Sin cos and tan of special angles :
There are two types of special triangles. These triangles are
A 30 - 60 - 90 triangle is a right triangle with a 30° degree angle and a 60° degree angle.
Let "x" be the length of shorter side. The hypotenuse side is twice as long as the shorter side. The longer side is √3 times as long as shorter side.
By using the above special triangle we can find the values of 30 and 60 degrees all six trigonometric ratios.
sin 30° = x /2x = 1/2 cos 30° = (x√3)/(2x) = √3/2 tan 30° = x/x√3 = 1/√3 sin 60° = (x√3)/(2x) = √3/2 cos 60° = x /2x = 1/2 tan 60° = x√3/x = √3 |
cosec 30° = 2x/x = 2 sec 30° = (2x)/(x√3) = 2/√3 cot 30° = x√3/x = √3/1 cosec 60° = (2x)/(x√3) = 2/√3 sec 60° = 2x/x = 2 cot 60° = x/x√3 = 1/√3 |
Let "x" be the length of equal side. The hypotenuse side is √2 times the length of either side.
sin 45° = x /x√2 = 1/√2 cos 45° = x /x√2 = 1/√2 tan 45° = x/x = 1 |
cosec 45° = x√2/x = √2 sec 45° = x√2/x = √2 cot 45° = x/x = 1 |
Trigonometric ratios of 0° and 90°
Consider the figure given below which shows a circle of radius 1 unit centered at the origin.
Let P be a point on the circle in the first quadrant with coordinates (x, y).
We drop a perpendicular PQ from P to the x-axis in order to form the right triangle OPQ.
Let <POQ = θ, then
sin θ = PQ / OP = y/1 = y (y coordinate of P)
cos θ = OQ / OP = x/1 = x (x coordinate of P)
tan θ = PQ / OQ = y/x
If OP coincides with OA, then angle θ = 0°.
Since, the coordinates of A are (1, 0), we have
If OP coincides with OB, then angle θ = 90°.
Since, the coordinates of B are (0, 1), we have
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