Question

If `cos^(2)20^@-sin^(2)20^@=p` then `sin80^@=`?

Answer 1

`sin80^@=1/2sqrt(2+sqrt(2+2p))`

Explanation:

As `cos2A=cos^2A-sin^2A=2cos^2A-1=1-2sin^2A`

`cos^2 20^@-sin^2 20^@=cos(2xx20^@)=cos40^@=p`

Also `cosA=sqrt((1+cos2A)/2)`

Hence `2cos^2 20^@-1=p` and `cos20^@=sqrt((1+p)/2)`

and `cos10^@=sqrt((1+cos20^@)/2)=sqrt((1+sqrt((1+p)/2))/2)`

Now `sin80^@=cos(90^@-10^@)=cos10^@`

= `sqrt((1+sqrt((1+p)/2))/2)`

= `sqrt((sqrt2+sqrt(1+p))/(2sqrt2))`

= `1/2sqrt(2+sqrt(2+2p))`