Deriving Values for Trignometric Ratios
Trigonometry is the solving of angler question without the use of neither calculator nor tables, but giving solutions in surd form.
Let us take an isosceles right angle triangle with base = height. Here the angle made by the hypotenuse with the base is 45 degrees. By the pythogoreas theorm the square of the hypotenuse is equal to the sum of the square of the base and the height. The square of the hypotenuse is thus sqrt(2) * base or sqrt(2) * height.
Sin(45) is hence height/length of hypotenuse = height / sqrt(2) * height = 1/ sqrt(2)
Cos (45) is defined as length of base / length of height and hence it is base / sqrt(2) * base which is equal to 1/sqrt(2).
Tan(45) is hence Sin(45)/Cos(45) which is equal to 1.
Let us derive the expression for Sin(60), Cosine(60) and Tan(60). Let us consider an equilateral triangle. In the equilateral triangle the three angles are equal to 60 degrees. Let us draw a perpendicular between one of the vertex to the opposite side. This will bisect the opposite side by exactly half as the perpendicular line will also be a perpendicular bisector. Let us consider any one of the two triangles created with the perpendicular bisector as the height. So the length of the perpendicular bisector is nothing but sqrt( l ** l - l * * l /4) = l * sqrt(3)/2. By definition Sin(60) is hence height of the triangle / hypotenuse, so Sin(60) can be calculated as l * sqrt(3/2) /l = sqrt(3)/2. Hence Cos(60) can be calculated as sqrt(1 - Sin(60) * Sin(60)) = sqrt(1 - 3/4) = 1/2.
In the same triangle the opposite angle is equal to 30 degrees. So Sin(30) = l/2 / l = 1/2 or 0.5. Using this Cos(30) can be calculated as sqrt(1 - 1/4) = sqrt(3)/2.
Let us go one step further and derive values for Cos(15). Cosine(A + B) is defined as CosineACosineB - SinASinB so when A = B then Cos(A + B) = Cos2A or in other words is equal to Cos (A) * Cos(A) - Sin(A) * Sin(A). Cos2A is equal to sqrt(3)/2 is equal to CosA * Cos A - Sin A * Sin A. Sin A * Sin A can be written as 1 - Cosine A * Cos A. So the expression becomes 2 Cosine A * Cosine A - 1 = sqrt(3)/2. So 2 Cos A * Cos A = (2 + sqrt(3))/2. Cos A * Cos A = (2 + sqrt(3))/2. So Cos 15 = Sqrt(2 + Sqrt(3))/2). Using this values for Sin 15, Sin 75, Cos 75, Sin 7.5. Sin 3,75, Cos 3.75 can be determined.
No comments
Post a Comment