Question

Find an exact value.

sine (-11pi/12)

Answer 1

is it\[(-\sqrt{6}-\sqrt{2})\div4\]

Answer 2

hmmm no exactly...

Answer 3

it's just a angle sum

Answer 4

let's take a peek at the rational only

-11/12 how can we split it in say, 2 fractions and have the result be -11/12

Answer 5

anyhow... so I used.. 9/12 - 20/12

Answer 6

pretty simple thus far

Answer 7

Your answer is almost correct, I think you just have a sign error in there somewhere.

Answer 8

\[\sqrt{2}-\sqrt{6}/4\]

Answer 9

is that it?

Answer 10

dunno

Answer 11

what you need is => http://hyperphysics.phy-astr.gsu.edu/hbase/imgmth/trid1.gif <-- the sum identities

Answer 12

so if I were to used 9/12 - 20/12 = -11/12
well then that leaves me with \(\bf sin\left(-\cfrac{11\pi}{12}\right)\\\quad \\
\cfrac{9\pi}{12} - \cfrac{20\pi}{12} \implies \cfrac{3\pi}{4} - \cfrac{5\pi}{3}\quad \\\quad \\
sin\left(-\cfrac{11\pi}{12}\right) \implies sin\left(\cfrac{3\pi}{4} - \cfrac{5\pi}{3}\right)\)

Answer 13

and those are the angles you can use to add up, using the trig sum identities

Answer 14

do you see the sum identity you need in the picture?

Answer 15

what the answer though

Answer 16

the answer is, whatever the sum gives you
you get the values for the angles from the Unit Circle

Answer 17

keep in mind the idea behind exercises is not to transcribe answers
is to do them so one can understand

it'd be of no use if a teacher hands out tests and also hands out the answers alonside, so students can just transcribe them

Answer 18

\[\sin \left( \frac{ -11\pi }{ 12 } \right)=-\sin \frac{ 11\pi }{ 12 }=-\sin \left( \pi-\frac{ \pi }{ 12 } \right)
\[=-\sin \frac{ \pi }{ 12 }\]
\[=-\sin \left( \frac{ \pi }{ 3 }-\frac{ \pi }{ 4 } \right)\]
\[=-\left[ \sin \frac{ \pi }{ 3 }\cos \frac{ \pi }{ 4 }-\cos \frac{ \pi }{3}\sin \frac{ \pi }{ 4 } \right]\]
\[=-\left[ \frac{ \sqrt{3} }{2 }*\frac{ 1 }{ \sqrt{2} }-\frac{ 1 }{2 }*\frac{ 1 }{ \sqrt{2} } \right]\]
\[=-\frac{ \sqrt{3}-1 }{ 2\sqrt{2} }\]

Answer 19

If you rationalize that^^^ den'r, notice that you get:

\(\Large =-\dfrac{ \sqrt{3}-1 }{ 2\sqrt{2} }\cdot\dfrac{ \sqrt{2} }{\sqrt{2} }=-\dfrac{ \sqrt{6}-\sqrt{2} }{ 4}=\dfrac{ -\sqrt{6}+\sqrt{2} }{ 4}\)

which is ALMOST what you answered up above, but for a sign error in the 2nd term in the num'r. So you almost had it, but you should trace your error to see where you went wrong with the sign, so that you won't make that same mistake again.