In-the-triangle-ABC-cos-B-C-1-3-Show-that-1-3cos-B-C-6sinBcosC-tanC-

Question Number 209560 by MM42 last updated on 15/Jul/24

I n t h e t r i a n g l e A B C; c o s (B - C) = \frac{1}{3}

S h o w t h a t : \frac{1 - 3 c o s (B + C)}{6 s i n B c o s C} = t a n C

Answered by Spillover last updated on 14/Jul/24

\frac{1 - 3 \times \frac{1}{3}}{6 s i n B c o s C} = t a n C

0 = \tan C

\dots \dots

s o m e t h i n g w r o n g

Commented by MM42 last updated on 15/Jul/24

c o r r e c t e d

Answered by Berbere last updated on 15/Jul/24

c o s (B - C) = \frac{1}{3} W i t h o u t l o s t o f G e n e r a l i t y B > C

B - C = a r c o s (\frac{1}{3})

A + B + C = π

B + C = a r c o s (\frac{1}{3}) + 2 C

s i n (B) c o s (c) = s i n (c + \cos^{- 1} (\frac{1}{3})) . c o s (c)

(\frac{1}{3} s i n (c) + \frac{2 \sqrt{2}}{3} c o s (c)) c o s C)

1 - 3 c o s (\cos^{- 1} (\frac{1}{3}) + 2 c) = 1 - c o s (2 c) + 3 . \frac{2 \sqrt{2}}{3} s i n (2 c)

= 2 {s i n}^{2} (c) - 4 \sqrt{2} s i n (c) c o s (c) = s i n (c) (2 s i n (c) + 4 \sqrt{2} c o s (c))

\frac{1 - 3 c o s (B + C))}{6 s i n (B) c o s (C)} = \frac{s i n (C) (2 s i n (C) + 4 \sqrt{2} c o s (C))}{(2 s i n (C) + 4 \sqrt{2} c o s (C)) c o s (C)} = \tan (C)

Commented by MM42 last updated on 15/Jul/24

\underset{―}{\underset{⏟}{⋚}}

Answered by MM42 last updated on 15/Jul/24

\frac{3 (\frac{1}{3} - c o s (B + C))}{6 s i n B c o s C} = \frac{3 (c o s (B - C) - c o s (B + C))}{6 s i n B c o s C}

= \frac{3 [(- 2) s i n (\frac{B - C - B - C)}{2}) s i n (\frac{B - C + B + C}{2})}{6 s i n B c o s C}

= \frac{6 s i n B s i n c}{6 s i n B c o s C} = t a n C ✓

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