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

sin [ n π + ( − 1 ) n 4 π ] = ( − 1 ) n sin [ ( − 1 ) n 4 π ] [ ∵ sin ( n π + θ ) = ( − 1 ) n sin θ ] ∴ sin = ( − 1 ) n ( − 1 ) n sin 4 π [ ( − 1 ) n θ ] = ( − 1 ) n sin θ ] = ( − 1 ) 2 n sin 4 π = sin 4 π = 2 1

Mathematics - Trigonometric Equation Question with Solution | TestHub

MathematicsTrigonometric EquationTrigonometric EquationEasy2 minPYQ_2020

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

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Solution:

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

Stream:BITSATSubject:MathematicsTopic:Trigonometric EquationSubtopic:Trigonometric Equation

⏱ 2mℹ️ Source: PYQ_2020

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