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

1. May 6, 2012

### mutineer123

1. The problem statement, all variables and given/known data
The exact question is in question one part ii of

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s10_qp_11.pdf

2. Relevant equations

3. 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)
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. May 6, 2012

### 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 $\text{cos}(x)$.

3. May 6, 2012

### mutineer123

I dont 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 wont that just introduce more variables?

4. May 6, 2012

### Steely Dan

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.

5. May 6, 2012

### mutineer123

Im terribly sorry Mr.Dan, but I have avery thick head, and im trying, but I dont 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) !

6. May 6, 2012

### Steely Dan

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

7. May 6, 2012

8. May 6, 2012

### mutineer123

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

9. May 6, 2012

### mutineer123

10. May 6, 2012

### Mentallic

Noo...

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

11. May 7, 2012

### Staff: Mentor

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) ....

12. May 7, 2012

### mutineer123

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

13. May 7, 2012

### mutineer123

we havent learnt that yet.

14. May 8, 2012

### Mentallic

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