हिंदी

In ΔABC, prove that sin(B−C2)=(b−ca)cos A2.

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प्रश्न

In ΔABC, prove that `sin(("B" − "C")/2) = (("b" − "c")/"a")cos  "A"/(2)`.

योग
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उत्तर

By the sine rule,

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

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

RHS = `(("b" − "c")/"a")cos  "A"/(2)`

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

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

RHS = `(2cos(("B" +  "C")/2).sin(("B" − "C")/2))/(2sin  "A"/(2) . cos  "A"/(2)).cos  "A"/(2)`

RHS = `(cos(("B" + "C")/(2)).sin(("B" − "C")/(2)))/(sin  "A"/(2)`

RHS = `(cos(π/2 − "A"/2).sin(("B" − "C")/(2)))/(sin  "A"/(2))`  ...[ ∵ A + B + C = π ]

RHS = `(sin  "A"/(2). sin(("B" − "C")/(2)))/(sin  "A"/(2)`

RHS = `sin(("B" − "C")/(2))`

LHS = `sin(("B" − "C")/(2))`

RHS = LHS

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

Hence proved.

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अध्याय 3: Trigonometric Functions - Exercise 3.2 [पृष्ठ ८८]

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बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 3 Trigonometric Functions
Exercise 3.2 | Q 4 | पृष्ठ ८८

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