Trigonometric ratios of some specific angles :
For certain specific angles such as 30°, 45° an d 60°, which are frequently seen in applications, we can use geometry to determine the trigonometric ratios.
Let ABC be an equilateral triangle whose sides have length a (see the figure given below). Draw AD perpendicular to BC, then D bisects the side BC.
So, BD = DC = a/2 and < BAD = <DAC = 30°.
Now, in right triangle ADB, <BAD = 30° and BD = a/2.
In right triangle ADB, by Pythagorean theorem,
AB² = AD² + BD²
a² = AD² + (a/2)²
AD² = a² - (a²/4)
AD² = 3a²/4
AD = √(3a²/4)
AD = √3 x (a/2)
Hence, we can find the trigonometric ratios of angle 30° from the right triangle ADB.
In right triangle ADB, <ABD = 60°. So, we can determine the trigonometric ratios of angle 60°.
If an acute angle of a right triangle is 45°, then the other acute angle is also 45°.
Thus the triangle is isosceles. Let us consider the triangle ABC with <B = 90°, <A = <C = 45°
Then AB = BC. Let AB = BC = a.
By Pythagorean theorem,
AC² = AB² + BC²
AC² = a² + a²
AC² = 2a²
AC = √2 x a
Hence, we can find the trigonometric ratios of angle 45° from the right triangle ABC.
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
The six trigonometric ratios of angles 0°, 30°, 45°, 60° and 90° are provided in the following table.
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