Question

If(A-B)=60° , verify that
(i) Cos(A-B)= CosACosB+SinASinB
(ii) Sin(A-B)= SinACosB - CosASinB

Answer 1

Let A be 90 and B be 30
(A-B)=90-30=60°..

(i)
Cos(A-B)=CosACosB+SinASinB
=Cos60°=1/2
Cos90.Cos30+Sin90.Sin30
=0×√3/2+1×1/2
=0+1/2
=1/2
Hence..LHS=RHS

(ii)
Sin(A-B)=SinA.CosB-CosA.SinB
Sin60°=√3/2
=Sin90.Cos30-Cos90.Sin30°
=1×√3/2-0×1/2
=√3/2-0
=√3/2
Hence...LHS=RHS
Therefore,proved!
hope it helps...
cheers! (:

Answer 2

HELLO DEAR,

HELLO DEAR,
it is possible when we take

the A=90 and B=30

given that:-

A-B=60

90-30=60

60=60

now

(1) cos (A-B)=COS60°=1/2------------(1)

=> CosACosB+SinASinB

=>cos90cos30+sin90sin30

=> 0 × √3/2 + 1 ×1/2

=0+1/2=2----------(2)

from(1)and(2)

Cos(A-B)= CosACosB+SinASinB

(2) sin(A-B) = sin60° = √3/2----------(3)

=> SinACosB - CosASinB

=>sin90° cos30° - cos90° sin30°

=>1×√3/2-0×1/2

=>√3/2-0

=>√3/2--------(4)

from--(3)and---(4)

we get

Sin(A-B)= SinACosB - CosASinB


I HOPE ITS HELP YOU DEAR,
THANKS