If A, B, C are angle of a triangle. Show that cos (1/2)C + cos (1/2)(A − B) = 2 sin (1/2)A sin (1/2)B


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Question Number 60533 by Tawa1 last updated on 21/May/19

$If A, B, C are angle of a triangle . Show that$ $\cos \frac{1}{2} C + \cos \frac{1}{2} (A - B) = 2 \sin \frac{1}{2} A \sin \frac{1}{2} B$
Commented by malwaan last updated on 21/May/19

$A + B + C = 180^{°}$ $c o s (x + y) - c o s (x - y) = - 2 s i n (x) s i n (y)$ $∴ 2 s i n (x) s i n (y) = c o s (x - y) - c o s (x + y)$ $R . H . S .$ $2 s i n \frac{1}{2} A s i n \frac{1}{2} B = c o s \frac{1}{2} (A - B) - c o s \frac{1}{2} (A + B)$ $= c o s \frac{1}{2} (A - B) - c o s \frac{1}{2} (180 - C)$ $= c o s \frac{1}{2} (A - B) - c o s (90 - \frac{C}{2})$ $= c o s \frac{1}{2} (A - B) - s i n \frac{1}{2} C$ $\neq L . H . S .$ $w h a t s w r o n g ?$ $t h e q u e s t i o n o r m e ?$
Commented by Tawa1 last updated on 21/May/19

$God bless you sir$
	Answered by MJS last updated on 21/May/19

	$let A = 60 ° B = 45 ° C = 75 °$ $\cos 37.5 ° + \cos 7.5 ° \approx 1.78$ $2 \sin 30 ° \sin 22.5 ° \approx 0.38$ $formula is wrong$
	Commented by Tawa1 last updated on 21/May/19

	$God bless you sir$
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