# Compound Angle

• Nov 9th 2009, 07:30 PM
Johnny123321
Compound Angle
Angle a lies in the second quadrant and angle b lies in the third quadrant such ath cos a = -3/5 and tan b = 24/7

determine an exact value for
cos(a+b)
and
sin (a-b)

I got cos(a+b) = cosacosb - sinasinb
=(-3/5)(-7/25) - (4/5)(-24/25)
= 117/125

i am not sure if this is right, but i am also completely lost for sin (a-b)
• Nov 9th 2009, 07:55 PM
Prove It
Quote:

Originally Posted by Johnny123321
Angle a lies in the second quadrant and angle b lies in the third quadrant such ath cos a = -3/5 and tan b = 24/7

determine an exact value for
cos(a+b)
and
sin (a-b)

I got cos(a+b) = cosacosb - sinasinb
=(-3/5)(-7/25) - (4/5)(-24/25)
= 117/125

i am not sure if this is right, but i am also completely lost for sin (a-b)

It will help if you recall the 3-4-5 Right-angle triangle, and the 7-24-25 Right-angle triangle.
• Nov 9th 2009, 07:56 PM
Johnny123321
check
can you help me check if i did cos (a+b) right
• Nov 9th 2009, 07:59 PM
Prove It
Quote:

Originally Posted by Johnny123321
Angle a lies in the second quadrant and angle b lies in the third quadrant such ath cos a = -3/5 and tan b = 24/7

determine an exact value for
cos(a+b)
and
sin (a-b)

I got cos(a+b) = cosacosb - sinasinb
=(-3/5)(-7/25) - (4/5)(-24/25)
= 117/125

i am not sure if this is right, but i am also completely lost for sin (a-b)

If $\displaystyle a$ is in quadrant 2, then $\displaystyle \sin{a} > 0, \cos{a} < 0$ and $\displaystyle \tan{a} < 0$.

If $\displaystyle b$ is in quadrant 3, then $\displaystyle \sin{b} < 0, \cos{b} < 0$ and $\displaystyle \tan{b} > 0$.

If you remember this as well as the 3-4-5 Right-angle triangle and the 7-24-25 Right-angle triangle, you will be fine.
• Nov 9th 2009, 08:43 PM
windir
The first part looks correct to me.

$\displaystyle sin(A-B)=sinAcosB-sinBcosA$
$\displaystyle (4/5)(-7/25)-(-24/25)(-3/5)$
$\displaystyle (-21/25)-(72/125)$
$\displaystyle (-93/125)$