## chels 4 years ago provide that this equation is an identity. sin(x)cos(x) / tan(x) = 1 - sin^2(x)

1. lokisan

$\frac{\sin x \cos x}{\tan x } =\frac{\sin x \cos x}{\frac{\sin x }{\cos x}}=\frac{\cos x}{\frac{1}{\cos x}}=\cos^2 x = 1-\sin^2 x$

2. chels

i can't scroll..

3. lokisan

Try refreshing the page, or log out and log back in.

4. beens

sinx cosx/tanx=sinx cosx cosx/sinx =cosxcosx =cos^2x =1-sin^x

5. beens

sorry...1-sin^2x

6. chels

how does sinxcosx/tanx = sinx?

7. lokisan

It doesn't. It equals cos^2x

8. chels

just kidding, i understand how you wrote it now

9. lokisan

I like how you stalked me then...lol

10. chels

haha sorry. i got confused.

11. chels

1 / sec(x) - 1 minus 1 / sec(x) + 1 = 2cot^2(x)

12. lokisan

Can you type out the question using the equation editor?

13. chels

$1\div \sec(x-1) - 1 \div \sec(x+1) = 2\cot^2x$

14. lokisan

okay

15. chels

in the denominator it's supposed to be sec(x) - 1 and sec(x) + 1

16. lokisan

Yeah...I was thinking that.

17. chels

yeahh.

18. lokisan

Okay, it's easy now...

19. lokisan

$\frac{1}{\sec x -1}-\frac{1}{\sec x +1}=\frac{\sec x +1 - (\sec x -1)}{\sec^2 x-1}=\frac{2}{\tan^2 x}$

20. lokisan

$=2\cot^2 x$

21. lokisan

Afte$1+\tan^2 x=\sec^2 x \rightarrow 1-\sec^2x = \tan^2 x$r putting it over a common denominator, I use the fact that

22. lokisan

Then cot^2(x) is just from the definition, $\cot x = \frac{1}{\tan x}$

23. lokisan

Once you're done, let me know if it works for you or not.

24. chels

okay, i understand that one

25. chels

$\cos(\pi/2 - x)\cos(-x) = \sin(2x) \div 2$

26. lokisan

Okay, you have to identify a few things here. (1)$\cos(\frac{\pi}{2}-x)=\sin x$(2)$\cos(-x)=\cos x$Then the left-hand side becomes,$\cos(\frac{\pi}{2}-x)\cos(-x)=\sin x \cos x$If you multiply both sides by 2, you get$2\sin x \cos x = \sin 2x$which is true (since this is the expansion of sin(2x)).

27. chels

how did you identify (1)?

28. lokisan

If you want to see why cos(pi/2 - x) = sin(x), draw coordinate axes, the unit circle, and swing the radial arm from 0 to pi (180 degrees) and then reverse a bit (that bit is equal to x). The angle in the triangle ...I'm doing that now...

29. lokisan

is x. I'll draw a pic.

30. lokisan

Actually, instead of a geometrical explanation (which will take forever online) do you know your double angle formulae?

31. chels

yeah, i get it now. i just had to go through it all

32. lokisan

Okay

33. chels

$\tan(x \div 2) \cos^2(x) - \tan(x \div 2) = \sin(x) \div \sec(x) minus \sin (x)$

34. chels

sin(x) is subtracted from sin(x) / sec(x) in the second equation.

35. lokisan

Okay, sorry, had to do something. Here we go...

36. lokisan

Recognize, on the left-hand side, you can factor as$\tan^2\frac{x}{2}(\cos^2 x-1)=\tan^2 x (-\sin^2 x)$

37. chels

ok, yep.

38. lokisan

Also, on the right-hand side, $\frac{\sin x }{\sec x}-\sin x = \sin x (\cos x - 1)$We have then$\tan^2 \frac{x}{2}(-\sin^2 x)=\sin x (\cos x -1)$

39. chels

got it so far..

40. lokisan

Divide both sides by (-sin^2(x)) to get$\tan^2\frac{x}{2}=\frac{1-\cos x}{\sin x}=\frac{1-(\cos^2 \frac{x}{2}- \sin ^2 \frac{x}{2})}{2\sin \frac{x}{2}\cos \frac{x}{2}}$

41. lokisan

But, in the numerator, 1-cos^2 (x/2) = sin^2(x/2). So after expanding the numerator and doing this, we have$\tan \frac{x}{2}=\frac{2\sin ^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$which is, tan(x/2) by definition.

42. lokisan

Forget the tan^2(x/2) thing I started with - transcribing notes to this site leads to screw-ups.

43. chels

ok

44. chels

find exact value of (-pi/12)

45. lokisan

function?

46. chels

sin(-pi/12)

47. chels

sorry

48. lokisan

Use the fact that 2 x pi/12 = pi/6 which is 30 degrees. We know sin(-30 deg) = -sin(30 deg) is. You can use the double angle formula to link the two.

49. chels

square root of 3 over 4?

50. lokisan

Do you want to try that and see how you go. I need a shower.

51. lokisan

The sine of 30 degrees is 1/2. Were you ever shown how to construct the basic angles from triangles? You can always work out 30, 45 and 60, and various combinations, using two triangles. One is a right-angled one with sides 1, 1 and hypotenuse sqrt(2). The angles are 90, 45 and 45. You can construct the 30 and 60 from taking an equilateral triangle of side length 2. Then cut it in half. You'll two right-angled triangles. The angles in each will be 30, 60 and 90. The sides will be 1, sqrt(3) and 2 (2 is the hypotenuse).

52. lokisan

53. lokisan

chels, you can do this with the double angle formula for sine:$\sin(-\frac{\pi}{12})=\sin(\frac{\pi}{6}-\frac{\pi}{4})$$=\sin \pi/6 \cos \pi/4 - \cos \pi/6 \sin \pi/4$$=\frac{1}{2}\frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{2}\frac{1}{\sqrt{2}}=\frac{1-\sqrt{3}}{2\sqrt{2}}=\frac{\sqrt{2}-\sqrt{6}}{4}$

54. lokisan

You could also use the combination,$\frac{\pi}{4}-\frac{\pi}{3}$