## anonymous one year ago What is this TRIG identity called. see attachment.

1. anonymous

2. anonymous

@dan815

3. anonymous

@LynFran

4. anonymous

@jim_thompson5910

5. jim_thompson5910

I'm not sure if there is a specific name for this identity

6. anonymous

how would I explain it though

7. jim_thompson5910

I guess you could say "composition of a trig function and another inverse" ? I'm not sure

8. jim_thompson5910

oh you want to know how they got that identity?

9. anonymous

yeah

10. anonymous

here let me give you an example

11. jim_thompson5910

tan(arccos(x)) let theta = arccos(x) so cos(theta) = x = x/1 agreed?

12. anonymous

$\tan(\cos^-1(1/2)$

13. anonymous

yes agreed

14. jim_thompson5910

so we can create this right triangle |dw:1438215201249:dw|

15. jim_thompson5910

|dw:1438215224912:dw|

16. jim_thompson5910

cos(theta) = x/1 cosine deals with adjacent over hypotenuse

17. jim_thompson5910

so we can add these labels |dw:1438215259342:dw|

18. jim_thompson5910

what is the missing side equal to (in terms of x) ?

19. anonymous

1^2-x^2

20. jim_thompson5910

it will be the square root of that

21. jim_thompson5910

|dw:1438215360088:dw|

22. anonymous

@jim_thompson5910 how do we gate the base of absolute value of x

23. jim_thompson5910

tan(arccos(x)) turns into tan(theta) this is because I let theta = arccos(x) calculating tan(theta) gives $\Large \frac{\text{opposite}}{\text{adjacent}} = \frac{\sqrt{1-x^2}}{x}$

24. jim_thompson5910

for some reason, and I don't know why, they made the denominator be |x| instead of just x. My guess is that they wanted to force tan(arccos(x)) to be positive. However, tan(arccos(x)) is only positive when x is positive

25. jim_thompson5910

So it should be $\Large \tan(\arccos(x)) = \frac{\sqrt{1-x^2}}{x}$