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# Can anyone help me with C2 maths on raidans?

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1. Find in radians correct to 2 decimal places, the two smallest positive values of θ for which sin(2θ +1/3π) = 0.123. I've worked out 2.68 but I figure out how to get to the other answer.
If you could help that would be great.
2. Hi, these are the steps I used:
(I'll have to use 'a' for theta as I cannot get that symbol)

sin(2a+1/3pi)=0.123
2a+1/3pi=sin^-1(0.123)
2a+1/3pi=0.123
2a=-0.925
a=-0.462

You then use the symmetry of the sin(a) graph:
(-pi)+0.462= -2.68

I hope two things: that's correct and that it helps you if it is. I cannot see how you got +2.68 (unless there is a typo in the original message or my working out)
3. (Original post by Jasminea)
Find in radians correct to 2 decimal places, the two smallest positive values of θ for which sin(2θ +1/3π) = 0.123. I've worked out 2.68 but I figure out how to get to the other answer.
If you could help that would be great.
If you call and then solve then we have and then .

Back-sub: (solve for theta) for the first solution and for the second solution.
4. (Original post by ErraticPhysicist)
Hi, these are the steps I used:
(I'll have to use 'a' for theta as I cannot get that symbol)

sin(2a+1/3pi)=0.123
2a+1/3pi=sin^-1(0.123)
2a+1/3pi=0.123
2a=-0.925
a=-0.462

You then use the symmetry of the sin(a) graph:
(-pi)+0.462= -2.68

I hope two things: that's correct and that it helps you if it is. I cannot see how you got +2.68 (unless there is a typo in the original message or my working out)
It asks for positive values.
5. (Original post by Zacken)
It asks for positive values.
Ah yes, I see. Your answer looks correct, well done
6. (Original post by Zacken)
If you call and then solve then we have and then .

Back-sub: (solve for theta) for the first solution and for the second solution.
Ok I can solve it but how did you get 2Pi? Other than that, thanks for help.
7. (Original post by Jasminea)
Ok I can solve it but how did you get 2Pi? Other than that, thanks for help.
Well, trigonometric functions are two-pi periodic, i.e: they repeat themselves after every two pi radians.

So if x is a solution to a trig equation, then so is x + 2pi, so is x + 2pi + 2pi, so is x + 2pi + 2pi + 2pi, etc... because sin (x) = sin (x + 2pi) since the function just repeats itself after 2pi, so if I add 2pi, it changes nothing.
8. Ok. Yeah that makes sense thanks.
9. (Original post by Jasminea)
Ok. Yeah that makes sense thanks.
No problem.
10. (Original post by Zacken)
No problem.
Hi, did u ever get 0.99 as a value, bcoz that's the other answer in the back of my textbook? I can't get that answer. It may be wrong.
11. (Original post by Jasminea)
Hi, did u ever get 0.99 as a value, bcoz that's the other answer in the back of my textbook? I can't get that answer. It may be wrong.
It's nor a mistake.

.

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