Question

Help with trigonometry:

Given the equations:
a = cot(x) + cos(x)
b = cot(x) - cos(x)

Cancel x to find an expression which relates "a" and "b".

We´re supposed to use trigonometric identities in order to cancel x

Any idea?. Thanks.

Answer 1

Since majority of the choices contain (a^2-b^2)^2
first try to find that.
whats a^2-b^2= ... ?
use : \(a^2-b^2=(a+b)(a-b)\)

Answer 2

Well, a+b = 2cot(x)
And a-b = 2cos(x)
Then a^2-b^2 = 4cot(x)cos(x)

How do we use that?

Answer 3

yes,
so, (a^2-b^2)^2 = 16 cot^2 x cos^2 x
remember this.
we can now try to find ab= ..... ?

Answer 4

Well,
ab = [cot(x)+cos(x)][cot(x)-cos(x)]
cot^2(x) - cos^2(x)

Answer 5

yes, good going. :)
ab = cot^2 x- cos^2 x
now convert cot in terms of sin and cos,
then you will be able to see a simplification step .....

Answer 6

So

\[\cot^2(x)-\cos^2(x)= \frac{ \cos^2x }{ \sin^2x }-\cos^2x\]
\[=\frac{ \cos^2x-\sin^2xcos^2x}{ \sin^2x}\]
\[=\frac{ \cos^2x(1-\sin^2x) }{ \sin^2x }\]

I think 1-sin^2x is equivalent to ohter expression right?

Answer 7

good!
to know whats 1-sin^2,
use
\(\Large \color{red}{\sin^2x+\cos^2x=1}\)

Answer 8

also you see cos^2x/ sin^2 x
what will it simplify to ?

Answer 9

Fine, then we have
\[\cot^2x-\cos^2 = \tan^2xcos^2x\]
\[= \sin^2x\]

So ab = sin^2x
Right?,

Answer 10

you are correct about cos^2 x
but tan^2x ??? :O
whats cos /sin ?
then whats cos^2/sin^2 =.... ?

Answer 11

Oops

\[ab= \cot^2xcos^2x\]
=\[\frac{ \cos^4x }{ \sin^2x }\]

Answer 12

keep ab as cot^2 x cos^2 x
(a^2-b^2)^2 was 16cot^2 x cos^2 x
now isn't the relation between them obvious ?

Answer 13

Great,

16ab = (a^2-b^2)^2

Thanks!

Answer 14

welcome ^_^