If cos^(2)20^@-sin^(2)20^@=pcos220∘−sin220∘=p then sin80^@=sin80∘=?

1 Answer
Shwetank Mauria
Dec 2, 2017

sin80^@=1/2sqrt(2+sqrt(2+2p))sin80∘=12√2+√2+2p

Explanation:

As cos2A=cos^2A-sin^2A=2cos^2A-1=1-2sin^2Acos2A=cos2A−sin2A=2cos2A−1=1−2sin2A

cos^2 20^@-sin^2 20^@=cos(2xx20^@)=cos40^@=pcos220∘−sin220∘=cos(2×20∘)=cos40∘=p

Also cosA=sqrt((1+cos2A)/2)cosA=√1+cos2A2

Hence 2cos^2 20^@-1=p2cos220∘−1=p and cos20^@=sqrt((1+p)/2)cos20∘=√1+p2

and cos10^@=sqrt((1+cos20^@)/2)=sqrt((1+sqrt((1+p)/2))/2)cos10∘=√1+cos20∘2= ⎷1+√1+p22

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

= sqrt((1+sqrt((1+p)/2))/2) ⎷1+√1+p22

= sqrt((sqrt2+sqrt(1+p))/(2sqrt2))√√2+√1+p2√2

= 1/2sqrt(2+sqrt(2+2p))12√2+√2+2p

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