1. ## inverse

If sin^-1(a)=%pi/6 and cos^-1(%pi/4)=b, then the exact value of sin(%pi*a+b) is:

i think you get rid of the inverse by going

sin %pi/6 = a so then a = 1/2

cosb = %pi/4

but im not sure how you solve for b or get the value the question is asking for.....

2. Originally Posted by samtheman17
If sin^-1(a)=%pi/6 and cos^-1(%pi/4)=b, then the exact value of sin(%pi*a+b) is:

i think you get rid of the inverse by going

sin %pi/6 = a so then a = 1/2

cosb = %pi/4

but im not sure how you solve for b or get the value the question is asking for.....
Good start ,

$\displaystyle \sin^{-1} a=\frac{\pi}{6}$

$\displaystyle a=\frac{1}{2}$

$\displaystyle \cos^{-1} \frac{\pi}{4}=b$

$\displaystyle \cos b=\frac{\pi}{4}$

$\displaystyle =\sin (\pi\cdot a+b)=\sin (\frac{\pi}{2}+b)$

$\displaystyle =\sin \frac{\pi}{2}\cos b+\cos \frac{\pi}{2}\sin b$

$\displaystyle =\cos b=\frac{\pi}{4}$

3. ahh okay thanks,

but then how do you put it into the form of

sin(%pi*a + b)?

i think you have to use

sin(a+b)=cos(a)sin(b)+sin(a)cos(b)

but i can't seem to get it right.....

4. Originally Posted by samtheman17
ahh okay thanks,

but then how do you put it into the form of

sin(%pi*a + b)?

i think you have to use

sin(a+b)=cos(a)sin(b)+sin(a)cos(b)

but i can't seem to get it right.....
$\displaystyle \sin (\frac{1}{2}\pi+b)$ ... substitute the a with 1/2

after that , that's what i did , expand using the sin formulas .