cos (7 π/8)+ cos (π/4)+ cos ((-π/8))-1=

Question

cos (7 π/8)+ cos (π/4)+ cos ((-π/8))-1= (A) 4 cos (π/16) cos (3 π/4) cos (5 π/8) (B) 4 cos (π/16) cos (π/8) sin (5 π/8) (C) 4 cos (π/16) cos (3 π/8) cos (9 π/16) (D) -4 cos (π/16) cos (5 π/8) cos (π/16)

Tardigrade Admin · Accepted Answer

cos ((7 π/8))+ cos ((π/4))+ cos ((-π/8))-1 = cos ((7 π/8))+ cos ((π/8))+ cos ((π/4))-1 =2 cos ((π/2)) cos ((3 π/8))+ cos ((π/4))-1=(1/√2)-1 Option (a) 4 cos ((π/16)) cos ((3 π/4)) cos ((5 π/8)) =2 ⋅ 2 cos ((π/16)) cos ((10 π/16)) cos ((3 π/4)) =2 ⋅[ cos ((11 π/16))+ cos ((9 π/16))] cos ((3 π/4)) = cos ((23 π/16))+ cos ((π/16))+ cos ((21 π/16))+ cos ((3 π/16)) Option (b) 4 cos ((π/16)) cos ((π/8)) cos ((5 π/8)) =2 ⋅[ cos ((3 π/16))+ cos ((π/16))] cos ((5 π/8)) = cos ((13 π/16))+ cos ((7 π/16))+ cos ((9 π/16))+ cos ((11 π/16)) Option (c) 4 cos ((π/16)) cos ((3 π/8)) cos ((9 π/16)) =2 ⋅[ cos ((7 π/16))+ cos ((5 π/16))] cos ((9 π/16)) = cos ((16 π/16))+ cos ((2 π/16))+ cos ((4 π/16))+ cos ((14 π/16)) =-1+ cos ((π/8))+(1/√2)+ cos ((7 π/8))=-1+(1/√2)

Q. $cos \frac{7 π}{8} + cos \frac{π}{4} + cos (\frac{- π}{8}) - 1 =$

$4 cos \frac{π}{16} cos \frac{3 π}{4} cos \frac{5 π}{8}$

$4 cos \frac{π}{16} cos \frac{π}{8} sin \frac{5 π}{8}$

$4 cos \frac{π}{16} cos \frac{3 π}{8} cos \frac{9 π}{16}$

$- 4 cos \frac{π}{16} cos \frac{5 π}{8} cos \frac{π}{16}$

Q. cos87π​+cos4π​+cos(8−π​)−1=

4cos16π​cos43π​cos85π​

4cos16π​cos8π​sin85π​

4cos16π​cos83π​cos169π​

−4cos16π​cos85π​cos16π​

Q. $cos \frac{7 π}{8} + cos \frac{π}{4} + cos (\frac{- π}{8}) - 1 =$

$4 cos \frac{π}{16} cos \frac{3 π}{4} cos \frac{5 π}{8}$

$4 cos \frac{π}{16} cos \frac{π}{8} sin \frac{5 π}{8}$

$4 cos \frac{π}{16} cos \frac{3 π}{8} cos \frac{9 π}{16}$

$- 4 cos \frac{π}{16} cos \frac{5 π}{8} cos \frac{π}{16}$