Trigonometric/Circular/Angle/Goniometric Functions

are real functions which relate an angle of a right-angled triangle to ratios of two side lengths
an analog to hyperbolic functions

Trigonometric Functions - Geometric View

Geometric view of trigonometric functions

Function	Syntax	Description		Relationship	Relationships
Function	Syntax	Description		Relationship	Using Radians	Using Degrees
sine	𝑠𝑖𝑛(𝜃)	opposite / hypotenuse	SOH	if hypotenuse = 1, then it’s just opposite	$s in (𝜃) = cos (\frac{𝜋}{2} - 𝜃) = \frac{1}{csc ( 𝜃 )}$	$s in (x) = cos (90° - x) = \frac{1}{csc ( x )}$
cosine	𝑐𝑜𝑠(𝜃)	adjacent / hypotenuse	CAH	if hypotenuse = 1, then it’s just adjacent	$cos (𝜃) = s in (\frac{𝜋}{2} - 𝜃) = \frac{1}{sec ( 𝜃 )}$	$cos (x) = s in (90° - x) = \frac{1}{sec ( x )}$
tangent	𝑡𝑎𝑛(𝜃)	opposite / adjacent	TOA	if adjacent = 1, then it’s just opposite	$t an (𝜃) = \frac{s in ( 𝜃 )}{cos ( 𝜃 )} = co t (\frac{𝜋}{2} - 𝜃) = \frac{1}{co t ( 𝜃 )}$	$t an (x) = \frac{s in ( x )}{cos ( x )} = co t (90° - x) = \frac{1}{co t ( x )}$
cosecant	𝑐𝑠𝑐(𝜃)	hypotenuse / opposite	CO SEC HO	if opposite = 1, then it’s just hypotenuse	$csc (𝜃) = sec (\frac{𝜋}{2} - 𝜃) = \frac{1}{s in ( 𝜃 )}$	$csc (x) = sec (90° - x) = \frac{1}{s in ( x )}$
secant	𝑠𝑒𝑐(𝜃)	hypotenuse / adjacent	SEC HA	if adjacent = 1, then it’s just hypotenuse	$sec (𝜃) = csc (\frac{𝜋}{2} - 𝜃) = \frac{1}{cos ( 𝜃 )}$	$sec (x) = csc (90° - x) = \frac{1}{cos ( x )}$
cotangent	𝑐𝑜𝑡(𝜃)	adjacent / opposite	CO TAO	if opposite = 1, then it’s just adjacent	$co t (𝜃) = \frac{cos ( 𝜃 )}{s in ( 𝜃 )} = t an (\frac{𝜋}{2} - 𝜃) = \frac{1}{t an ( 𝜃 )}$	$co t (x) = \frac{cos ( x )}{s in ( x )} = t an (90° - x) = \frac{1}{t an ( x )}$
cosine squared	𝑐𝑜𝑠²(𝜃)	𝑐𝑜𝑠²(𝜃) is running 𝑐𝑜𝑠(𝜃) twice. For example: 𝑐𝑜𝑠(𝜃) = adjacent / hypotenuse if hypotenuse = 𝑐𝑜𝑠(𝜃), then: 𝑐𝑜𝑠(𝜃) = adjacent / 𝑐𝑜𝑠(𝜃) thus, 𝑐𝑜𝑠²(𝜃) = adjacent		𝑐𝑜𝑠²(𝜃) is the area of the square with side 𝑐𝑜𝑠(𝜃)	1 = 𝑠𝑖𝑛²(𝜃) + 𝑐𝑜𝑠²(𝜃)	trigonometry-geometric-view.drawio
sine squared	𝑠𝑖𝑛²(𝜃)	𝑠𝑖𝑛²(𝜃) is running 𝑠𝑖𝑛(𝜃) twice. For example: 𝑠𝑖𝑛(𝜃) = opposite / hypotenuse if hypotenuse = 𝑠𝑖𝑛(𝜃), then: 𝑠𝑖𝑛(𝜃) = opposite / 𝑠𝑖𝑛(𝜃) thus, 𝑠𝑖𝑛²(𝜃) = opposite		𝑠𝑖𝑛²(𝜃) is the area of the square with side 𝑠𝑖𝑛(𝜃)	1 = 𝑠𝑖𝑛²(𝜃) + 𝑐𝑜𝑠²(𝜃)	trigonometry-geometric-view.drawio