# Easy trig problem

• Jul 2nd 2010, 06:36 AM
dapore
Easy trig problem
• Jul 2nd 2010, 10:31 PM
eumyang
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:
$cos (a-b) = cos\, a\: cos\, b + sin\, a\: sin\, b \quad$ (Eq1)

Use the Law of sines to write sin b in terms of sin a:
$sin\, b = \frac{3}{4}sin\, a \quad$ (Eq2)

Use the pythagorean identity to write cosine in terms of sine:
$cos\, a = \sqrt{1-sin^2\, a}$ (Eq3a)
$cos\, b = \sqrt{1-sin^2\, b}$ (Eq3b)

Plug (2), (3a), and (3b) into (1), plus the fact that cos (a-b) = 3/4:
$\frac{3}{4} = \sqrt{1-sin^2\, a}\: \sqrt{1-\left(\frac{3}{4}sin\, a\right)^2} + sin\, a\: \left(\frac{3}{4}sin\, a\right)$ (Eq4)

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

I'm sure I did something wrong here.
• Jul 3rd 2010, 12:04 AM
simplependulum
In general , if $|ac| = B ~,~ |bc| = A ~,~ \cos(a-b) = X$

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

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

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

Then we have

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

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

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

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