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Trigonometric Identities

Name: Anonymous 2008-12-02 16:31

sin²x - sinx + cos²x + 1 = 3 - sinx - sec²x + sin²x/cos²x

How do I prove this /sci/ ?

Name: 4tran 2008-12-02 16:41

Get rid of obvious stuff first; you'll get
0 = 1 - sec²x + tan²x

Name: Anonymous 2008-12-02 17:23

OP: What? I need steps!

Name: Anonymous 2008-12-02 17:44

>>3
On the right side, sin^2(x)/cos^2(x) = tan^2(x), and tan^2(x) - sec^2(x) = -1, so you have 2 - sin(x). Replace 2 with 1 + sin^2(x) + cos^2(x) and you're done.

Name: Anonymous 2008-12-02 18:46

Protip: EVERY trig identity can be proved by writing everything in sines and cosines and using sin^2 x + cos^2 x = 1

Name: Anonymous 2008-12-02 18:47

Name: Anonymous 2008-12-02 18:49

>>6
>Hardcore Schoolgirls 14

Dammit, that was total false advertising >:(

Name: Anonymous 2008-12-02 19:20

trig identities are fun so is log.

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