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- Jul 2nd 2010, 05:36 AMdaporeEasy trig problem
- Jul 2nd 2010, 09:31 PMeumyang
Maybe it's me, but this didn't seem easy!?!? I'm getting an answer of sin a = 1; is this correct?

Here's a summary of what I did:

You have the cosine of a difference identity:

$\displaystyle cos (a-b) = cos\, a\: cos\, b + sin\, a\: sin\, b \quad$ (Eq1)

Use the Law of sines to write sin b in terms of sin a:

$\displaystyle sin\, b = \frac{3}{4}sin\, a \quad$ (Eq2)

Use the pythagorean identity to write cosine in terms of sine:

$\displaystyle cos\, a = \sqrt{1-sin^2\, a}$ (Eq3a)

$\displaystyle cos\, b = \sqrt{1-sin^2\, b}$ (Eq3b)

Plug (2), (3a), and (3b) into (1), plus the fact that cos (a-b) = 3/4:

$\displaystyle \frac{3}{4} = \sqrt{1-sin^2\, a}\: \sqrt{1-\left(\frac{3}{4}sin\, a\right)^2} + sin\, a\: \left(\frac{3}{4}sin\, a\right)$ (Eq4)

Solved for sin a and I got sin a = 1.

I'm sure I did something wrong here. - Jul 2nd 2010, 11:04 PMsimplependulum
In general , if $\displaystyle |ac| = B ~,~ |bc| = A ~,~ \cos(a-b) = X $

The method to find the value of $\displaystyle \sin(a)$ is as follows :

First determine which of the angles $\displaystyle a,b $ is larger by comparing their opposite sides , let's say $\displaystyle a > b $ .

Let $\displaystyle d $ be the other point on line $\displaystyle ab$ such that $\displaystyle |ac| = |cd| = B $

Then we have

$\displaystyle \angle bcd = a-b $ , by using cosine formula ,

$\displaystyle |bd| = \sqrt{A^2 + B^2 - 2ABX} $ ,

Therefore , $\displaystyle \sin(a) : A = \sin(a-b) : |bd| $

$\displaystyle \sin(a) =\frac{A \sin(a-b)}{ \sqrt{A^2 + B^2 - 2ABX} } $