Question

prove geometrically that cos(X+Y)=cosX . cosY-sinX . sinY

Answer 1

ANSWER WITHOUT WORDS IN PIC :
STILL I WILL WRITE :

STEPS:
1)Let ABCD be a rectangle with given constructions as shown in figure.
DE = 1 units (say)

2) After some manipulations, we get
We use,
$\sin(a) = \frac{Opposite \: side}{Hypotenuse} \\ \cos(a) = \frac{Adjacent \: side}{Hypotenuse }$

$EF = \sin(a) , FD = \cos(a) \\ AF = \cos(a) \sin(b), AD = \cos(a) \cos(b) \\ BE = \sin(a) \sin(b) , BF = \sin(a) \cos(b) \\ CE = \cos(a+b) , CD = \sin(a+b)$

3) Since, Opposite sides of rectangle are equal.
$BC = AD \\ \cos(a) \cos(b) = \sin(a) \sin(b) + \cos(a+b) \\ => \cos(a+b) = \cos(a) \cos(b) - \sin(a) \sin(b)$

Hence ,got our Desired Proof :