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I know all the formulas for these type of proofs. I just have trouble getting one side to equal the other, and these two are giving me trouble.


1.) sin 3x (over) sin x - cos 3x (over) cos x = 2

2.) (1-sin^2x)(1-tan^2x) = cos2x

Get left to equal right on both ones.

Thanks for the help!

2007-10-31 09:07:49 · 3 answers · asked by neverisenough 1 in Science & Mathematics Mathematics

3 answers

Prove the identities.

1)
[sin 3x] / (sin x) - [cos 3x] / (cos x) = 2

Left Hand Side = [sin 3x] / (sin x) - [cos 3x] / (cos x)

= [sin(2x + x)] / (sin x) - [cos(2x + x)] / (cos x)

= [(sin 2x)(cosx) + (cos 2x)(sinx)] / (sin x)
- [(cos 2x)(cosx) - (sin 2x)(sinx)] / (cos x)

= [2(sinx)(cos²x) + (cos²x - sin²x)(sinx)] / (sin x)
- [(cos²x - sin²x)(cosx) - 2(sin²x)(cosx)] / (cos x)

= [2(cos²x) + (cos²x - sin²x)] - [(cos²x - sin²x) - 2(sin²x)]

= (3cos²x - sin²x) - (cos²x - 3sin²x)

= 2cos²x + 2sin²x

= 2(cos²x + sin²x) = 2*1 = 2 = Right Hand Side
_____________

2)
(1 - sin²x)(1 - tan²x) = cos 2x

Left Hand Side = (1 - sin²x)(1 - tan²x)

= (cos²x)(1 - sin²x/cos²x) = cos²x - sin²x

= cos 2x = Right Hand Side

2007-10-31 20:28:30 · answer #1 · answered by Northstar 7 · 0 0

1) write sin(3x) = sin(2x + x), do the same for cos(3x) and expand it out, it should simplify nicely.

2)
(1-sin²(x)) * (1 - tan²(x)) = cos(2x)
replace 1-sin²(x) and change tangent to sine/cosine
(cos²(x)) * (1 - sin²(x) / cos²(x)) = cos(2x)
simplify
cos²(x) - sin²(x) = cos(2x)
replace with the double angle formula for cosine
cos(2x) = cos(2x)

2007-10-31 16:11:56 · answer #2 · answered by Mαtt 6 · 0 0

1) sin(3x)/sin(x) - cos(3x)/cos(x) = [sin(3x) cos(x) - cos(3x) sin(x)](sin(x) cos(x)) = (sin (3x - x))/((1/2) sin(2x)) = 2 sin(2x)/sin(2x) = 2, for every x such that sin (x) and cos (x) are diffrenet from 0.


2) (1-sin^2x)(1-tan^2x) = cos^2(x) (1- tan^2(x)) = cos^2(x) - cos^2(x) sen^2(x)/cos^2(x) = cos^(x) - sen^2(x) = cos(2x).

2007-10-31 16:32:43 · answer #3 · answered by Steiner 7 · 0 0

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