## Jhulian one year ago Solve within interval [pi,2pi] -sin(squared)theta+cos(squared)theta=1-sin(theta)

1. anonymous

well.... recall that $$\bf sin^2(\theta)+cos^2(\theta)=1$$ so... if you solve that for say... $$\bf cos^2$$ what does that give you?

2. anonymous

$\cos ^2\theta-\sin ^2\theta=\1-sin \theta$ $1-\sin ^2\theta-\sin ^2\theta=1-\sin \theta$ $1-2\sin ^2\theta=1-\sin \theta$ $2 \sin ^2 \theta-\sin \theta=0$

3. Jhulian

1-sin^2 (theta)

4. Jhulian

okay...now what do we have to do next?

5. anonymous

well... @surjithayer kinda expanded it above.... lemme show you what he did

6. anonymous

$\sin \theta \left( 2\sin \theta-1 \right)=0$

7. anonymous

Parenthesis would have been nice.

8. anonymous

$$\color{blue}{\text{Originally Posted by}}$$ @surjithayer $(\color{red}{\cos ^2\theta})-\sin ^2\theta=1-\sin \theta$ $(\color{red}{1-\sin ^2\theta})-\sin ^2\theta=1-\sin \theta$ $1-2\sin ^2\theta=1-\sin \theta$ $2 \sin ^2 \theta-\sin \theta=0$ $$\color{blue}{\text{End of Quote}}$$

9. Jhulian

10. anonymous

Ok? Just solve for $$\sin(\theta)$$

11. anonymous

Don't use brown @jdoe0001 haha :\

12. anonymous

hmmm hold the mayo... the ones should cancel out

13. anonymous

hehe

14. anonymous

lemme redo it some :)

15. anonymous

$$\bf -sin^2(\theta)+cos^2(\theta)=1 \\ \quad \\ sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)={\color{blue}{ 1-sin^2(\theta) }}\qquad thus \\ \quad \\ -sin^2(\theta)+cos^2(\theta)=1\implies -sin^2(\theta)+[{\color{blue}{ 1-sin^2(\theta) }}]=1 \\ \quad \\ 1-sin^2(\theta)-sin^2(\theta)=1\implies 1-2sin^2(\theta)=1 \\ \quad \\ 0=2sin^2(\theta)\implies 0=sin^2(\theta)\implies 0=sin(\theta) \\ \quad \\ sin^{-1}(0)=sin^{-1}\left[ sin(\theta) \right]\implies sin^{-1}(0)=\theta$$ blue then =)

16. anonymous

much better $$\checkmark$$

17. Jhulian

lol...so the answers are {0 , pi , pi/6 , 5pi/6 , }...is that right?

18. anonymous

well... hmm what angles does it give you a sine of 0?

19. Jhulian

0 and pi

20. Jhulian

i am not really sure so help me out

21. anonymous

hmmm hold... shoot... just notice... missed somethjing, OR it was added after I read it :/

22. anonymous

anyhow, @surjithayer's is correct either way :)

23. Jhulian

okay...i really just wanna get to the correct answer..lol...i have only two chances left

24. anonymous

well... I just noticed it was = 1-sine.... thus

25. anonymous

which @surjithayer has correct :)

26. Jhulian

okay..bit it said wrong when i put it in...the hint says use your unit circle for final answer

27. anonymous

yeap... you should lemme see if I can redo all that, with the correct equal value quick but @surjithayer is correct, is a quadratic

28. anonymous

$$\bf -sin^2(\theta)+{\color{brown}{ cos^2(\theta)}}=1-sin(\theta) \\ \quad \\ -sin^2(\theta)+{\color{brown}{ 1-sin^2(\theta)}}=1-sin(\theta)\implies 1-2sin^2(\theta)=1-sin(\theta) \\ \quad \\ \cancel{1-1}+sin(\theta)-2sin^2(\theta)=0\implies -2sin^2(\theta)+sin(\theta)=0 \\ \quad \\ 2sin^2(\theta)-sin(\theta)=0\implies 2sin^2(\theta)-sin(\theta)+0=0$$ notice the quadratic

29. Jhulian

yeah

30. anonymous

then just solve it like you'd any quadratic then take the inverse sine of it

31. anonymous

$either \sin \theta=0=\sin n \pi$ in the given interval $\theta=\pi ,2 \pi$

32. anonymous

hmmm actually, just a quick common factor should do

33. Jhulian

i get two equations sin Theta = 0...and sin Theta = 1/2 when I used my unit circle ti find the final answers it says wrong...lol...This is my last chance

34. anonymous

$$\large { 2sin^2(\theta)-sin(\theta)=0\implies {\color{brown}{ sin(\theta)}}[2sin(\theta)-1]=0 \\ \quad \\ \begin{cases} sin(\theta)=0\to &\theta=sin^{-1}(0)\\ \quad \\ 2sin(\theta)-1=0\to sin(\theta)=\frac{1}{2}\to &\theta=sin^{-1}\left( \frac{1}{2} \right) \end{cases} }$$

35. anonymous

$or \sin \theta=\frac{ 1 }{ 2 }$ rejected because in [pi,2 pi] sin is negative.

36. Jhulian

yeah that too...okay so i will put in {pi , pi/6 and 5pi/6}...i cant get this wrong again

37. Jhulian

omg!!...it says wrong again!!! i am done...:-((

38. anonymous

only pi and 2 pi

39. Jhulian

its void now...but i would like to know how you got your answers from the unit circle

40. anonymous

|dw:1444534632816:dw|

41. Jhulian

okay what about sin (theta) = 1/2

42. anonymous

i say it is positive ,so not in [pi,2 pi] it is in [0,2pi] which you don't want.

43. Jhulian

0

44. Jhulian

lol...thank you very much @surjithayer

45. anonymous

correction$\sin\ theta$=1/2, it is in [0,pi]

46. anonymous

yw