Question of sin 23pi/12
which is 345 degrees which is 300 degrees + 45..
so the 300 degrees would be negative right? -sqrt3/2?
sin 300 will be negative as it is in 4th quadrant
yup -sqrt3/2! :D
so i plug into the sum sin formula as negative?
Well we are dealing with radians, so keep it in radians. \(\sf 23\pi /12 = 2\pi-pi/12\)
the answer ends up in radians..?
23pi/12 is radian already
still not sure how to solve with sinAcosB+cosAsinB
\(\sf 23pi/12 = 5pi/3 + pi/4\)
wouldn't you have to simplify out the radians?
Yes, we would have to. \(\sf sin(5pi/3+pi/4)=sin (5pi/3) cos (pi/4) + cos(5pi/3)sin(pi/4)\)
|dw:1428981185396:dw|
i'm getting \[(\frac{ -\sqrt{3} }{ 2 })(\frac{ \sqrt{2} }{ 2 })+(\frac{ 1 }{ 2 })(\frac{ \sqrt{2} }{ 2 })\]
\(\color{blue}{\text{Originally Posted by}}\) @TheSmartOne Yes, we would have to. \(\sf sin(5pi/3+pi/4)=sin (5pi/3) cos (pi/4) + cos(5pi/3)sin(pi/4)\) \(\color{blue}{\text{End of Quote}}\) \(\Large\sf = \frac{-\sqrt{3}}{2}(\frac{ \sqrt{2} }{ 2 })+(\frac{ 1 }{ 2 })(\frac{ \sqrt{2} }{ 2 })\)
correct :)
So at the end you should get \(\sf\Large\frac{\sqrt{2}-\sqrt{6}}{4}\)
- sqrt 6 + sqrt 2 = sqrt2 - sqrt 6
can the answer be \[\frac{ 2\sqrt{2} }{ 4 }\] that's what the answer key says
because i got the same thing as you
\[\frac{ -\sqrt{6}+\sqrt{2} }{ 4 }\] which is the same as yours
The answer is key is soo wrong.
i agree, not even gonna bother with it. gonna go with our answers.
Our answer will be negative since sqrt 6 is bigger than sqrt 2 and tiny number = big number = negative number The answer key has an answer that is positive. And this also helps our case: http://prntscr.com/6thvjd (made sure it was radians too always have too :P)
lol yep, gonna stick with what we got. thank you
Thank you! This was fun practice for my trig class :D
haha np. I got one more that seems a bit harder. will post shortly.
I gtg now, I'll check it later tomorrow :P
alright np.
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