Relations between trigonometric functions

Sine, cosine and tangent are in different relationships. A distinction is made between the complement relationships and the supplement relationships.

Complement relationships

$\sin (90^{\circ} - α) = \cos (α)$
$\cos (90^{\circ} - α) = \sin (α)$
$\tan (90^{\circ} - α) = \frac{1}{\tan (α)}$

Since in a triangle the sum of the interior angles is always $180 °$ , the following applies in a right triangle $β = 90 ° - α$ .

Using the sine, cosine and tangent formulas, you can see:

$\sin (90 ° - α) = \frac{opposite cathetus}{hypotenuse} = \frac{b}{c}$

$\cos (α) = \frac{adjacent cathetus}{hypotenuse} = \frac{b}{c} .$

Hence, $\sin (90 ° - α) = \cos (α)$ .

The other equations can be explained in the same way.

Example

Consider the given triangle. Calculate $\cos (α)$ in the same way as above.

With the complement relation you can equate $\cos (α)$ and $\sin (90 ° - α)$ .

$\cos (α) = \sin (90 ° - α)$

Because of the sum of the interior angles, the following equation applies.

$\sin (90 ° - α) = \sin (β)$

Insert the value of $β$ , calculate the result and round it to $2$ decimal places.

$\sin (β) = \sin (40 °) \approx 0.59.$

Therefore, $\cos (α) \approx 0.59.$

Supplement relationships

Sine	Kosine	Tangent
$\sin (180 ° + α) = - \sin (α)$	$\cos (180^{\circ} + α) = - \cos (α)$	$\tan (180^{\circ} + α) = + \tan (α)$
$\sin (180 ° - α) = + \sin (α)$	$\cos (180 ° - α) = - \cos (α)$	$\tan (180^{\circ} - α) = - \tan (α)$
$\sin (360^{\circ} - α) = - \sin (α)$	$\cos (360^{\circ} - α) = + \cos (α)$	$\tan (360^{\circ} - α) = - \tan (α)$

Visualization

$\sin (180 ° + α) = - \sin (α)$ and $\cos (180 ° + α) = - \cos (α)$ can be tested, here

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