The value of (1+cos (π/10))(1+cos (3π /10))(1+cos (7π /10))(1+cos (9π /10)) is

Question

The value of (1+cos (π/10))(1+cos (3π /10))(1+cos (7π /10))(1+cos (9π /10)) is (A) (1/8) (B) (1/16) (C) (1/32) (D) None of these

Tardigrade Admin · Accepted Answer

The expression = (1+cos (π/10))(1+cos (3π /10))(1-cos (3π /10))(1-cos (π /10)) [∵ cos (7π /10) = cos(π-(3π /10)) = -cos (3π /10) and cos (9π /10) = cos(π -(π /10)) =- cos (π /10)] = (1-cos2 (π /10))(1-cos2 (3π /10)) = sin2 (π /10). sin2 (3π /10) = sin2 18°. = sin2 54° = ((√5-1/4).(√5+1/4))2 = (1/16)

Q. The value of
$(1 + cos \frac{π}{10}) (1 + cos \frac{3 π}{10}) (1 + cos \frac{7 π}{10}) (1 + cos \frac{9 π}{10})$ is

$\frac{1}{8}$

$\frac{1}{16}$

$\frac{1}{32}$

None of these

Q. The value of (1+cos10π​)(1+cos103π​)(1+cos107π​)(1+cos109π​) is

81​

161​

321​

None of these

The expression =(1+cos10π​)(1+cos103π​)(1−cos103π​)(1−cos10π​) [∵cos107π​=cos(π−103π​)=−cos103π​andcos109π​=cos(π−10π​)=−cos10π​] =(1−cos210π​)(1−cos2103π​)=sin210π​.sin2103π​ =sin218∘.=sin254∘=(45​−1​.45​+1​)2=161​

Q. The value of
$(1 + cos \frac{π}{10}) (1 + cos \frac{3 π}{10}) (1 + cos \frac{7 π}{10}) (1 + cos \frac{9 π}{10})$ is

$\frac{1}{8}$

$\frac{1}{16}$

$\frac{1}{32}$