Trig proof

[thatstheguy9]

If tan x = t, show that sinxcosx = 1/(1+t^2)


So far:
tanx = t,
--> sinx/cosx = t
--> sinx = t*cosx


therefore sinxcosx = tcosxcosx = tcosx^2



From this point I am unsure how to proceed

 

[Deleted member 4993]

If tan x = t, show that sinxcosx = 1/(1+t^2)


So far:
tanx = t, --> sinx/cosx = t --> sinx = t*cosx

therefore sinxcosx = tcosxcosx = tcosx^2

From this point I am unsure how to proceed

I would use different sets of identities. Those being:

sec^2(x) = 1 + tan^2(x)

cos^2(x) = 1/sec^2(x)

sin^2(x) = 1 - cos^2(x)

However, I think there is a typo in your OP. Please review it.

 

[thatstheguy9]

I would use different sets of identities. Those being:

sec^2(x) = 1 + tan^2(x)

cos^2(x) = 1/sec^2(x)

sin^2(x) = 1 - cos^2(x)

However, I think there is a typo in your OP. Please review it.

Great, thank you for the response. Good spot on the typo, that'll teach me for posting while half asleep!

 

[Steven G]

Great, thank you for the response. Good spot on the typo, that'll teach me for posting while half asleep!

Can you inform the rest of us what the typo is?!

 

[Steven G]

If tan x = t, show that sinxcosx = 1/(1+t^2)


So far:
tanx = t,
--> sinx/cosx = t
--> sinx = t*cosx


therefore sinxcosx = tcosxcosx = tcosx^2



From this point I am unsure how to proceed

I like to draw a triangle as it has all the information there.
So draw a right triangle where <x is such that tanx = t=t/1. So opposite angle x put a t and adjacent to angle x to a 1. Then ask Pythagoras for help for the 3rd side of the triangle.
Armed with all sides of the triangle you just need to compute sinx and cosx and multiply them together.

 

[thatstheguy9]

Can you inform the rest of us what the typo is?!

I assume "tcosx^2" should be "tcos^2(x).

Thanks for the response, always helpful to see different approaches!

 

[Deleted member 4993]

I assume "tcosx^2" should be "tcos^2(x).

Thanks for the response, always helpful to see different approaches!

Have you "shown" the required identity in the OP to be true!

 

[pka]

If tan x = t, show that sinxcosx = 1/(1+t^2)

In the right \(\Delta ABC\), \(m(\angle C)=\frac{\pi}{2}\) so that \(m\left(\overline{BC}\right)=1,~ \&~m\left(\overline{AC}\right)=\sqrt{3}\).
Hence, \(\tan\left(\angle CBA\right)=\sqrt 3\)
So \( \sin\left(\angle CBA\right)\cdot\cos\left(\angle CBA\right)=\bf\large{?}\).

 

[Steven G]

I assume "tcosx^2" should be "tcos^2(x).

No, there was a typo in the statement of your problem ( and with your work writing tcosx^2 for tcos^2(x) ).

 

[Steven G]

In the right \(\Delta ABC\), \(m(\angle C)=\frac{\pi}{2}\) so that \(m\left(\overline{BC}\right)=1,~ \&~m\left(\overline{AC}\right)=\sqrt{3}\).
Hence, \(\tan\left(\angle CBA\right)=\sqrt 3\)
So \( \sin\left(\angle CBA\right)\cdot\cos\left(\angle CBA\right)=\bf\large{?}\).

Are you reading the same problem that I read?

 

[pka]

Are you reading the same problem that I read?

Well I always quote the post I am addressing. Is that what you are reading.
It may help you to use \(m(\angle CBA)=m(\angle x)~\&~\tan(\angle x)=\sqrt 3\)
in a right triangle with legs \(1~\&~\sqrt 3\) SEE HERE

 

[Steven G]

Well I always quote the post I am addressing. Is that what you are reading.
It may help you to use \(m(\angle CBA)=m(\angle x)~\&~\tan(\angle x)=\sqrt 3\)
in a right triangle with legs \(1~\&~\sqrt 3\) SEE HERE

Can you please show where it says in this thread that tan x = sqrt(3). On my computer it say that tan x = t

 

[thatstheguy9]

No, there was a typo in the statement of your problem ( and with your work writing tcosx^2 for tcos^2(x) ).

Does this help?

 

[pka]

Can you please show where it says in this thread that tan x = sqrt(3). On my computer it say that tan x = t

If the original statement is universally true, \(\forall t\), it must be true for \(\large t=\sqrt 3\). Well is it?
Again SEE HERE.

 

[Steven G]

If the original statement is universally true, \(\forall t\), it must be true for \(\large t=\sqrt 3\). Well is it?
Again SEE HERE.

I don't buy that for a minute. t can equal sqrt(3) but t does not equal sqrt(3).

 

[pka]

If tan x = t, show that sinxcosx = 1/(1+t^2)

I don't buy that for a minute. t can equal sqrt(3) but t does not equal sqrt(3).

Are you, Sir Jomo, really saying that if \(\tan(x)=t\) implies that \(\sin(x)\cdot\cos(x)=\dfrac{1}{1+t^2}\)
EVEN if \(\large t=\sqrt 3\) then the Super Moderators have their work cut out for them.
Need I ask, do you understand counter-examples, I certainly hope that you do given your many contractions here.
Did you read Mr, Kahn's post #7?

 

[Steven G]

Are you, Sir Jomo, really saying that if \(\tan(x)=t\) implies that \(\sin(x)\cdot\cos(x)=\dfrac{1}{1+t^2}\)
EVEN if \(\large t=\sqrt 3\) then the Super Moderators have their work cut out for them.
Need I ask, do you understand counter-examples, I certainly hope that you do given your many contractions here.
Did you read Mr, Kahn's post #7?

No sir, I am not. In my earlier post I stated that the problem was not correctly written. Just look at post #9

 

[pka]

No sir, I am not. In my earlier post I stated that the problem was not correctly written. Just look at post #9

So what is your point? What about the mathematics do you not follow?
You have wasted a good deal of my time!
The original post is incorrect if \(t=\sqrt 3\).
The O.P. should have been: if \(\large\tan(x)=t\) then \(\large \sin(x)\cdot\cos(x)=\dfrac{t}{1+t^2} \)
Can you even show that?

 

[Steven G]

Yes

So what is your point? What about the mathematics do you not follow?
You have wasted a good deal of my time!
The original post is incorrect if \(t=\sqrt 3\).
The O.P. should have been: if \(\large\tan(x)=t\) then \(\large \sin(x)\cdot\cos(x)=\dfrac{t}{1+t^2} \)
Can you even show that?

Yes, I can show that. My method is clearly stated in post #5