Question

if A+B+C=2S then prove that cos(s-a)+cos(s-b)+cos(c)=4cos(s-a/2).cos(s-b/2).cosc/2-1

Answer 1

HELLO DEAR,

Your questions is ------------> if A + B + C = 2S then prove that $\sf{cos(S - A) + cos(S - B) + cosc = -1 + 4cos(S - A/2)*cos(S - B/2)*cos(c/2).}$

$\sf{GIVEN:-A + B + C = 2S}$

we know :- $\sf{cosA + cosB = 2cos(A + B)/2 * cos(A - B)/2}$

Now, L.H.S = $\sf{cos(A - A) + cos(S - B) + cosc =2cos\frac{S - A + S - B}{2}*cos\frac{S - A - S + B}{2} + cosc}$

⇒$\sf{2cos(2S - A - B)/2*cos(B - A)/2 + cosc}$

⇒$\sf{2cos(A + B + C - A - B)/2 * cos(B - A)/2 + cosc}$
$\to\to\to\to\to\to\to\boxed{\bold{A + B + C = 2S}}$​

⇒$\sf{2cos(C/2)*cos(B - A)/2 + 2cos^2(c/2) - 1}$

⇒$\sf{2cos(c/2)[cos(B - A)/2 + cos(c/2)] - 1}$

⇒$\sf{2cos(c/2)[2cos\frac{(B - A + C)/2}{2} * cos\frac{(B - A - C)/2}{2}] - 1}$

⇒$\sf{4cos(c/2)[cos\frac{2S - A - A}{4}*cos\frac{2S - B - B}{2}] - 1}$
$\to\to\to\to\to\to\to\boxed{\bold{A + B + C = 2S}}$​

⇒$\sf{4cos(C/2)*cos(S - A)/2*cos(S - B)/2 - 1}$

Hence, $\boxed{\sf{cos(S - A) + cos(S - B) + cosc = -1 + 4cos(S - A/2)*cos(S - B/2)*cos(c/2).}}$

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Answer:

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