Question

tan(x) is a linear function ?

Answer 1

ya!

Answer 2

We first compute f '(0)

f '(x) = sec 2 x

f '(0) = sec 2 (0) = 1

Hence the linear approximation fl(x) is given by

fl(x) = f(0) + f '(0) (x - 0) = x

The above result means that tan x ≈ x for x close to 0 when x is in RADIANS.

Put your calculator to RADIANS and calculate tan x for the following values of x.

x = 0 , x = 0.001 , x = 0.01, x = 0.1, x = 0.2, x = 0.3 and x = 0.5

Note compare tan x and x. Conclusion.

Answer 3

analyze math.com

Answer 4

check out for these two properties.
http://en.wikipedia.org/wiki/Linear_function

Answer 5

linear - linelike

Answer 6

@Rohangrr you just linearized tan(x) at zero.
by the same logic sin(x) will also become linear function because
f(x)=sin(x)
f'(x)=cos(x)
f(0)=0
f'(0)=1
fl(x) = f(0) + f '(0) (x - 0) =
=x
while sin(x) is not a linear function .

Answer 7

tan x is also not a linear function

Answer 8

@akash123 i answered the same yesterday in my Quiz and got 0 :(
since
\[\sin(x) =x-\frac{x^3}{3!}+\frac{x^5}{5!}+...\]
so it is non linear, how to do the same for tangent and show that it is non linear.?

Answer 9

series expansion for tan x is x+x^3/3 + 2x^5/15+....
clearly its non-linear,
but for small angles, it can be approximated as x as already proved.
bus since the question didn't mention that, i think the answer should be non-linear...

Answer 10

@hartnn thanks for your reply. but was wondering
since
\[\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}+...\]
\[\cos(x)=1-\frac{x^2}{2!} +\frac{x^4}{4!}+...\]
since
\[\tan(x)=\frac{\sin(x)}{\cos(x)}\]
by dividing both series wouldn't we get linear series for tangent ??

Answer 11

@mukushla

Answer 12

tan(ax+by) = a tan(x) + b tan(y) where a and b are two constants ... if this property is satisfied by any function then it's called linear function. Tan does not satisfy these relations so this is not linear function.

Answer 13

by dividing both series wouldn't we get linear series for tangent ?? No. It is complicated to do this division --- it is not term by term.

the easiest way to show it is not linear is to show a counter example. You should know that
tan(30º)= 1/sqrt(3)
tan(60º)= sqrt(3)
tan(90º)= undefined
if linear, tan(30º+60º) should = tan(30º)+tan(60º)
or
tan(90º)= tan(30º)+tan(60º) ?
It isn't, so tan is not linear.

Answer 14

ya, i cannot think of any traditional way to divide those infinite polynomials... and maybe u have seen this actual graph of tan x, which shows it goes to infinity at +/- npi/2

Answer 15

by dividing both series (sin x and cos x) wouldn't we get linear series for tangent ??

Yes...we can get the tan series by long division. but it wont be a linear..
tanx= x+ x^3/3+ 2/15 x^5+.....

but we can approximate it for small x so that higher order terms can be neglected
so tanx ~ x.
hence for small angles(in radian) tan is linear.(approximation only)