Sin/Cos Question (1 Viewer)

[frenzal_dude]

If x is between 180 and 270 degrees and sinx=p, express cosx in terms of p.

I know the answer is this but have no idea how to get there, hope you guys can help: cosx= -sqrt(1 - p^2)

 

[madsam]

As 180< x < 270
0 > cosx > -1
Therefore

 

[frenzal_dude]

thanx heaps!

 

[shaon0]

thanx heaps!

-sqrt(1-p^2)?

 

[madsam]

oh right, sorry when i square rooted both sides it should be +-, and then while x is between 180 and 270, cos x is negative, and as it is in the Real field, the sqrt has to be positive, making the only possibility -sqrt

 

[adomad]

woow... thats trippy.. would of have never though of that. but wouldn't u get a square root of a -ve number as it is p^2? just a query

 

[shaon0]

woow... thats trippy.. would of have never though of that. but wouldn't u get a square root of a -ve number as it is p^2? just a query

No because -1 =< p =< 1