If A+B+C=(π/4), then 4 cos (A/2) cos (B/2) cos (C/2)- cos (π/8)=

Question

If A+B+C=(π/4), then 4 cos (A/2) cos (B/2) cos (C/2)- cos (π/8)= (A)  cos ((π/4)-A)+ cos ((π/4)-B)+ cos ((π/4)-C) (B)  cos ((π/8)-A)+ cos ((π/8)-B)+ cos ((π/8)-C) (C)  sin ((π/4)-A)+ sin ((π/4)-B)+ sin ((π/4)-C) (D)  sin ((π/8)-A)+ sin ((π/8)-B)+ sin ((π/8)-C)

Tardigrade Admin · Accepted Answer

2(2 cos (A/2) cos (B/2)) cos (c/2)- cos (π/8) =2[ cos ((A/2)+(B/2))+ cos ((A/2)-(B/2))] cos (c/2) =2[ cos ((π/8)-(C/2))+ cos ((A/2)-(B/2))] cos (C/2)- cos (π/8) =2 cos ((π/8)-(C/2)) cos (C/2)+2 cos ((A/2)-(B/2)) cos (C/2)- cos (π/8) = cos ((π/8))+ cos ((π/8)-C)+ cos ((A/2)-(B/2)+(C/2)) + cos ((A/2)-(B/2)-(C/2))- cos (π/8) = cos ((π/8)-c)+ cos ((π/8)-(B/2)-(B/2))+ cos [(A/2)-((π/8)-(A/2))] = cos ((π/8)-C)+ cos ((π/8)-B)+ cos ((π/8)-A)

Q. If $A + B + C = \frac{π}{4}$ , then $4 cos \frac{A}{2} cos \frac{B}{2} cos \frac{C}{2} - cos \frac{π}{8} =$

$cos (\frac{π}{4} - A) + cos (\frac{π}{4} - B) + cos (\frac{π}{4} - C)$

$cos (\frac{π}{8} - A) + cos (\frac{π}{8} - B) + cos (\frac{π}{8} - C)$

$sin (\frac{π}{4} - A) + sin (\frac{π}{4} - B) + sin (\frac{π}{4} - C)$

$sin (\frac{π}{8} - A) + sin (\frac{π}{8} - B) + sin (\frac{π}{8} - C)$

Q. If A+B+C=4π​, then 4cos2A​cos2B​cos2C​−cos8π​=

cos(4π​−A)+cos(4π​−B)+cos(4π​−C)

cos(8π​−A)+cos(8π​−B)+cos(8π​−C)

sin(4π​−A)+sin(4π​−B)+sin(4π​−C)

sin(8π​−A)+sin(8π​−B)+sin(8π​−C)

Q. If $A + B + C = \frac{π}{4}$ , then $4 cos \frac{A}{2} cos \frac{B}{2} cos \frac{C}{2} - cos \frac{π}{8} =$

$cos (\frac{π}{4} - A) + cos (\frac{π}{4} - B) + cos (\frac{π}{4} - C)$

$cos (\frac{π}{8} - A) + cos (\frac{π}{8} - B) + cos (\frac{π}{8} - C)$

$sin (\frac{π}{4} - A) + sin (\frac{π}{4} - B) + sin (\frac{π}{4} - C)$

$sin (\frac{π}{8} - A) + sin (\frac{π}{8} - B) + sin (\frac{π}{8} - C)$