sarihmanlapaz 2 years ago If cosA= -15/17 and A is in quadrant III, find cos 1/2A

1. AravindG

ya you have the value of cos A try to use the formula of cos 2A to get to the answer!!

2. AravindG

@sarihmanlapaz did you try that?

3. sarihmanlapaz

what will i use in the 3 formula of cos2a?

4. AravindG

just look at the equations and look at what we are trying to get the answer in terms of

5. AravindG

we want cos A/2 isnt it?

6. AravindG

and we have cos A , now you tell me which equation to use!!

7. sarihmanlapaz

i think cos^2 - 1

8. AravindG

$2\cos^2 x-1$

9. sarihmanlapaz

so how will i substitute it to that equation?

10. hartnn

cosA=2cos^2(A/2)-1 find cos(A/2) from here.

11. AravindG

@sarihmanlapaz gt it?

12. sarihmanlapaz

got it! thanks :)

13. sarihmanlapaz

but its none of the choices. >.< how can i solve it?

14. hartnn

what did u get ?

15. sarihmanlapaz

0.2435

16. hartnn

i m getting -0.2435, is that the option ?

17. sarihmanlapaz

oh no. haha my answer 0.2435 is in the options. i did'nt notice it.

18. hartnn

but A/2 will be in 2nd quadrant where cos is negative. so we put minus -0.2435

19. sarihmanlapaz

its written in the question that the A is in quadrant 3

20. hartnn

yes, A in 3rd,A/2 in 2nd. is -0.2435 in options?

21. sarihmanlapaz

the options are. A).0.24054 B).0.24125 C) 0.24254 D) 0.24354

22. sarihmanlapaz

pls can you help me?

23. hartnn

since all options are positive, i guess A/2 goes in 1st quadrant...then 0.2435 is correct.

24. sarihmanlapaz

no. its wrong my friend. i just check the half angle formula. haha there is a formula for cosA/2 = squareroot of 1/2(1+A)

25. hartnn

u mean ....(1+cos A) ?

26. sarihmanlapaz

yes. yes :) i got the right answer 0.2425

27. hartnn

how? the formula we gave and what u wrote are exactly the same...

28. sarihmanlapaz
29. sarihmanlapaz

there is the formula for half angle :)

30. hartnn

so using that u got different answer? we gave u that formula only..... cosA=2cos^2(A/2)-1 so cos(A/2)=sqrt{0.5*(1+cos A)}

31. sarihmanlapaz

yeah. thanks i got the right answer.