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View Full Version : Proving trigonometric identities..


blair.1969
Oct 10, 2010, 09:12 AM
1. Cotx cosx = cscx - sinx
2. (tanx - sinx)^2 + (1 - cosx) = (1 - secx)^2
3. cotx + tanx = cscx^2+ sec^2 x/ cscx secx
4. 1 - 2 cos^2x / cotx cosx = tanx - cotx
5. sin^4x - cos^4x = sin^2x - cos^2x
6. sin^6x + cos^6 = 1 - 3 sin^2x cos^2x
7. cscx/cscx-1 + cscx/ cscx + 1 = 2sec^2x

please help

Unknown008
Oct 10, 2010, 09:53 AM
That's just too much. You need to show us what you tried so far.

A hint for any trig proof, convert everything to cos and sin if possible. Also, you might find it easier to go from several terms to single terms. For the first one as example, you would be easier off starting from (csc x - sin x).

blair.1969
Oct 10, 2010, 10:04 AM
ex.
tan^2x - sin^2x = tan^2x sin^2x
proof:
=tan^2x - sin^2x
= (sin^2x/cos^2x) - sin^2x
= (sin^2x - c0s^2x sin^2x)/ cos^2x
=(sin^2x (1 - cos^2x))/ cos^2x
= (sin^2x/cos^2x)(1-cos^2x)
= tan^2x sin^2x

Unknown008
Oct 10, 2010, 10:12 AM
Right! :)

paulini
May 3, 2011, 02:49 PM
prove by identity cosx+cosxtan^2x=secx

ebaines
May 3, 2011, 04:57 PM
prove by identity cosx+cosxtan^2x=secx

Do as Unk suggested in his earlier post - convert tan^2x to sin^2x/cos^2x and go from there.