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AIEEE 2005

JEE Main 2005 Mathematics question on Trigonometry

In a △ABC\triangle ABC, among the following which one is true?
  1. A.
    (b+c)cos⁡A2=asin⁡(B+C2)(b+c)\cos\dfrac{A}{2} = a\sin\left(\dfrac{B+C}{2}\right)
  2. B.
    (b+c)cos⁡(B+C2)=asin⁡A2(b+c)\cos\left(\dfrac{B+C}{2}\right) = a\sin\dfrac{A}{2}
  3. C.
    (b−c)cos⁡(B−C2)=acos⁡(A2)(b-c)\cos\left(\dfrac{B-C}{2}\right) = a\cos\left(\dfrac{A}{2}\right)
  4. D.
    (b−c)cos⁡A2=asin⁡(B−C2)(b-c)\cos\dfrac{A}{2} = a\sin\left(\dfrac{B-C}{2}\right)
Show answer and solution

Correct option: D

Solution

Using the projection formula and sine rule, by the law of sines b = 2R sinB, c = 2R sinC, a = 2R sinA.

b - c = 2R(sinB - sinC) = 2R · 2cos((B+C)/2)sin((B-C)/2)

Since A + B + C = π, (B+C)/2 = (π - A)/2, so cos((B+C)/2) = sin(A/2).

Thus b - c = 4R · sin(A/2) · sin((B-C)/2).

Now (b - c)cos(A/2) = 4R · sin(A/2) · cos(A/2) · sin((B-C)/2) = 2R · sinA · sin((B-C)/2) = a · sin((B-C)/2).

This confirms option 4: (b - c)cos(A/2) = a·sin((B-C)/2).

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