The definition of tan(x) = sin(x)/cos(x). By this property, cos(x)tan(x) = sin(x).
(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
You use the identity sin2x + cos2x = 1 (to simplify the expression in parentheses), and convert all functions to sines and cosines. sec x tan x (1 - sin2x) = (1/cos x) (sin x / cos x) (cos2x) = (sin x / cos2x) cos2x = sin x
If y = sin(cos(tan(x))) Using the chain rule: (f(g(x)))' = f'(g(x)).g'(x) Then dy/dx = cos(cos(tan(x))).-sin(tan(x)).sec2(x) = -cos(cos(tan(x))).sin(tan(x)).sec2(x) Unfortunately I don't think this can be simplified much more. ( sec = 1/cos )
(sin x + cos x) / cosx = sin x / cos x + cosx / cos x = tan x + 1
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
The definition of tan(x) = sin(x)/cos(x). By this property, cos(x)tan(x) = sin(x).
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)
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)
(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
You use the identity sin2x + cos2x = 1 (to simplify the expression in parentheses), and convert all functions to sines and cosines. sec x tan x (1 - sin2x) = (1/cos x) (sin x / cos x) (cos2x) = (sin x / cos2x) cos2x = sin x
Start on the left-hand side. cos(x) + tan(x)sin(x) Put tan(x) in terms of sin(x) and cos(x). cos(x) + [sin(x)/cos(x)]sin(x) Multiply. cos(x) + sin2(x)/cos(x) Make the denominators equal. cos2(x)/cos(x) + sin2(x)/cos(x) Add. [cos2(x) + sin2(x)]/cos(x) Use the Pythagorean Theorem to simplify. 1/cos(x) Since 1/cos(x) is the same as sec(x)- the right-hand side- the proof is complete.
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)]
If y = sin(cos(tan(x))) Using the chain rule: (f(g(x)))' = f'(g(x)).g'(x) Then dy/dx = cos(cos(tan(x))).-sin(tan(x)).sec2(x) = -cos(cos(tan(x))).sin(tan(x)).sec2(x) Unfortunately I don't think this can be simplified much more. ( sec = 1/cos )
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.
(sin x + cos x) / cosx = sin x / cos x + cosx / cos x = tan x + 1