If sin 2 A -1-2 and sin 2 B -1-2- then which one of the following is falseA- sin A-B may be 0B- cos A - B may be zeroC- sin A-B or cos A-B is zeroD- sin A-B-0

The correct option is D sin (A + B) = 0sin 2A = 1/2,sin 2B = 1/2 sin2A + sin 2B = 0 2sin(A + B) cos (A B) = 0 Sin(A + B) = 0 or cos(A B) = 0 ...

Point of Inflection

If sin 2 A =1...

Question

If $sin 2 A = \frac{1}{2}$ and $sin 2 B = - \frac{1}{2}$ , then which one of the following is false

sin (A + B) may be 0

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cos (A - B) may be zero

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sin(A + B) or cos (A - B) is zero

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sin (A + B) = 0

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Solution

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Similar questions

If sinA=sinB and cosA=cosB, then

[EAMCET 1994]

\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

\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

\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

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)

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)

\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}

\frac{sin (A + B) + sin (A - B)}{cos (A + B) + cos (A - B)} = tan A

sin (A + B) + sin (A - B) = sin A

cos (A + B) + cos (A - B) = cos A

\frac{s i n A + c o s A}{s i n A - c o s A} + \frac{s i n A - c o s A}{s i n A + c o s A} = \frac{2}{s i n^{2} A - c o s^{2} A} = \frac{2}{2 s i n^{2} A - 1} = \frac{2}{1 - 2 c o s^{2} A}

A B C

sin A = sin (B + C)

sin 2 A = - sin (2 B + 2 C)

cos A = - cos (A + B)

tan \frac{A + B}{2} = - cot \frac{C}{2}

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