plse solve this

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plse solve this

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[\theta=\cos^{-1}\frac{-\sqrt3}{2}\] \[\cos\theta=\frac{-\sqrt3}{2}\] So vat is the value of \(\theta\)? @msingh
cos60=1/2
well ya. but there is another value.

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Other answers:

-cos30
why do you need to find theta? It's just cos(A+B)
hw abt 5pi/6?
i mean cos(5pi/6)
okay
\[\cos(\frac{5\pi}{6}+\frac{\pi}{4})=?\]
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
|dw:1367584877456:dw|
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.
okay
150=5pi/6
nw plug in those values and simplify
okay
i got (1/2root2)*(-root3-1)
yup right:)
@ajprincess thank you
welcome:)

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