If sin(A−B)=12 and cos(A+B)=12 then (A,B)=

sin(A−B)=12 , cos(A+B)=12 A−B=30°→(1) A+B=60°→(2) (1)+(2) 2A=90° A=45°=π4 B=15°=π12

If sin(A−B)=12 and cos(A+B)=12 then (A,B)=

sin(A−B)=12 , cos(A+B)=12 A−B=30°→(1) A+B=60°→(2) (1)+(2) 2A=90° A=45°=π4 B=15°=π12

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If $\sin (A - B) = \frac{1}{2}$ and $\cos (A + B) = \frac{1}{2}$ then $(A, B) =$

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(\frac{π}{4}, \frac{π}{12})

(\frac{π}{12}, \frac{π}{4})

(\frac{π}{3}, \frac{π}{9})

(\frac{π}{4}, \frac{π}{6})

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Detailed Solution

\sin (A - B) = \frac{1}{2}

\cos (A + B) = \frac{1}{2}

$A - B = 30 ° \to (1)$

$A + B = 60 ° \to (2)$

$(1) + (2)$

$2 A = 90 °$

$A = 45 ° = \frac{π}{4}$

$B = 15 ° = \frac{π}{12}$

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If $\sin (A - B) = \frac{1}{2}$ and $\cos (A + B) = \frac{1}{2}$ then $(A, B) =$