If tan x = k(where k is a positive constant),How is tan( pie/2 − x)=1/k

[mutineer123]

Homework Statement

The exact question is in question one part ii of

http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s10_qp_11.pdf

Homework Equations

The Attempt at a Solution

For the first one I just pluged in a value for x, and got -k. But I got stuck a two which tells me my approach is perhaps not entirely what cie is looking for, is there a better way of getting the answer?(like drawing quadrants or something)

Homework Statement

Homework Equations

The Attempt at a Solution

 

[Steely Dan]

The standard way would be to simply write the tangent function as the ratio of sine and cosine, and then use the relationship between those two to rewrite, say, \text{sin}(\frac{\pi}{2}-x) in terms of \text{cos}(x).

 

[mutineer123]

The standard way would be to simply write the tangent function as the ratio of sine and cosine, and then use the relationship between those two to rewrite, say, \text{sin}(\frac{\pi}{2}-x) in terms of \text{cos}(x).

I don't quite get you, although I know tanx=sinx/cosx I am not sure how I express ∏/2 -x, in terms of sin and cos. Do you mean, I should write sinx/tanx= K and then proceed? But won't that just introduce more variables?

 

[Steely Dan]

Do you mean, I should write sinx/tanx= K and then proceed? But won't that just introduce more variables?

I mean you should write \text{tan}(\frac{\pi}{2}-x) = \frac{\text{sin}(\frac{\pi}{2}-x)}{\text{cos}(\frac{\pi}{2}-x)} and continue from there, simplifying the numerator and denominator.

 

[mutineer123]

I mean you should write \text{tan}(\frac{\pi}{2}-x) = \frac{\text{sin}(\frac{\pi}{2}-x)}{\text{cos}(\frac{\pi}{2}-x)} and continue from there, simplifying the numerator and denominator.

I am terribly sorry Mr.Dan, but I have avery thick head, and I am trying, but I don't seem to understand, I did as you said, I got sin(∏-2x)/2 divided by cos(∏-2x). But that just leads me back to sin(∏-2x)/cos(∏-2x) !

 

[Steely Dan]

Im terribly sorry Mr.Dan, but I have avery thick head, and I am trying, but I don't seem to understand, I did as you said, I got sin(∏-2x)/2 divided by cos(∏-2x). But that just leads me back to sin(∏-2x)/cos(∏-2x) !

It's actually simpler than that. I'm not talking about, say, the double angle formula, I'm talking about how, for example \text{sin}(\frac{\pi}{2}-x) = \text{cos}(x).

 

[Bohrok]

It's a good idea to be familiar with the identities labeled as co-function identities:
http://www.sosmath.com/trig/Trig5/trig5/trig5.html

 

[mutineer123]

It's actually simpler than that. I'm not talking about, say, the double angle formula, I'm talking about how, for example \text{sin}(\frac{\pi}{2}-x) = \text{cos}(x).

Alright, okay. Thanks a lot, I thnk I am gettin it. Thanks a tonne

 

[mutineer123]

It's a good idea to be familiar with the identities labeled as co-function identities:
http://www.sosmath.com/trig/Trig5/trig5/trig5.html

you're right, thnx for the sheet, really helpful!

 

[Mentallic]

I did as you said, I got sin(∏-2x)/2

Noo...

\sin\left(A+B\right)\neq \frac{\sin\left(2A+2B\right)}{2}

 

[NascentOxygen]

got stuck a two which tells me my approach is perhaps not entirely what cie is looking for, is there a better way of getting the answer?

Hi mutineer123! Draw a right-angled triangle, and mark one of the acute angles x. Label the lengths of the two sides appropriate to this question such that the ratio of these sides = k.

The remaining angle in the triangle is (π/2 -x) ...

 

[mutineer123]

Hi mutineer123! Draw a right-angled triangle, and mark one of the acute angles x. Label the lengths of the two sides appropriate to this question such that the ratio of these sides = k.

The remaining angle in the triangle is (π/2 -x) ...

Holy CRAP! I did not think of a right angled triangle! It made it so much easier! THANKS!

 

[mutineer123]

Noo...

\sin\left(A+B\right)\neq \frac{\sin\left(2A+2B\right)}{2}

we haven't learned that yet.

 

[Mentallic]

we haven't learned that yet.

Ok, but that looks to be essentially what you've done earlier:

Im terribly sorry Mr.Dan, but I have avery thick head, and I am trying, but I don't seem to understand, I did as you said, I got sin(∏-2x)/2 divided by cos(∏-2x). But that just leads me back to sin(∏-2x)/cos(∏-2x) !

You either meant \sin\left(\frac{\pi}{2}-x\right)=\frac{\sin\left(\pi-2x\right)}{2} or \frac{\sin\left(\frac{\pi}{2}-x\right)}{\cos\left(\frac{\pi}{2}-x\right)}=\frac{\sin\left(\frac{\pi-2x}{2}\right)}{\cos\left(\frac{\pi-2x}{2}\right)}=\frac{\sin\left(\pi-2x\right)}{\cos\left(\pi-2x\right)}

Which are both incorrect!