# Find the value of `2sin^(2)pi/6 + cosec^(2) 7pi/6 cos^(2) pi/3?` ``

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dear alex1981, pls. note that cos(pi/3) = 1/2 and not sqrt(3)/2 as you mentioned.

(sin (pi/3) = sqrt(3)/2).

Using the identity, cosec x =1/sin x , we get

2 sin^2 (pi/6) + (cos^2 (pi/3)) /(sin^2(7pi/6))

using the identity, sin (u+v) = sin u cos v +- cos u sin v

and writing 7pi/6 as (pi + pi/6), we get

2 sin^2(pi/6) + (cos^2(pi/3))/[sin pi cos (pi/6) +- sin (pi/6) cos pi]^2

using the values of sin and cos as sin (pi/6) = 1/2; cos (pi/3) = 1/2; sin pi = 0; cos pi = -1; we get

2 (1/2)^2 + [(1/2)^2]/[-+ 1/2]^2 = 2/4 + (1/4)/(1/4) = **3/2**

sin(pi/6)=1/2

sin^2(pi/6)=1/4

2sin^2(pi/6)=2/4=1/2

Then: cosec(7pi/6)=-2

So: cosec^2(7pi/6)=(-2)^2=4

And cos(pi/3)=sqrt(3)/2

So cos^2(pi/3)=3/4

And so: 2sin^2(pi/6)+cosec^2(7pi/6)cos^2(pi/3)= 1/4 + 4*3/4=1/4 + 3 =13/4

Please ask if in doubt!