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Exercise 3.2 · Q35

Q.With usual notations prove that 2[asin⁡2C2+csin⁡2A2]=a−b+c2\left[a\sin^2\dfrac{C}{2} + c\sin^2\dfrac{A}{2}\right] = a - b + c.

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L.H.S. =2asin⁡2C2+2csin⁡2A2=a(1−cos⁡C)+c(1−cos⁡A)=(a+c)−(acos⁡C+ccos⁡A)=2a\sin^2\dfrac C2+2c\sin^2\dfrac A2=a(1-\cos C)+c(1-\cos A)=(a+c)-(a\cos C+c\cos A). By the Projection Rule, acos⁡C+ccos⁡A=ba\cos C+c\cos A=b. So L.H.S. =(a+c)−b=a−b+c==(a+c)-b=a-b+c= R.H.S. …

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