## msingh 2 years ago plse solve this

1. ajprincess

$\theta=\cos^{-1}\frac{-\sqrt3}{2}$ $\cos\theta=\frac{-\sqrt3}{2}$ So vat is the value of $$\theta$$? @msingh

2. ajprincess

cos60=1/2

3. ajprincess

well ya. but there is another value.

4. msingh

-cos30

5. apple_pi

why do you need to find theta? It's just cos(A+B)

6. ajprincess

hw abt 5pi/6?

7. ajprincess

i mean cos(5pi/6)

8. msingh

okay

9. ajprincess

$\cos(\frac{5\pi}{6}+\frac{\pi}{4})=?$

10. ajprincess

nw as @apple_pi mentioned u can use $$\cos(A+B)=\cos A\cos B-\sin A\sin B$$.

11. apple_pi

@ajprincess why did you find theta? You could just leave it as acos(...), then cos(acos(...)) would cancel and then use pythagorus to find sin(acos(...))

12. ajprincess

well i found it easy to find theta first nd then do..:)

13. apple_pi

ok whatever suits you :p

14. msingh

@ajprincess can u plse tell me exact answer

15. ajprincess

Well i am really sorry. I am nt supposed to give direct answers. pls check the code of conduct

16. ajprincess

|dw:1367584877456:dw|

17. ajprincess

plug in the values nd simplify

18. msingh

yes, after this, i m getting cos(5pi/6)-sin(5pi/6)

19. ajprincess

what is cos(pi/4)?

20. msingh

oh again i did wroong i place 1 as it is 1/ root2

21. ajprincess

what are cos(5pi/6),sin(pi/4) and sin(5pi/6)?

22. msingh

-root3/2 1/root2 root3/2

23. ajprincess

first two values are right. chck the last one

24. msingh

- root 3/2

25. ajprincess

nope. sin150=1/2.

26. msingh

okay

27. ajprincess

150=5pi/6

28. ajprincess

nw plug in those values and simplify

29. msingh

okay

30. msingh

i got (1/2root2)*(-root3-1)

31. ajprincess

yup right:)

32. msingh

@ajprincess thank you

33. ajprincess

welcome:)