Question

solve this question...............

Answer 1

HELLO DEAR,

let ( phi )= alpha

we know that:-
sin ²alpha+cos²alpha=1

and :-
2sin (alpha)×cos (alpha)

=sin2 (alpha)

$( { \sin \alpha - \cos \alpha }^{2} ) \\ = > { \sin \alpha }^{2} + { \cos \alpha }^{2} - 2 \sin \alpha \times \cos\alpha \\ = > ({ \sin \alpha }^{2} + { \cos \alpha }^{2}) - 2 \sin \alpha \times \cos\alpha \\ \\ \: \: USING \: \: ABOVE \: \: IDENTITIES \: \: WE \: GET \\ = > 1 - \sin2 \alpha$
I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Hey !!

◆ Basic Trigonometry ◆

L.H.S :-
$(sin x - \cos x)^{2} \\ = sin {}^{2} x + cos ^{2} x \: - 2sinx.cosx \\ as \: we \: know \: \\ sin ^{2} x \: + cos^{2} x \: = 1 \\ and \: sin2x = 2sinx.cosx \\ \\ hence \: above \: equation \: equates \: to \\ 1 - sin2x \: = \: rhs$
Hence , Proved.

Hope it helps.