## anonymous one year ago If sin(theta)=X, then express sec(theta) in terms of x

1. anonymous

The answer I got was $\sqrt{1-(1/x)^2}$ Would this be correct?

2. freckles

|dw:1438813890812:dw| use pythagorean theorem to find the other side

3. freckles

and recall for secant use hyp/adj

4. anonymous

I'm sorry, I'm a bit confused where to go from there

5. freckles

can you find the adjacent side by use of Pythagorean theorem ?

6. freckles

$(\text{ adjacent })^2+x^2=1^2 \\ \text{ solve for adjacent }$

7. freckles

let me know if you don't know how maybe I can give you a step

8. anonymous

9. freckles

$( \text{ adjacent } )^2+x^2=1 \\ \text{ subtract} x^2 \text{ on both sides } \\ ( \text{ adjacent })^2=1-x^2 \\ \text{ take square root of both sides } \\ \text{ adjacent } = \pm \sqrt{1-x^2}$ you cannot simplify this ... if that is what you tried to do

10. freckles

|dw:1438814521770:dw| now find sec(theta) using the right triangle

11. anonymous

The whole negative thing kinda threw me off. Thanks

12. anonymous

What I tried to do initially was rewrite the identity sin^2+Cos^2=1 and go from there

13. freckles

you can use that too

14. freckles

$\sin^2(\theta)+\cos^2(\theta)=1 \\ \text{ we are given } \sin(\theta)=x \text{ so } \sin^2(\theta)=x^2 \\ x^2+\cos^2(\theta)=1 \\ \\ \text{ subtract} x^2 \text{ on both sides } \\ \cos^2(\theta)=1-x^2 \\ \text{ take square root of both sides } \\ \cos(\theta)=\pm \sqrt{1-x^2}$ then just flip both sides to find sec(theta)

15. freckles

its the same thing really

16. anonymous

Ooh, I tried to use the recirpocals first, hence why I had (1/x)^2 Thanks a bunch mate (:

17. freckles

np

18. freckles

well csc^2(x) would actually be (1/x)^2

19. anonymous

Yeah, I figured if the identity works with sin and cos, it'd work with their reciprocals. I see the error

20. anonymous

Do you think you might be able to help me with one more question?

21. freckles

i can try

22. anonymous

Evaluate cos1+cos2+cos3+...+cos357+cos358+cos359 (all measurements are in degrees) The only thing I can imagine is that it repeats at a certain point or something

23. freckles

$\cos(x)+\cos(180-x) \\=\cos(x)+\cos(180)\cos(x)-\sin(180)\sin(x) \\ =\cos(x)-1\cos(x)-0\sin(x)\\ =0$ So I think maybe we can use cos(x)+cos(180-x)

24. freckles

So I think maybe we can use cos(x)+cos(180-x)=0*

25. freckles

$\cos(1)+\cos(179)=0 \\ \cos(2)+\cos(178)=0 \\ \cos(3)+\cos(177)=0 \\ \cos(4)+\cos(176) =0 \\ \cdots \\$ and I have to come back to finish this i'm being called away unfortunately

26. anonymous

Alright

27. freckles

notice above I put a minus but it doesn't matter that one part is still 0 $\cos(x)+\cos(180-x) \\=\cos(x)+\cos(180)\cos(x)+\sin(180)\sin(x) \\ =\cos(x)-1\cos(x)+0\sin(x)\\ =0$ hmm... $\cos(180)=-1 \\ \text{ hmmm... we need to think now for the rest of the finite series }$ like we already know cos(1)+cos(2)+...+cos(180)=-1 we need to figure out would to do with cos(181)+cos(182)+cos(183)+...+cos(359) we should be able to think of something similar

28. freckles

how about for that other sign try to see if something like cos(180+x)+cos(360-x)=0 works.. cos(180)cos(x)-sin(180)sin(x)+cos(360)cos(x)+sin(360)sin(x) =-cos(x)+cos(x)=0