Question

prove geometrically cos(x+y)=cosx×cosy-sinx×siny and hence show that cos 2x=cos²x-sin²x​

Answer 1

Proof of cos(x+y) = cosx cosy - sinx siny:

Consider unit circle with centre O at the origin let A be the point (1 , 0). Let P , Q and R be the points on the circle such that arc AP = x , arc PQ = y and arc AR = - y.

Then Arc AQ = arc AP + arc PQ = x + y.

Therefore, the co-ordinates of point P , Q and R are

$\sf (cos \: x ,sin \: x),(cos(x + y) ,sin(x + y)) \\ \bf and \\ \sf (cos( - y),sin ( - y)) \: respectively$

We have ,

arc PQ = arc RA

⟼ arc PQ + arc AP = arc RA + arc AP

⟼ arc AQ = arc RP

⟼ Length of chord AQ = Length of chord RP

(∵ In a circle , equal arcs cut off equal chords)

⟼ AQ = RP = $\sf AQ^2=RP^2$

$⟼ \sf( \cos(x + y) - 1) {}^{2} + (\sin(x + y - 0 {)}^{2} ) \\ \sf = ( \cos \:x - \cos( - y)) {}^{2} + (\sin \: x - \sin( - y) ) {}^{2}$

$⟼ \sf { \cos }^{2} (x + y) + 1 - 2 \cos(x + y) + \sin {}^{2} (x + y) \\ \sf = {( \cos \: x - \cos \: y)}^{2} + ( { \sin \: x + \sin \: y)}^{2}$

$\bf sin ( - y) = - sin \: y \: \: and \: \: cos( - y) = cos \: y$

$⟼ \sf \{\cos {}^{2} (x + y) + \sin {}^{2} (x + y) \} \\ \sf+ 1 - 2 \cos(x + y) \\ \sf = { \cos}^{2} x + \cos {}^{2} y - 2 \cos \: x \: \cos \: y \\ \sf + \: { \sin}^{2} x + { \sin}^{2} y + 2 \sin \: x \: \sin \: y$

$⟼\sf1 + 1 - 2 \cos(x + y) \\ \sf = {( \cos }^{2} x + { \sin}^{2} x) + {( \cos }^{2} y + { \sin}^{2} y) \\ \sf - 2 \cos x \cos y + 2 \sin x \sin y$

$\sf⟼2 - 2 \cos(x + y) = \\ \sf1 + 1 - 2 \cos x \cos y + 2 \sin x \sin y \\ \\ \sf⟼ - 2 \cos(x + y) = - 2 \cos x \cos y + 2 \sin x \sin y \\ \\ \purple{   \underline {\boxed{{\bf⟼ \cos(x + y) = \cos x \cos y - \sin x \sin y } }}}$

Hence Proved !

___________________________

Proof of cos 2x = cos²x - sin²x :

$\sf \cos(2x) = \cos(x + x) \\ \\ \bf \{using \: above \: formula \} \\ \\ = \sf \cos x \cos x - \sin x \sin x \\ \\ \large\purple{ \implies  \underline {\boxed{{\bf cos 2x=cos {}^{2} x-sin {}^{2} x} }}}$

Hence Proved !