In triangle $A B C$ prove that
i. $sin A = sin (B + C)$ ii. $sin 2 A = - sin (2 B + 2 C)$ iii. $cos A = - cos (A + B)$ iv $tan \frac{A + B}{2} = - cot \frac{C}{2}$

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Solution

$A + B + C = π$
i) $sin A = sin (π - (B + C)) = sin (B + C)$
ii) $sin 2 A = sin (2 π - 2 (B + C)) = - sin (2 B + 2 C)$
iii) $cos A = cos (π - (B + C)) = - cos (B + C)$
iv) $tan (\frac{A + B}{2}) = tan (\frac{π - C}{2}) = cot (\frac{C}{2})$

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Similar questions

Q. Prove that:
(i)

\frac{s i n (A + B) + s i n (A - B)}{c o s (A + B) + c o s (A - B)} = t a n A

(ii)

\frac{s i n (A - B)}{c o s A c o s B} + \frac{s i n (B - C)}{c o s B c o s C} + \frac{s i n (C - A)}{c o s C c o s A} = 0

(iii)

\frac{s i n (A - B)}{s i n A s i n B} + \frac{s i n (B - C)}{s i n B s i n C} + \frac{s i n (C - A)}{s i n C s i n A} = 0

(iv)

s i n^{2} B = s i n^{2} A + s i n^{2} (A - B) - 2 s i n A c o s B s i n (A - B)

(v)

c o s^{2} + c o s^{2} B - 2 c o s A c o s B c o s (A + B) = s i n^{2} (A + B)

(vi)

\frac{t a n (A + B)}{c o t (A - B)} = \frac{t a n^{2} A - t a n^{2} B}{1 - t a n^{2} A t a n^{2} B}

Q. In any

△ A B C

prove the following:

(i) a s i n A - b s i n B = c s i n (A - B)

(i i) \frac{c - b c o s A}{b - c c o s A} = \frac{c o s B}{c o s C}

(i i i) a^{2} s i n (B - C) = (b^{2} - c^{2}) s i n A

(i v) \frac{(a - b)}{(a + b)} = \frac{tan (\frac{A - B}{2})}{t a n (\frac{A + B}{2})}

(v) a c o s B - b c c o s A = (a^{2} - b 2)

If A, B, C are the angles of a triangle, then
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Prove that:
(i) $\frac{s i n A + s i n 3 A}{c o s A - c o s 3 A} = c o t A$
(ii) $\frac{s i n 9 A - s i n 7 A}{c o s 7 A - c o s 9 A} = c o t 8 A$
(iii) $\frac{s i n A - s i n B}{c o s A - c o s B} = t a n \frac{A - B}{2}$
(iv) $\frac{s i n A + s i n B}{s i n A - s i n B} t a n (\frac{A + B}{2}) c o t \frac{A + B}{2}$
(v) $\frac{c o s A + c o s B}{c o s B - c o s A} = c o t \frac{A + B}{2} c o t \frac{A - B}{2}$

Q.
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A + B + C = 180^{\circ}

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cos A + cos B + cos C

2 + 2 cos A cos B cos C

{sin}^{2} A + {sin}^{2} B + {sin}^{2} C

4 cos \frac{A}{2} cos \frac{B}{2} cos \frac{C}{2}

sin 2 A + sin 2 B + sin 2 C

1 + 4 sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}

sin A + sin B + sin C

4 sin A sin B sin C

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In triangle ABC prove that i. sinA=sin(B+C) ii.sin2A=−sin(2B+2C) iii.cosA=−cos(A+B) iv tanA+B2=−cotC2

A+B+C=πi) sinA=sin(π−(B+C))=sin(B+C)ii) sin2A=sin(2π−2(B+C))=−sin(2B+2C)iii) cosA=cos(π−(B+C))=−cos(B+C)iv) tan(A+B2)=tan(π−C2)=cot(C2)

In triangle $A B C$ prove that
i. $sin A = sin (B + C)$ ii. $sin 2 A = - sin (2 B + 2 C)$ iii. $cos A = - cos (A + B)$ iv $tan \frac{A + B}{2} = - cot \frac{C}{2}$

$A + B + C = π$
i) $sin A = sin (π - (B + C)) = sin (B + C)$
ii) $sin 2 A = sin (2 π - 2 (B + C)) = - sin (2 B + 2 C)$
iii) $cos A = cos (π - (B + C)) = - cos (B + C)$
iv) $tan (\frac{A + B}{2}) = tan (\frac{π - C}{2}) = cot (\frac{C}{2})$