Question

If sin^16 theta =1/5 then the value of 1/cos^2 theta + 1/1+sin^2 theta +2/1+sin^4 theta +4/1+sin^8 theta is

Answer 1

Given : Sin¹⁶θ  = 1/5

To find : 1/Cos²θ   + 1/(1 + Sin²θ)  +  2/(1 + Sin⁴θ)  +  4/(1 + Sin⁸θ)  

Solution:

1/Cos²θ   + 1/(1 + Sin²θ)  +  2/(1 + Sin⁴θ)  +  4/(1 + Sin⁸θ)  

Cos²θ  = 1 - Sin²θ

=  1/(1 - Sin²θ)  + 1/(1 + Sin²θ)  +  2/(1 + Sin⁴θ)  +  4/(1 + Sin⁸θ)  

= (1 + Sin²θ + 1 - Sin²θ)/(1 -  Sin⁴θ)   + 2/(1 + Sin⁴θ)  +  4/(1 + Sin⁸θ)  

=  2/(1 -  Sin⁴θ)   +  2/(1 + Sin⁴θ)  +  4/(1 + Sin⁸θ)  

= (2 + 2Sin⁴θ + 2 - 2Sin⁴θ) /(1 -Sin⁸θ)    + 4/(1 + Sin⁸θ)  

= ( 4)/(1 -Sin⁸θ)    + 4/(1 + Sin⁸θ)  

=   4 ( 1/(1 -Sin⁸θ)    + 1/(1 + Sin⁸θ))

=4 ( 1 + Sin⁸θ + 1 - Sin⁸θ)/( 1 - Sin¹⁶θ)

= 4 (2) /  ( 1 - Sin¹⁶θ)

= 8/ ( 1 - Sin¹⁶θ)

= 8/(1 - 1/5)

= 8/(4/5)

= 10

1/Cos²θ   + 1/(1 + Sin²θ)  +  2/(1 + Sin⁴θ)  +  4/(1 + Sin⁸θ)    = 10

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