Need help with trig identities/determine of sin(pi/12)

[Pingu]

The question in my text book is:

These two expressions simplify to pi/12: ((pi/3)-(pi/4))and((pi/4)-(pi/6)).
Determine the value of sin pi/12 be expanding each expression.
b) sin((pi/4)-(pi/6))

here is what I did:
I expand it sin(pi/4) - sin(pi/6)
using identities ((Root2)-1)/2

The answer book says that I am wrong and that the real answer is:

((root6)-(root2))/4

Please help me answer this math question so I can do my homework.

 

[Unco]

G'day, Pingu.

. . \(\displaystyle \L \sin{\left(\frac{\pi}{4} - \frac{\pi}{6}\right)} \, \neq \, \sin{\frac{\pi}{4}} - \sin{\frac{\pi}{6}}\)

Rather, we need to use the compound angle formula:

. . \(\displaystyle \L \sin{(A - B)} = \sin{A}\cos{B} - \cos{A}\sin{B}\)

So we have
. . \(\displaystyle \L \sin{(\frac{\pi}{4} - \frac{\pi}{6})} = \sin{\frac{\pi}{4}}\cos{\frac{\pi}{6}} - \cos{\frac{\pi}{4}}\sin{\frac{\pi}{6}}\)

Now continue.

 

[Pingu]

Why does sin((pi/4)-(pi/6)) not equal sin(pi/4) - sin(pi/6)?

 

[Unco]

Just as if we had a function \(\displaystyle f(x)\):
. where \(\displaystyle f(x) = x^2\)

. . \(\displaystyle f(3 - 1) = f(2) = 2^2 = 4\)

. . but \(\displaystyle f(3) - f(1) = 3^2 - 1^2 = 9 - 1 = 8\)

We can conclude that
. \(\displaystyle f(a - b)\) is not necessarily equal to \(\displaystyle f(a) \, - \, f(b)\).

The same applies to the sine function:
. \(\displaystyle \sin{(a - b)}\) is not necessary equal to \(\displaystyle \sin{(a)} - \sin{(b)}\).

You can always plug values into your calculator to check.

 

[Pingu]

Ok thank you

So I now have sin(pi/4)*cos(pi/6) - cos(pi/4)*sin(pi/6)
what do I do now?

 

[Unco]

Plug in their exact values, as you appeared to show you are familiar with.

 

[Pingu]

Thank you for helping me
I figure it out and I was able to do the rest of my work.