School Survival Forums
Pythagorean Identities? - Printable Version

+- School Survival Forums (http://forums.school-survival.net)
+-- Forum: Learning, Youth Rights and School Survival (/forumdisplay.php?fid=3)
+--- Forum: Homework Help & Answers (/forumdisplay.php?fid=30)
+--- Thread: Pythagorean Identities? (/showthread.php?tid=5096)



Pythagorean Identities? - Dreamer567 - 03-25-2010 12:27 PM

Can someone please verify these problems for me? I will be eternally grateful! (The little 2 after the "tan + 1" means squared)

1) tan + cot= sec(csc)

2) (tan + 1)2 - sec= 2tan


Re: Pythagorean Identities? - genuine anarchist - 03-27-2010 03:19 PM

Sorry if this is overdue; I learned this last year in precalculus:

1) tan + cot = sec (csc)
tan=sin/cos, cot=cos/sin, sec=1/cos, csc=1/sin

(sin/cos) + (cos/sin) = (1/cos)(1/sin) >substitute values from other identities

((sin/cos) + (cos/sin) = (1/cos)(1/sin)) (cos) >multiply whole equation by cos

(sin2/sincos) + (cos2/sincos) = (1/cos)(1/sin) [the 2 means "squared]

(sincos) ((sin2/sincos) +(cos2/sincos) = (1/sincos)) >multiply whole equation by sincos to cancel out the denominator

answer: you are left with sin2 + cos2 = 1, a fundamental pythagorean identity

2) (tan +1)2 - sec = 2tan
tan2+1=sec2, tan=sin/cos, sec-1/cos

tan2 + 2tan + 1 -sec = 2tan <use FOIL method to multiply out (tan + 1)2

sec2 - 1 + 2tan + 1 -sec = 2tan <substitute tan2 with (sec2 - 1)

sec2-sec=0
sec2=sec
tan2 + 1 = 1/cos <sec2=tan2+1, sec=1/cos
(sin/cos)2 + 1 = (1/cos) <tan=sin/cos, remember to square whole fraction
(cos2) ((sin/cos)2 + 1 = (1/cos)) <multiply whoe equation by cos2

answer: sin2 + cos2 = 1, a basic pythagorean identity

Hope this helps!