# Compound Angle Forumla

• Feb 4th 2007, 07:06 AM
classicstrings
Compound Angle Forumla
If A and B are angles between 0 and 90, and sinA = 3/5 and tanB = 7/24, find the value of

sin(A-B).

How do I tackle this question? I just need to know the pathway. Thanks.
• Feb 4th 2007, 07:37 AM
Soroban
Hello, classicstrings!

Quote:

If $A$ and $B$ are angles between $0^i$ and $90^o$,
and $\sin A = \frac{3}{5}$ and $\tan B = \frac{7}{24}$,
find the value of $\sin(A-B)$

We have the identity: . $\sin(A - B) \:=\:\sin(A)\cos(B) - \sin(B)\cos(A)$

We know that: . $\sin(A) = \frac{3}{5}$

We need the values of: . $\cos(A),\:\sin(B),\:\cos(B)$

From $\sin(A) = \frac{3}{5} = \frac{opp}{hyp}$ and Pythagorus, we find that: . $adj = 4$
. . Now you can find $\cos(A).$

From $\tan(B) = \frac{7}{24} = \frac{opp}{adj}$ and Pythagorus, we find that: . $hyp = 25$
. . Now you can find $\sin(B)$ and $\cos(B).$

Got it?

• Feb 4th 2007, 07:47 AM
CaptainBlack
Quote:

Originally Posted by classicstrings
If A and B are angles between 0 and 90, and sinA = 3/5 and tanB = 7/24, find the value of

sin(A-B).

How do I tackle this question? I just need to know the pathway. Thanks.

First you need to know that:

sin(A-B)=sin(A)cos(B)-cos(A)sin(B).

Then as A and B are both in the range 0 to 90 degrees,

if sin(A)=3/5, then cos(A)=4/5 (there are a number of ways of getting this
result one of which is recognising that A is an angle of a 3,4,5 right triangle
which imediatly tells you that cos(A)=4/5).

if tan(B)=7/24, then sin(B)=7/25, and cos(B)=24/25 (as 25 is the hypotenuse
of a right triangle who's other two sides are 7 and 24).

So:

sin(A-B)=(3/5)(24/25)-(4/5)(7/25)=44/125

RonL