Maharashtra State BoardHSC Arts 12th Board Exam
Advertisement Remove all ads

In ∆ABC, prove that sin (A-B2)=(a-bc)cos(C2) - Mathematics and Statistics

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Sum

In ∆ABC, prove that `sin  (("A" - "B")/2) = (("a" - "b")/"c") cos ("C"/2)` 

Advertisement Remove all ads

Solution

In ∆ABC by sine rule, we have

`"a"/(sin "A") = "b"/(sin "B") = "c"/(sin "C")` = k

∴ a = k sin A, b = k sin B, c = k sin C

Consider R.H.S. = `(("a" - "b")/"c") cos ("C"/2)`

= `(("k"sin"A" - "k"sin"B")/("k"sin"C")) cos("C"/2)`

= `((sin"A" - sin"B")/sin"C") cos("C"/2)`

= `(2cos(("A" + "B")/2)sin(("A" - "B")/2))/(sin "C") cos("C"/2)`  .......(i)

But A + B + C = π

∴ A + B = π − C

∴ `("A" + "B")/2 = pi/2 - "C"/2`

∴ `cos(("A"  "B")/2) = cos(pi/2 - "C"/2)`

= `sin("C"/2)`    .......(ii)

Substituting (ii) in (i), we get

R.H.S. = `((2sin  "C"/2 cos  "C"/2)sin(("A" - "B")/2))/(sin "C")`

= `(sin"C" sin(("A" - "B")/2))/(sin "C")`

= `sin(("A" - "B")/2)`

= L.H.S.

∴ `sin  (("A" - "B")/2) = (("a" - "b")/"c") cos ("C"/2)` 

Concept: Solutions of Triangle
  Is there an error in this question or solution?
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×