Question

Find the exact radical value of sin 18 by the following method. First note that for x =18 the equation cos 3x = sin 2x is true. Express each member of this equation in terms of function of x itself then solve for sin x.

Answer 1

Do you remember your trig identities? These two may be useful:

\[\sin(2x) = 2\sin(x)\cos(x)\]\[\cos(u+v) = \cos(u)\cos(v) - \sin(u)\sin(v)\]
The latter can be used to express \(\cos(3x)\) in terms of \(x\) by choosing \(u=2x\) and \(v = x\).

Set up your equation equating \(\cos(3x)\) and \(\sin(2x)\) using those identities, then solve for \(\sin(x)\). You may want to do a substitution such as \(a = \sin(x)\) to make the algebra easier. At the end you'll have a quadratic to solve, and only one of the solutions will have the correct sign to be the value of \(\sin(18^\circ)\).

Answer 2

I did that but I did not get the right answer and I don't what I'm doing wrong

Answer 3

Can you show your work? What did you get for the initial equation?

Answer 4

\[\cos2xcosx-\sin2xsinx=2coxsinx\]\[cosx(\cos ^{2}x-\sin ^{2}x)-sinx(2sinxcosx)=2cosxsinx\]\[cosx(-3\sin ^{2}x-2sinx+\cos ^{2}x)=0\]

Answer 5

then I just used the quadratic formula

Answer 6

I ended up with
\[2 \sin (x) \cos (x)=\left(1-2 \sin^2(x)\right) \cos (x)-2 \sin ^2(x) \cos (x)\]
Then I divided through by \(\cos(x)\) and expanded the right hand side to get
\[2\sin(x) = 1-4\sin^2(x)\]substituting \(a = \sin(x)\) gives
\[2a = 1-4a^2\]