For any angle θ, the expression ( 2 cos 8 θ + 1 / 2 cos θ + 1 ) =

Question

For any angle θ, the expression ( 2 cos 8 θ + 1 / 2 cos θ + 1 ) =  (A) (2 cos θ + 1) (2 cos 2θ + 1) (2 cos 4θ + 1) (B) (cos θ - 1) (cos 2θ - 1) (cos 4θ - 1) (C) (2 cos θ - 1) (2 cos 2θ - 1) (2 cos 4θ - 1) (D) (2 cos θ + 1) (2 cos 2θ + 1) (2 cos 4θ - 1),

Tardigrade Admin · Accepted Answer

(2 cos 8θ + 1 /2 cos θ +1) = (2(2 cos2 4 θ - 1) - 1/2 cos θ + 1 ) = ((2 cos 4 θ -1)(2 cos 4 θ +1 )/2 cos θ +1) = ((2 cos4 θ - 1 )(4 cos2 2θ -1)/2 cos θ +1) =((2 cos 4 θ -1)(2 cos 2θ -1)(2 cos 2 θ +1)/2 cos θ +1) =((2 cos 4θ -1)(2 cos 2 θ-1) (4 cos2 θ -1) /2 cos θ +1)

Q. For any angle $θ$ , the expression $\frac{2 c o s 8 θ + 1}{2 c o s θ + 1} =$

(2 cos $θ$ + 1) (2 cos 2 $θ$ + 1) (2 cos 4 $θ$ + 1)

(cos $θ$ - 1) (cos 2 $θ$ - 1) (cos 4 $θ$ - 1)

(2 cos $θ$ - 1) (2 cos 2 $θ$ - 1) (2 cos 4 $θ$ - 1)

(2 cos $θ$ + 1) (2 cos 2 $θ$ + 1) (2 cos 4 $θ$ - 1),

Q. For any angle θ, the expression 2cosθ+12cos8θ+1​=

(2 cos θ + 1) (2 cos 2θ + 1) (2 cos 4θ + 1)

(cos θ - 1) (cos 2θ - 1) (cos 4θ - 1)

(2 cos θ - 1) (2 cos 2θ - 1) (2 cos 4θ - 1)

(2 cos θ + 1) (2 cos 2θ + 1) (2 cos 4θ - 1),

2cosθ+12cos8θ+1​=2cosθ+12(2cos24θ−1)−1​ =2cosθ+1(2cos4θ−1)(2cos4θ+1)​ =2cosθ+1(2cos4θ−1)(4cos22θ−1)​ =2cosθ+1(2cos4θ−1)(2cos2θ−1)(2cos2θ+1)​ =2cosθ+1(2cos4θ−1)(2cos2θ−1)(4cos2θ−1)​

Q. For any angle $θ$ , the expression $\frac{2 c o s 8 θ + 1}{2 c o s θ + 1} =$

(2 cos $θ$ + 1) (2 cos 2 $θ$ + 1) (2 cos 4 $θ$ + 1)

(cos $θ$ - 1) (cos 2 $θ$ - 1) (cos 4 $θ$ - 1)

(2 cos $θ$ - 1) (2 cos 2 $θ$ - 1) (2 cos 4 $θ$ - 1)

(2 cos $θ$ + 1) (2 cos 2 $θ$ + 1) (2 cos 4 $θ$ - 1),