Question

HELP ME FILL IN THE BLANK FOR THIS EQUATION!!
sin pi/4 sin pi/6 = 1/2 (cos pi/12 "blank" 5pi/12)

Answer 1

btw I know we have to simplify!
so sin pi/4 = sqrt2/2
sin pi/6 = 1/2
and cos pi/12 = a little harder to find lol

Answer 2

We have
\[\sin {\frac{\pi}{4}}\times \sin{\frac{\pi}{6}}=\frac{1}{2}(\cos{\frac{\pi}{12}}\times ..............(\frac{5\pi}{12})\]

Answer 3

do you know the relation of half angle formula? for cos

Answer 4

no x between the two sins on the left side!

Answer 5

it's multiply sign

Answer 6

Uhh I think its in my notes, oh okay:)

Answer 7

go through your notes and try to find the formula. If you aren't able to, I will help..

Answer 8

cos(x/2)=sqrt((1+cosx)/(2))

Answer 9

right?

Answer 10

good, now we can use this here to find cos (pi/12)
we know cos (pi/6)
so:
\[cos {\frac{\pi}{12}}=\sqrt{\frac{1+cos {\frac{\pi}{6}}}{2}}\]

could you substitute the values and find cos 15 degrees?

Answer 11

cos pi/12 = sqrt ((1+sqrt3/2) / (2))

Answer 12

let's simplify it a bit
\[cos {\frac{\pi}{12}}=\sqrt {\frac{2+\sqrt3}{4}}\]

ok, could you put this in the given expression and simplify?

Answer 13

did you just multiply everything by 2?
Hmm I am a little lost, so now the "cospi/12" will be replaced with the equation we just came up with?

Answer 14

yep, I just multiplied both numerator and denominator in square root by 2.
yes, we need to plugin the value in the question given

Answer 15

\[ ( \sqrt2/2) (1/2) = \sqrt(2+\sqrt3)/4) blank 5\pi/12\]

Answer 16

\[\sqrt2/4 = \sqrt(2+3)/(4) blank 5\pi/12\]

Answer 17

oops that 3 should be sqrt3!

Answer 18

correct so we have
\[\frac{\sqrt 2}{4}=\frac{\sqrt{2+\sqrt 3}}{2} X\]

I have replaced blank 5pi/12 by X

Answer 19

alright, good idea!

Answer 20

Oh i didn't notice, but let's go ahead and can you solve this further: have X one side and other things on other side

Answer 21

shouldnt the number under the fraction on the right side be 4?

Answer 22

so divide both sides by the coeeficient on the right side.

Answer 23

I took the 4 out of square root, notice only the numerator is in the square root...

Answer 24

oh i see!

Answer 25

yes, divide both sides by the right side expression

Answer 26

Yeah that is really difficult, I am stumped on how to divide the left side by such a weird number

Answer 27

okay, let me show you. We will work in step by step manner
\[\frac{\sqrt 2}{4} \times \frac{2}{\sqrt{2+\sqrt 3}}= X\]

do you follow this?

Answer 28

I would probably recommend using the Cosine Angle Addition/Subtraction Formula instead of the Half-Angle, so you don't end up with nested roots like that.\[\large\rm \cos\left(\frac{\pi}{12}\right)\quad=\cos\left(\frac{4\pi}{12}-\frac{3\pi}{12}\right)\quad=\cos\left(\frac{\pi}{3}-\frac{\pi}{4}\right)\]\[\large\rm =\cos\left(\frac{\pi}{3}\right)\cos\left(\frac{\pi}{4}\right)+\sin\left(\frac{\pi}{3}\right)\sin\left(\frac{\pi}{4}\right)\]

I'm still really confused by this problem though.
Is the "blank" referring to an operation?
Or a trig function?
Is the 5pi/12 supposed to be included in the cosine angle on the right?

Answer 29

This problem doesn't make a lot of sense.
Do you have a screen shot or something?

Answer 30

@zepdrix blank is supposed to be a trigonometric function..

Answer 31

Yes @ash2326 I think that is correct, the blank should be a trig function.

Answer 32

@cutiecomittee123 Did you follow my last step?

Answer 33

Oh, there is a minus in front of the blank...

Answer 34

which doesnt necessarily change anything thus far? right?

Answer 35

i think they were expecting us to use
\(\Large \cos y - \cos x = 2 [\sin \dfrac{(x+y)}{2} \sin \dfrac{(x-y)}{2}] \)

Answer 36

I didn't calculate, but i think with
(x+y)/2 = pi/4
and
(x-y)/2 = pi/6

we would get
x = pi/12
and y = 5pi/12

Answer 37

oops, we actually get
x = 5pi/12
and y = pi/12

Answer 38

got what i did there?

Answer 39

@hartnn is right, I will let him help you @cutiecomittee123

I will be here to help in case it's needed but I don't think that would be the case

Answer 40

okay I am trying to follow, it is a little challenging but I think I got it so far.

Answer 41

I dont know what the equation that you represented for us to use is?

Answer 42

since we had this formula
\( \cos y - \cos x = 2 [\sin \dfrac{(x+y)}{2} \sin \dfrac{(x-y)}{2}]\)
and we had a 2[sin pi/4 sin pi/6 ]

so comparing we can say that
(x+y)/2 = pi/4
(x-y)/2 = pi/6

once we get the value of x and y

we can substitute in cos y - cos x
to get our final answer

Answer 43

solving
(x+y)/2 = pi/4
(x-y)/2 = pi/6 is purely algebraic

Answer 44

but how is this relevant to our origional equation?

Answer 45

the original question has similar form as that of the formula