2 \frac{a}{2} } + 1 } $$

We further simplify the denominator with some algebraic steps:

$$ \sin a = \frac{ \frac{ 2 \cdot \sin \frac{a}{2} }{ \cos \frac{a}{2} } }{ \frac { \sin^2 \frac{a}{2} }{ \cos^2 \frac{a}{2} } + \frac { \cos^2 \frac{a}{2} }{ \cos^2 \frac{a}{2} } } $$

$$ \sin a = \frac{ \frac{ 2 \cdot \sin \frac{a}{2} }{ \cos \frac{a}{2} } }{ \frac { \sin^2 \frac{a}{2} }{ \cos^2 \frac{a}{2} } + 1 } $$

Since the sine-to-cosine ratio is the tangent, we arrive at:

$$ \sin a = \frac{ 2 \cdot \tan \frac{a}{2} }{ \tan^2 \frac{a}{2} + 1 } $$

Thus, we’ve derived the parametric sine formula.

Proof of the parametric formula for cosine

The cosine of an angle a

$$ \cos a $$

can also be rewritten in terms of a/2 using the cosine doubling formula:

$$ \cos a = \cos^2 \frac{a}{2} - \sin^2 \frac{a}{2} $$

Again, we explicitly include the denominator (1) for clarity:

$$ \cos a = \frac{ \cos^2 \frac{a}{2} - \sin^2 \frac{a}{2} }{1} $$

Using the fundamental trigonometric identity, where sin2 + cos 2 = 1, we rewrite the denominator as:

$$ \cos a = \frac{ \cos^2 \frac{a}{2} - \sin^2 \frac{a}{2} }{ \sin^2 \frac{a}{2} + \cos^2 \frac{a}{2} } $$

Next, we divide both the numerator and denominator by cos2(a/2):

$$ \cos a = \frac{ \frac{ \cos^2 \frac{a}{2} - \sin^2 \frac{a}{2} }{ \cos^2 \frac{a}{2} } }{ \frac{ \sin^2 \frac{a}{2} + \cos^2 \frac{a}{2} }{ \cos^2 \frac{a}{2} } } $$

Simplifying both the numerator and denominator yields:

$$ \cos a = \frac{ \frac{ \cos^2 \frac{a}{2} }{ \cos^2 \frac{a}{2} } - \frac{ \sin^2 \frac{a}{2} }{ \cos^2 \frac{a}{2} } }{ \frac{ \sin^2 \frac{a}{2} }{ \cos^2 \frac{a}{2} } + \frac{ \cos^2 \frac{a}{2} }{ \cos^2 \frac{a}{2} } } $$

$$ \cos a = \frac{ 1 - \frac{ \sin^2 \frac{a}{2} }{ \cos^2 \frac{a}{2} } }{ \frac{ \sin^2 \frac{a}{2} }{ \cos^2 \frac{a}{2} } + 1 } $$

Since the sine-to-cosine ratio is the tangent, we derive:

$$ \cos a = \frac{ 1 - \tan^2 \frac{a}{2} }{ \tan^2 \frac{a}{2} + 1 } $$

Thus, we’ve derived the parametric cosine formula.

And so on.

 

 
 

Please feel free to point out any errors or typos, or share suggestions to improve these notes. English isn't my first language, so if you notice any mistakes, let me know, and I'll be sure to fix them.

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