The value of sin [n π+(-1)n (π/4)], n ∈ I is

Question

The value of sin [n π+(-1)n (π/4)], n ∈ I is (A) 0 (B) (1/√2) (C) -(1/√2) (D) None of these

Tardigrade Admin · Accepted Answer

sin [n π+(-1)n (π/4)]=(-1)n sin [(-1)n (π/4)] [∵ sin (n π+θ)=(-1)n sin θ] =(-1)n(-1)n sin (π/4) .∴ sin [(-1)n θ]=(-1)n sin θ]=(-1)2 n sin (π/4) = sin (π/4)=(1/√2)

Q. The value of $sin [nπ + (- 1)^{n} \frac{π}{4}], n \in I$ is

$0$

$\frac{1}{2}$

$- \frac{1}{2}$

None of these

$sin [nπ + (- 1)^{n} \frac{π}{4}] = (- 1)^{n} sin [(- 1)^{n} \frac{π}{4}]$
$[∵ sin (nπ + θ) = (- 1)^{n} sin θ]$
$= (- 1)^{n} (- 1)^{n} sin \frac{π}{4}$
$∴ sin [(- 1)^{n} θ] = (- 1)^{n} sin θ] = (- 1)^{2 n} sin \frac{π}{4}$
$= sin \frac{π}{4} = \frac{1}{2}$

Q. The value of sin[nπ+(−1)n4π​],n∈I is

0

2​1​

−2​1​

None of these

sin[nπ+(−1)n4π​]=(−1)nsin[(−1)n4π​] [∵sin(nπ+θ)=(−1)nsinθ] =(−1)n(−1)nsin4π​ ∴sin[(−1)nθ]=(−1)nsinθ]=(−1)2nsin4π​ =sin4π​=2​1​

Q. The value of $sin [nπ + (- 1)^{n} \frac{π}{4}], n \in I$ is

$0$

$\frac{1}{2}$

$- \frac{1}{2}$

$sin [nπ + (- 1)^{n} \frac{π}{4}] = (- 1)^{n} sin [(- 1)^{n} \frac{π}{4}]$
$[∵ sin (nπ + θ) = (- 1)^{n} sin θ]$
$= (- 1)^{n} (- 1)^{n} sin \frac{π}{4}$
$∴ sin [(- 1)^{n} θ] = (- 1)^{n} sin θ] = (- 1)^{2 n} sin \frac{π}{4}$
$= sin \frac{π}{4} = \frac{1}{2}$