If A+B+C = 4S, then (2S-A)+(2S-B)-(2S-C)-(2S) =

Correct Answer is : 4sin⁡A2sin⁡B2cos⁡C24/2/2/24sin2Asin2Bcos2C

Correct Answer is : 4sin⁡A2sin⁡B2cos⁡C24/2/2/24sin2Asin2Bcos2C

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TS EAMCET 5-Aug-2021 Shift 2 Question Paper

Section: Mathematics

Question : 23 of 160

Marks: +1, -0

A+B+C

4S

\cos(2S-A)+\cos(2S-B)-\cos(2S-C)-\cos(2S)

Solution: 👈: Video Solution

Given,

A+B+C=4S \Rightarrow\;\frac{A+B+C}{2}=2S

To find :

\cos(2S-A)+\cos(2S-B)-\cos(2S-C)

-\cos 2S

\cos\left(\frac{A+B+C}{2}-A\right)+\cos\left(\frac{A+B+C}{2}-B\right)

-\cos\left(\frac{A+B+C}{2}-C\right)-\cos\left(\frac{A+B+C}{2}\right)

=\cos\left(\frac{B+C-A}{2}\right)+\cos\left(\frac{A+C-B}{2}\right)

-\cos\left(\frac{A+B-C}{2}\right)-\cos\left(\frac{A+B+C}{2}\right)

=\cos\left(\frac{B+C-A}{2}\right)-\cos\left(\frac{B+C+A}{2}\right)

+\cos\left(\frac{A-B+C}{2}\right)-\cos\left(\frac{A+B-C}{2}\right)

=2\sin\left(\frac{B+C}{2}\right)\cdot\sin\left(\frac{A}{2}\right)+2\sin\left(\frac{A}{2}\right)\cdot\sin\left(\frac{B-C}{2}\right)

=2\sin\left(\frac{A}{2}\right)\left\{\sin\left(\frac{B+C}{2}\right)+\sin\left(\frac{B-C}{2}\right)\right\}

{\begin{array}{l} \because \cos C-\cos D=2\sin\frac{C+D}{2}\cdot\sin\frac{D-C}{2} \\ \text{ and } \sin C+\sin D=2\sin\frac{C+D}{2}\cdot\cos\frac{C-D}{2} \end{array}}

=\sin\left(\frac{A}{2}\right)\left\{2\sin\left(\frac{B}{2}\right)\cdot\cos\left(\frac{C}{2}\right)\right\}

=4\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\cdot\cos\left(\frac{C}{2}\right)

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