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Use the periodic and even–odd properties.

If f(θ)=cscθand f(a)=2, find the exact value of :

(a) f(-a)

(b) localid="1647096128554" f(a)+f(a+2π)+f(a+4π).

Short Answer

Expert verified

(a) The value of f(-a)is-2.

(b) The value of localid="1647095937703" f(a)+f(a+2π)+f(a+4π)is 6.

Step by step solution

01

Step 1. Given Information   

We have given that following function :-

f(θ)=cscθand f(a)=2.

We have to find the value of f(-a)and value of f(a)+f(a+2π)+f(a+4π).

To find value of f(-a)we will use even-odd properties and to find the value of f(a)+f(a+2π)+f(a+4π)we will use periodic properties.

02

Step 2. Part (a). To find value of  f(-a)

We have given that :-

f(θ)=cscθand f(a)=2.

We know that :-

csc(-θ)=-cscθ.

Now put θ=-a, in f(θ)=cscθ, then we have :-

f(-a)=csc(-a)f(-a)=-csc(a)f(-a)=-f(a)

Now put f(a)=2, then we have :-

f(-a)=-2.

This is the required value.

03

Step 3. Part (b). To find value of  f(a)+f(a+2π)+f(a+4π)

We have given that :-

f(θ)=cscθ.

We know that cosecant function is of period 2π.

This gives us :-

csc(θ+2πk)=cscθ, for any integer k.

Now :-

role="math" localid="1647130898644" f(a)+f(a+2π)+f(a+4π)=csc(a)+csc(a+2π)+csc(a+4π)f(a)+f(a+2π)+f(a+4π)=csc(a)+csc(a)+csc(a)f(a)+f(a+2π)+f(a+4π)=3csc(a)f(a)+f(a+2π)+f(a+4π)=3f(a)

Put f(a)=2, then we have :-

f(a)+f(a+2π)+f(a+4π)=3(2)f(a)+f(a+2π)+f(a+4π)=6

This is the required value.

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