sin, cos and tan are used in trigonometry and to calculate angles in right angle triangles
in a right angle triangle there is the hypotenuse, opposite and adjacent
sin= O/H
cos=A/H
tan=O/A
the slash being the dividing thingy, dont know wat its called :P
there are also certain formulas where u can work out angles, and length of sides, etc such as the sine rule and the cos rule
if you want a more detailed answer its best if you look it up in a maths textbook because you cant draw pictures with this wiki answer thing
in pratical applications it can be used to determine a ships bearing (location from a certain point, i.e lighthouse or dock) , inclined planes in engineering
dont worry trigonometry is very useful and theres a reason why its being taught
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
To show that (cos tan = sin) ??? Remember that tan = (sin/cos) When you substitute it for tan, cos tan = cos (sin/cos) = sin QED
'csc' = 1/sin'tan' = sin/cosSo it must follow that(cos) (csc) / (tan) = (cos) (1/sin)/(sin/cos) = (cos) (1/sin) (cos/sin) = (cos/sin)2
(sin(x)cot(x) - cos(x))/tan(x)(Multiply by tan(x)/tan(x))sin(x) - cos(x)tan(x)(tan(x) = sin(x)/cos(x))sinx - cos(x)(sin(x)/cos(x))(cos(x) cancels out)sin(x) - sin(x)0
The definition of tan(x) = sin(x)/cos(x). By this property, cos(x)tan(x) = sin(x).
The identity for tan(theta) is sin(theta)/cos(theta).
To show that (cos tan = sin) ??? Remember that tan = (sin/cos) When you substitute it for tan, cos tan = cos (sin/cos) = sin QED
tan(9) + tan(81) = sin(9)/cos(9) + sin(81)/cos(81)= {sin(9)*cos(81) + sin(81)*cos(9)} / {cos(9)*cos(81)} = 1/2*{sin(-72) + sin(90)} + 1/2*{sin(72) + sin(90)} / 1/2*{cos(-72) + cos(90)} = 1/2*{sin(-72) + 1 + sin(72) + 1} / 1/2*{cos(-72) + 0} = 2/cos(72) since sin(-72) = -sin(72), and cos(-72) = cos(72) . . . . . (A) Also tan(27) + tan(63) = sin(27)/cos(27) + sin(63)/cos(63) = {sin(27)*cos(63) + sin(63)*cos(27)} / {cos(27)*cos(63)} = 1/2*{sin(-36) + sin(90)} + 1/2*{sin(72) + sin(36)} / 1/2*{cos(-36) + cos(90)} = 1/2*{sin(-36) + 1 + sin(36) + 1} / 1/2*{cos(-36) + 0} = 2/cos(36) since sin(-36) = -sin(36), and cos(-36) = cos(36) . . . . . (B) Therefore, by (A) and (B), tan(9) - tan(27) - tan(63) + tan(81) = tan(9) + tan(81) - tan(27) - tan(63) = 2/cos(72) – 2/cos(36) = 2*{cos(36) – cos(72)} / {cos(72)*cos(36)} = 2*2*sin(54)*sin(18)/{cos(72)*cos(36)} . . . . . . . (C) But cos(72) = sin(90-72) = sin(18) so that sin(18)/cos(72) = 1 and cos(36) = sin(90-36) = sin(54) so that sin(54)/cos(36) = 1 and therefore from C, tan(9) – tan(27) – tan(63) + tan(81) = 2*2*1*1 = 4
Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(theta).
No. Tan(x)=Sin(x)/Cos(x) Sin(x)Tan(x)=Sin2(x)/Cos(x) Cos(x)Tan(x)=Sin(x)
When tan A = 815, sin A = 0.9999992 and cos A = 0.0012270 so that sin A + cos A*cos A*(1-cos A) = 1.00000075, approx.
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.
'csc' = 1/sin'tan' = sin/cosSo it must follow that(cos) (csc) / (tan) = (cos) (1/sin)/(sin/cos) = (cos) (1/sin) (cos/sin) = (cos/sin)2
A useful property in Trigonometry is: tan(x) = sin(x) / cos(x) So, cos(x) tan(x) = cos(x) [ sin(x) / cos (x)] = sin(x)
Sine sum identity: sin (x + y) = (sin x)(cos y) + (cos x)(sin y)Sine difference identity: sin (x - y) = (sin x)(cos y) - (cos x)(sin y)Cosine sum identity: cos (x + y) = (cos x)(cos y) - (sin x)(sin y)Cosine difference identity: cos (x - y) = (cos x)(cos y) + (sin x)(sin y)Tangent sum identity: tan (x + y) = [(tan x) + (tan y)]/[1 - (tan x)(tan y)]Tangent difference identity: tan (x - y) = [(tan x) - (tan y)]/[1 + (tan x)(tan y)]
(sin(x)cot(x) - cos(x))/tan(x)(Multiply by tan(x)/tan(x))sin(x) - cos(x)tan(x)(tan(x) = sin(x)/cos(x))sinx - cos(x)(sin(x)/cos(x))(cos(x) cancels out)sin(x) - sin(x)0
The definition of tan(x) = sin(x)/cos(x). By this property, cos(x)tan(x) = sin(x).
The identity for tan(theta) is sin(theta)/cos(theta).