Hi friends in my first POST i have introduced trigonometry.  Now in this POST i will explain the formulas and some problems for better understanding.

SIN2A=2SINA.COS A

DERIVATION;

SIN (A+A) =SIN A COS A+COS A SIN A    (SINCE SIN(A+B)=SIN A COS B+COS A SIN B)

                =2SINA COS A…

COS 2A=COS²A-SIN²A=2 COS²A-1

DERIVATION;

COS 2A=COS²A-SIN²A

             =COS²A-(1-COS²A)

             =COS²A-1+COS²A 

             =2COS²A-1…

SIN3A=3SINA-4SIN³A

DERIVATION

SIN (2A+A)

=SIN2ACOSA+COS2A+SIN A

=2SIN ACOS²A+ (1-2SIN²A) SIN A

=2SIN A (1-SIN²A) + (1-2SIN²A) SIN A

=2SINA-2SIN³A+SINA-2SIN³A

=3SINA-4SIN³A…

SOME TRIGONOMETRIC RATIOS.

	00	300	450	600	900
SINE	0	½	1/√2	√3/2	1
COSINE	1	√3/2	1/√2	½	0
TANGENT	0	1/√3	1	√3	ND
COTANGENT	ND	√3	1	1/√3	0
CO SECANT	ND	2	√2	2/√3	1
SECANT	1	2/√3	√2	2	ND

FOR BETTER UNDERSTANDING OF USING FORMULAS I CAN SOLVE SOME PROBLEMS FOR YOU.

1)       2 SIN15.COS15

SOLUTION;

2SINACOSA=SIN2A

=2SIN (15) =SIN30=½

2)       COS²15-SIN²15

SOLUTION;

COS2A=COS²A-SIN²A

=COS2(15)=COS 30=√3/2

1)  W.OUGHTERED INTRODUCED NOTATIONS SIN FOR SINE AND COS FOR COSINE

2)JOHN BERNOULI GAVE TRIGONOMETRIC FORMULAE SIN2A=2SIN A.COS A.