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\[\theta=\cos^{-1}\frac{-\sqrt3}{2}\] \[\cos\theta=\frac{-\sqrt3}{2}\] So vat is the value of \(\theta\)? @msingh
well ya. but there is another value.

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Other answers:

why do you need to find theta? It's just cos(A+B)
hw abt 5pi/6?
i mean cos(5pi/6)
nw as @apple_pi mentioned u can use \(\cos(A+B)=\cos A\cos B-\sin A\sin B\).
@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(...))
well i found it easy to find theta first nd then do..:)
ok whatever suits you :p
@ajprincess can u plse tell me exact answer
Well i am really sorry. I am nt supposed to give direct answers. pls check the code of conduct
plug in the values nd simplify
yes, after this, i m getting cos(5pi/6)-sin(5pi/6)
what is cos(pi/4)?
oh again i did wroong i place 1 as it is 1/ root2
what are cos(5pi/6),sin(pi/4) and sin(5pi/6)?
-root3/2 1/root2 root3/2
first two values are right. chck the last one
- root 3/2
nope. sin150=1/2.
nw plug in those values and simplify
i got (1/2root2)*(-root3-1)
yup right:)
@ajprincess thank you

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