If ( θ _ 1 ) + ( θ _ 2 ) + . . . . . . + ( θ _ n ) = n . Then value of ( θ _ 1 ) + ( θ _ 2 ) + ( θ _ 3 ) + ( θ _ 4 ) . . . . . + ( θ _ n - 1 ) + ( θ _ n ) is equal to ( θ _ 1 , θ _ 2 , θ _ 3 , . . . . . . . . θ _ n lies

Correct answer: B. (n)/(2) Given,& ( θ _ 1 ) + ( θ _ 2 ) + . . + ( θ _ n ) = n ⇒ ( θ _ 1 ) = ( θ _ 2 ) = ( θ _ 3 ) = . . = ( θ _ n ) = 1 ⇒ θ _ 1 = θ _ 2 . . = θ _ n = 90 ^ o ∴ ( θ _ 1 ) + ( θ _ 2 ) + ( θ _ 3 ) + . . + ( θ _ n - 1 ) + ( θ _ n ) = ( θ _ 1 ) + ( θ _ 3 ) + . . + ( θ _ n - 1 ) + ( cos θ _ 2 + cos θ _ 4 + + cos θ _ n ) =(n)/(2) + 0 =(n)/(2)

JEE Main

Exam: JEE Main
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

JEE Main · Mathematics · Trigonometric Ratios & Identities

If ( θ _ 1 ) + ( θ _ 2 ) + . . . . . . + ( θ _ n ) = n . Then value...

\sin \left({θ}_{1}\right) + \sin \left({θ}_{2}\right) + . . . . . . + \sin \left({θ}_{n}\right) = n .

Then value of

\sin \left({θ}_{1}\right) + \cos \left({θ}_{2}\right) + \sin \left({θ}_{3}\right) + \cos \left({θ}_{4}\right) . . . . . +

\sin \left({θ}_{n - 1}\right) +

\cos \left({θ}_{n}\right) is equal to

( {θ}_{1}, {θ}_{2}, {θ}_{3}, . . . . . . . . {θ}_{n}

lies in

[ 0,90 ° ]

)

Options

A $n – 1$
B $\dfrac{n}{2}$
C $0$
D $n$

Correct answer

B. $\dfrac{n}{2}$

Step-by-step solution

Given,&

\sin \left({θ}_{1}\right) + \sin \left({θ}_{2}\right) + \ldots \ldots \ldots . . + \sin \left({θ}_{n}\right) = n

⇒ \sin \left({θ}_{1}\right) = \sin \left({θ}_{2}\right) = \sin \left({θ}_{3}\right) = \ldots \ldots \ldots . . = \sin \left({θ}_{n}\right) = 1

⇒ {θ}_{1} = {θ}_{2} \ldots \ldots \ldots . . = {θ}_{n} = {90}^{o}

∴ \sin \left({θ}_{1}\right) + \cos \left({θ}_{2}\right) + \sin \left({θ}_{3}\right) + \ldots \ldots \ldots . . + \sin \left({θ}_{n - 1}\right) + \cos ( {θ}_{n} )

= \sin \left({θ}_{1}\right) + \sin \left({θ}_{3}\right) + \ldots \ldots \ldots . . + \sin \left({θ}_{n - 1}\right) + \left(

cos{θ}_{2} + cos{θ}_{4} + \ldots + cos{θ}_{n}\right)

<br /><br />=\dfrac{n}{2} +

<br /><br />=\dfrac{n}{2}

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