# Thread: Equating two equations - basic algebra questions re exponents and trig functions

1. ## Equating two equations - basic algebra questions re exponents and trig functions

Hi,

I have the following:
Consider the following two expressions:

f(x) = e ^(2x) − (e^(x) cos x)^ (2)/x^2 and g(x) =  (e^xsin x/x)^2

I have to show that these equations are equal for x not equal to 0.

I have a couple of basic questions:

When an expression is expressed to the power 2 (say) is it that case that to multiply a term by an exponent you add exponents?

I am given the hint: cos^(2) x + sin^(2) x (cos squared x plus sin squared x) = 1

I have fiddled around a bit however can't work out how that hint helps...does anyone have any other ideas?

Any help welcome

Kind regards
Beetle

2. ## Re: Equating two equations - basic algebra questions re exponents and trig functions

When an expression is expressed to the power 2 (say) is it that case that to multiply a term by an exponent you add exponents?
you don't "multiply" a term by an exponent ... the two basic rules are $(a^x)^y = a^{xy}$ and $a^x \cdot a^y = a^{x+y}$

Consider the following two expressions:

f(x) = [e ^(2x) − (e^(x) cos x)^ (2)]/x^2 and g(x) = (e^xsin x/x)^2

I have to show that these equations are equal for x not equal to 0.
ok ... I'm gonna make a request that you start using brackets to make your expressions clear

as you've written it ... $f(x) = e^{2x} - \dfrac{(e^x\cos{x})^2}{x^2}$

what it really is ...

$f(x) = \dfrac{e^{2x}-(e^x\cos{x})^2}{x^2}$

proceeding ...

$f(x) = \dfrac{e^{2x}-e^{2x}\cos^2{x}}{x^2}$

$f(x) = \dfrac{e^{2x}(1-\cos^2{x})}{x^2}$

can you finish?

3. ## Re: Equating two equations - basic algebra questions re exponents and trig functions

Hi,

First of all my apologies for the bracket problem. I actually had more brackets and then following feedback last week took some of them out – to many it seems. I tried typing my posts in work and then copying and pasting however that still doesn't return the correct syntax - it reverts to what is used in this forum.

Thank you for your help – there is one step in your working I don’t understand that I have pointed out – following that I can get from your last line of working to the solution.

Would you be able to explain the missing step?

f(x)=e^(2x)-(e^(2x) cos(^2) x)/x(^2)

You gave me the expression with (e^(x)cos(^2)) multiplied out right?

So f(x) = e^(2x) - e^(2x) cos^(2)x/x^(2)

I don’t understand the next step:

Being f(x)=e^(2x)(1-cos^(2)x)/x^(2)

Are you able to explain how you got this step. Thank you again.

Kind regards

4. ## Re: Equating two equations - basic algebra questions re exponents and trig functions

The numerator is like:

$\displaystyle a - ab = a(1-b)$

where$\displaystyle a=e^{2x}$ is a common factor and $\displaystyle b=cos^2x$.