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

Pingu

Junior Member
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Dec 22, 2005
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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.
 
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.
 
Why does sin((pi/4)-(pi/6)) not equal sin(pi/4) - sin(pi/6)?
 
Just as if we had a function f(x)\displaystyle f(x):
. where f(x)=x2\displaystyle f(x) = x^2

. . f(31)=f(2)=22=4\displaystyle f(3 - 1) = f(2) = 2^2 = 4

. . but f(3)f(1)=3212=91=8\displaystyle f(3) - f(1) = 3^2 - 1^2 = 9 - 1 = 8

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

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

You can always plug values into your calculator to check.
 
Ok thank you

So I now have sin(pi/4)*cos(pi/6) - cos(pi/4)*sin(pi/6)
what do I do now?
 
Plug in their exact values, as you appeared to show you are familiar with.
 
Thank you for helping me
I figure it out and I was able to do the rest of my work.
 
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