basic trigonometric functions trigonometry

The circle whose radius is 1 unit whose centre is the origin of a rectangular co-ordinate system is called the unit circle. 1. cos q = x. 2. sin q = y. 3. tan q = y/x. 4. sec q = 1/x. 5. cosec q = 1/y. 6. cot q = x/y. The six functions of q defined by the above equation are..

The Values of Trigonometric Functions of 90 o and 0 o are: cos90 o = 0, sin90 o = 1, tan90 o = Not defined, sec90 o = Not defined, cot90 o = 0, cosec90 o = 1, cos0 o = 1, sin0 o = 0, here cosec0 o and cot0 o are not defined, sec0 o = 1, tan0 o = ..

Values of Trigonometric Functions of 30 o , 45 o , 60 o and 90 o - Let OA be the revolving ray starting from A. Let OA take the new position OP so that Draw PM perpendicular to OX and produce it to Q. Draw O..

For x [-1, 1], if q is an angle whose sine is x, then we say that sine inverse x is q and write sin -1 x = q..

In the principle value branches, the following formulae holds: - sin -1 (sin x) = x - cos -1 (cos x) = x - tan -1 (tan x) = x - cos -1 (cot x) = x - sec -1 (secx) = x - cosec -1 (cosecx) = x - sin -1 (-x) = -sin -1 x - cos -1 (-x) = p - cos -1 x - t..

A trigonometric equation is an equation involving the trigonometric functions of unknown angles. Example; sin x = 1/..

In the above topics many identities have been discussed. They are true for all values of the angles for which trigonometric functions are defined. In this section we prove identities, where a certain relationship exists among the angles considered. Many interesting a..

In the previous sections many identities have been discussed. They are true for all values of the angles for which trigonometric functions are defined. In this section we prove identities, where a certain relationship exists among the angles considered. Many interesting and impo..

Trigonometric Ratios of Multiple and Sub-multiple Angles - The following results have already been derived under Circular function..

Let OA be the revolving ray starting from A. Let OA take the new position OP so that Draw PM perpendicular to OX and produce it to Q. Draw OQ. Let 2l be the length of each side of the triangle. Then OP = OQ = QP = 2l. PM = MQ= l. From OMP ..