## anonymous 3 years ago If tanΘ= -4/3 and Θ lies in the second quadrant, then sinΘ-2cosΘ=?

1. anonymous

sinΘ-2cosΘ = (sin Θ/ cos Θ) -(2cos Θ/cosΘ) = tan Θ - 2 = -4/3 - 2 = ???

2. anonymous

why we need to divided by cosΘ??

3. hartnn

you can't just divide by cos theta, you need to multiply also.

4. hartnn

i would suggest you find sec theta first, from the identity sec^2 theta = 1+ tan^2 theta and then, cos = 1/ sec, sin = tan * cos.

5. anonymous

i have learnt sec yet, i have learnt sin cos tan only so far

6. anonymous

|dw:1365910549853:dw|

7. anonymous

there is a picture of an angle whose tangent is $$\frac{4}{3}$$

8. anonymous

you need the third side, which you get via pythagoras or by remembering the 3 - 4 - 5 right triangle

9. anonymous

|dw:1365910630356:dw|

10. anonymous

and?

11. anonymous

now you see that $$\sin(\theta)=\frac{4}{5}$$ and $$\cos(\theta)=\frac{3}{5}$$ except that since you are in quadrant II you have $$\cos(\theta)=-\frac{3}{5}$$ because in quadrant II cosine is negative

12. anonymous

you want $\sin(\theta)-2\cos(\theta)$ and you know the numbers that you need $\frac{4}{5}-2\times (-\frac{3}{5})=\frac{4}{5}+\frac{6}{5}$

13. anonymous

rather easy if you draw a triangle otherwise a pita

14. anonymous

2?

15. hartnn

Alternate way to look at the same thing, |dw:1365910821672:dw| so the angle inside triangle is $$\pi - \theta$$ so, $$\cos(\pi-\theta)=3/5, -\cos \theta=3/5 , \cos \theta = -3/5$$

16. anonymous

yeah,2

17. anonymous

thank you both of you

18. anonymous

and the next questions is : [1/cosθ + tanθ](1-sinθ)=?

19. hartnn

write tan = sin/cos.

20. anonymous

i got the answer -cosθ but there are only five choices: A. sinθ B. cosθ C. cos^(2)θ D. 1+sinθ E. sinθtanθ

21. hartnn

how you got -cos theta ?

22. hartnn

more specifically, how you got that - *minus*

23. anonymous

$(\frac{1}{a}+\frac{b}{a})(1-b)=\frac{1-b^2}{a}$

24. anonymous

$\large \tan \theta=-\frac{4}{3}$ $\large \theta =12652'$ $\sin \theta -2cos\theta=\sin 12652' -2\cos126`52'$ $\large =2$

25. anonymous

in this case $$a=\cos(\theta)$$ and $$b=\sin(\theta)$$ so you get $\frac{1-\sin^2(\theta)}{\cos(\theta)}$ since $$1-\sin^2(\theta)=\cos^2(\theta)$$ you end up with $\frac{\cos^2(\theta)}{\cos(\theta)}=\cos(\theta)$

26. anonymous

|dw:1365911248263:dw|

27. anonymous

ummm, may be i got something wrong

28. anonymous

the algebra is easier if you put $$\cos(\theta)=a$$ and $$\sin(\theta)=b$$

29. anonymous

then you won't get so confused when you multiply stuff out almost all the steps are algebra, there is very little trig here

30. anonymous

$(\frac{1}{\cos(\theta)}+\tan(\theta))(1-\sin(\theta))$ becomes the more simple to compute $(\frac{1}{a}+\frac{b}{a})(1-b)$

31. anonymous

or if you prefer $(\frac{1+b}{a})(1-b)$

32. anonymous

now you get $$\frac{1-b^2}{a}$$ almost instantly

33. anonymous

ohhh, i write sin^(2)θ-1 ....

34. anonymous

actually $$1-\sin^2(\theta)$$ for the numerator

35. anonymous

36. hartnn

yes^

37. hartnn

$$(a+b)(a-b)=a^2-b^2$$

38. anonymous

39. hartnn

yes, absolutely.

40. anonymous

and the next question: If cosθ=1/k and 0°≤θ≤90°, then -tan(90°-θ)=?

41. hartnn

|dw:1365911889934:dw| find the unknown side by using pythagoras theorem ?

42. anonymous

k^(2)-1

43. hartnn

yes.

44. hartnn

|dw:1365911994361:dw| now find (tan 90-theta) = opposite side / adjacent side

45. hartnn

sorry, that should read $$\large k^2-1$$

46. anonymous

tan(90-theta)=-1/tan theta?

47. hartnn

you can use that identity , its just another way to solve the same problem.

48. hartnn

you'll get the same answer using any of the 2 methods.

49. anonymous

$-\frac{ 1 }{ \sqrt{k ^{2}}-1 }$ ??

50. hartnn

if you meant , $$\Large -\frac{ 1 }{ \sqrt{k ^{2}-1} }$$ then you are correct.

51. anonymous

thanks

52. hartnn

welcome ^_^