i hate trig?

Question

write sin in terms of tan starting with sin^2+cos^2=1

Answer 1

check in your book for trig stuff, most likely on the inside cover of your math book. i can't remember exactly how to do it but use what you already know to break it down and simplify things.

Answer 2

You need to use some trig identities like tan(x) = sin(x)/cos(x) so if you solve for cos and replace in your equation you get

sin^2 + sin^2/tan^2 = 1 or

sin^2(1 + 1/tan^2) = 1 or

sin^2 = 1/(1 + 1/tan^2) so

sin = sqrt( 1/ (1 + 1/ tan^2))

Answer 3

(sin x)^2 = 1 - (cos x)^2
By identity 1 + (tan x)^2 = (sec x)^2 = 1/(cos x)^2, or
(cos x)^2 = 1/{1 + (tan x)^2}
Then substituting
(sin x)^2 = 1 - 1/{1 + (tan x)^2}
sin x = +-sqr[1 - 1/{1 + (tan x)^2}]

Answer 4

tan^2+1=1/cos^2 because tan=sin/cos so divide through by cos^2

Answer 5

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

1/(tan(x) )^2
= [1-(sin(x))^2)]/(sin(x))^2
= 1/(sin(x))^2 -1

Solve for sin(x),
sin(x) = ± tan(x)/[1+(tan(x) )^2 ]^(1/2)
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Don't miss "±" . In quadrants 2 and 3, you have to use the negative sign.

Answer 6

tan = sin/cos which gives:
cos = sin/tan

sin^2 + (sin/tan)^2 = 1
sin^2 (1 + 1/tan^2) = 1
sin^2 [(tan^2 + 1) / tan^2] = 1
sin^2 = tan^2 / (tan^2 + 1)
sin = tan / [square root (tan^2 + 1)]

Answer 7

cos ( csc^(a million)(x) ) to clean up this, enable t = csc^(-a million)(x). Taking the csc of the two factors, csc(t) = x. From right here, use SOHCAHTOA. bear in ideas that sin = opp/hyp, cos = adj/hyp, tan = opp/adj. It for this reason follows that csc = hyp/opp, sec = hyp/adj, cot = adj/opp. csc(t) = x/a million = hyp/opp. hyp = x opp = a million, so by using Pythagoras, adj = sqrt( hyp^2 - opp^2 ) = sqrt(x^2 - a million). for this reason, cos(t) = cos( csc^(-a million)(x) ) = adj/hyp = sqrt(x^2 - a million)/x

Answer 8

look at your other post! i just did it! :)