## chels Group Title provide that this equation is an identity. sin(x)cos(x) / tan(x) = 1 - sin^2(x) 3 years ago 3 years ago

1. lokisan Group Title

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

2. chels Group Title

i can't scroll..

3. lokisan Group Title

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

4. beens Group Title

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

5. beens Group Title

sorry...1-sin^2x

6. chels Group Title

how does sinxcosx/tanx = sinx?

7. lokisan Group Title

It doesn't. It equals cos^2x

8. chels Group Title

just kidding, i understand how you wrote it now

9. lokisan Group Title

I like how you stalked me then...lol

10. chels Group Title

haha sorry. i got confused.

11. chels Group Title

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

12. lokisan Group Title

Can you type out the question using the equation editor?

13. chels Group Title

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

14. lokisan Group Title

okay

15. chels Group Title

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

16. lokisan Group Title

Yeah...I was thinking that.

17. chels Group Title

yeahh.

18. lokisan Group Title

Okay, it's easy now...

19. lokisan Group Title

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

20. lokisan Group Title

$=2\cot^2 x$

21. lokisan Group Title

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 Group Title

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

23. lokisan Group Title

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

24. chels Group Title

okay, i understand that one

25. chels Group Title

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

26. lokisan Group Title

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 Group Title

how did you identify (1)?

28. lokisan Group Title

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 Group Title

is x. I'll draw a pic.

30. lokisan Group Title

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

31. chels Group Title

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

32. lokisan Group Title

Okay

33. chels Group Title

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

34. chels Group Title

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

35. lokisan Group Title

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

36. lokisan Group Title

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 Group Title

ok, yep.

38. lokisan Group Title

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 Group Title

got it so far..

40. lokisan Group Title

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 Group Title

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 Group Title

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

43. chels Group Title

ok

44. chels Group Title

find exact value of (-pi/12)

45. lokisan Group Title

function?

46. chels Group Title

sin(-pi/12)

47. chels Group Title

sorry

48. lokisan Group Title

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 Group Title

square root of 3 over 4?

50. lokisan Group Title

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

51. lokisan Group Title

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 Group Title

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