For a positive integer n- let fn-2cos-12cos- -12cos 2- -12cos 22- -1-2cos 2n-1- -1- Which one of the following holds not good-

The correct option is D f_5(π/128)=1+√(2) [ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Sum of Cosines of Angles in Arithmetic Progression

For a positiv...

Question

For a positive integer n, let $f_{n} (θ) = (2 cos θ + 1) (2 cos θ - 1) (2 cos 2 θ - 1) (2 cos (2^{2}) θ - 1) . . . . . . (2 cos (2^{n - 1}) θ - 1)$ . Which one of the following hold(s) not good?

f_{2} (π / 6) = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

f_{3} (π / 8) = - 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

f_{4} (π / 32) = 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

f_{5} (π / 128) = 1 + \sqrt{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $f_{5} (π / 128) = 1 + \sqrt{2}$
$\begin{matrix} f (θ) = (2 cos θ + 1) (2 cos θ - 1) (2 cos 2 θ - 1) (2 cos (2^{2}) θ - 1) . . . . . . (2 cos (2^{n - 1}) θ - 1) . = (2 cos θ + 1) + (2 cos θ - 1) . . . . (2 cos 2^{n - 1} θ + 1) N o w, f_{n} (θ) = 2 cos 2^{n} θ + 1 f_{2} (\frac{π}{6}) = 2 cos (4 \times \frac{π}{6}) + 1 = 2 cos \frac{2 π}{3} + 1 = 0 f_{3} (\frac{π}{8}) = 2 cos (8 \times \frac{π}{8}) + 1 = - 1 f_{4} (\frac{π}{32}) = 2 cos (16 \times \frac{π}{32}) + 1 = 1 f_{5} (\frac{π}{128}) = 2 cos (32 \times \frac{π}{128}) + 1 = 1 + \sqrt{2} H e n c e, t h e o p t i o n D i s t h e c o r r e c t a n s w e r . \end{matrix}$

Suggest Corrections

Similar questions

For a positive integer n, let
$f_{n} (θ) = (t a n \frac{θ}{2}) (1 + s e c θ) (1 + s e c 2 θ) (1 + s e c 4 θ) \dots (1 + s e c 2^{n} θ)$ .
Then

For a positive integer n, let $f_{n} (θ) = (t a n \frac{θ}{2}) (1 + s e c θ) (1 + s e c 2 θ) (1 + s e c 4 θ) . . . . . . (1 + s e c 2^{n} θ) . T h e n$

(2 c o s θ - 1) (2 c o s 2 θ - 1) (2 c o s 2^{2} θ - 1) . . . . . . (2 c o s 2^{n - 1} θ - 1) = \frac{2 c o s 2^{n} θ + 1}{2 c o s θ + 1}

Q. Show that

\frac{2 cos 2^{n} θ + 1}{2 cos θ + 1} = (2 cos θ - 1) (2 cos 2 θ - 1) (2 cos 2^{2} θ - 1) \dots (2 cos 2^{n - 1} θ - 1)

Q. For a positive integer

n,

let

f_{n} (θ) = (tan \frac{θ}{2}) (1 + sec θ) (1 + sec 2 θ) (1 + sec 4 θ) . . . (1 + sec 2^{n} θ) .

Then

Join BYJU'S Learning Program

For a positive integer n, let fn(θ)=(2cosθ+1)(2cosθ−1)(2cos2θ−1)(2cos(22)θ−1)......(2cos(2n−1)θ−1). Which one of the following hold(s) not good?

For a positive integer n, let $f_{n} (θ) = (2 cos θ + 1) (2 cos θ - 1) (2 cos 2 θ - 1) (2 cos (2^{2}) θ - 1) . . . . . . (2 cos (2^{n - 1}) θ - 1)$ . Which one of the following hold(s) not good?