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-

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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