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They are sine, cosine and tangent, three trigonometric functions. There is a mnemonic device to help you remember what these functions represent. Imagine a right triangle (a triangle containing a 90 degree or right angle). We are interested in the two angles that are NOT 90 degrees. When you imagine these angles you can see that on one side will be the hypotenuse, the long side opposite the right angle. The other side of the angle is the adjacent leg for that angle. So either of these angles is made up of the hypotenuse and its adjacent leg. The other side is the opposite leg.
Now imagine that a space alien named Soh-cah-toa is teaching you trigonometry.
SOH means that the sine is calculated by "opposite over hypotenuse"; the length of the opposite leg divided by the length of the hypotenuse".
CAH means that the cosine is calculated by "adjacent over hypotenuse"; the length of the adjacent leg divided by the length of the hypotenuse".
TOA means that the tangent is calculated by "opposite over adjacent"; the length of the opposite leg divided by the length of the adjacent leg.
If youd like a simpler method, check out these articles for a simple free tool and tutorial that will make trig simple enough for ANYBODY to understand!
http://www.ehow.com/how_5520340_memorize-trig-functions-losing-mind.html
http://www.ehow.com/how_5227490_pass-mind-part-unknown-sides.html
http://www.ehow.com/how_5428511_pass-part-ii-unknown-angles.html
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Sin x Tan x equals Sin x?
No. Tan(x)=Sin(x)/Cos(x) Sin(x)Tan(x)=Sin2(x)/Cos(x) Cos(x)Tan(x)=Sin(x)
How do you prove sin x tan x equals cos x?
You can't. tan x = sin x/cos x So sin x tan x = sin x (sin x/cos x) = sin^2 x/cos x.
How do you solve the following identity sec x - cos x equals sin x tan x?
sec x - cos x = (sin x)(tan x) 1/cos x - cos x = Cofunction Identity, sec x = 1/cos x. (1-cos^2 x)/cos x = Subtract the fractions. (sin^2 x)/cos x = Pythagorean Identity, 1-cos^2 x = sin^2 x. sin x (sin x)/(cos x) = Factor out sin x. (sin x)(tan x) = (sin x)(tan x) Cofunction Identity, (sin x)/(cos x) = tan x.
How do you prove tan x plus tan x sec 2x equals tan 2x?
tan x + (tan x)(sec 2x) = tan 2x work dependently on the left sidetan x + (tan x)(sec 2x); factor out tan x= tan x(1 + sec 2x); sec 2x = 1/cos 2x= tan x(1 + 1/cos 2x); LCD = cos 2x= tan x[cos 2x + 1)/cos 2x]; tan x = sin x/cos x and cos 2x = 1 - 2 sin2 x= (sin x/cos x)[(1 - 2sin2 x + 1)/cos 2x]= (sin x/cos x)[2(1 - sin2 x)/cos 2x]; 1 - sin2 x = cos2 x= (sin x/cos x)[2cos2 x)/cos 2x]; simplify cos x= (2sin x cos x)/cos 2x; 2 sinx cos x = sin 2x= sin 2x/cos 2x= tan 2x
Sec x times sin x divided by tan x?
1 (sec x)(sin x /tan x = (1/cos x)(sin x)/tan x = (sin x/cos x)/tan x) = tan x/tan x = 1
