assume that cos (a-b)=cos a cos b + sin a sin b, if cos a=1/7 and cos b =13/14 where a and b are positive acute angle prove that a-b =60 degree
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Answer:A2A
cos x = 13/14
Sin x = √ (1- cos^2 x) = √27 /14 =3√3 /14
Cos y =1/7
Sin y = √48 /7 = 4√3 /7
Cos(x-y) = cos x. cos y + sin x .sin y
= 13/14 .1/7 + 3√3 /14 . 4√3 / 7
=13/98 + 36 /98
=49/98
=1/2
(x - y ) = cos inverse (1/2)
= +60° or - 60°
x and y are acute ange so both lies in 1st quadrant
in 1st quadrant cos value decreases with increase in angle
E.G
Cos 0° = 1
Cos 60° = 1/2
Cos 90° = 0
So Concept is
lesser angle in cos give large value
Cos x = 13 / 14
Cos y = 1/7 = 2/ 14
Cos x > cos y
so from above concept
x < y
( x - y) < 0
x - y will give negative value
So +60° is ruled out
Hence
(x - y ) = -60°
Step-by-step explanation:
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