Question

Help with Trig question? Having to do with Angle sum and difference formulas. Find the exact value of sin[ sin^-1(5/13)-cos^-1 (-7/25)]. Any help would be appreciated!

Answer 1

sin(a-b)=sin(a)cos(b)-cos(a)sin(b)
\(\sin[ \arcsin(5/13)-\arccos (-7/25)]=\\
sin(\arcsin(5/13))cos(\arccos (-7/25))-cos(\arcsin(5/13))sin(\arccos (-7/25))\)

now the left hand cancels out pretty evenly \(\sin[sin^{-1}(\dfrac{5}{13})]=\dfrac{5}{13}\)

same thing for the -7/25

brb pee break

Answer 2

Ok, so the answer is just the value for sin ^-1

Answer 3

ok back
so now we have
\(
\sin[ \arcsin(5/13)-\arccos (-7/25)]=\\
sin(\arcsin(5/13))cos(\arccos (-7/25))-cos(\arcsin(5/13))sin(\arccos (-7/25))
\\\dfrac{5}{13}*\dfrac{-7}{25}-\cos(\arcsin(5/13))\sin(\arccos (-7/25))\)


\(\cos(\arcsin(5/13))=\dfrac{12}{13}\)

Answer 4

that leaves us with \(\\\dfrac{5}{13}*\dfrac{-7}{25}-\dfrac{12}{13}\sin(\arccos (-7/25))\)

one last thingie to simplify

Answer 5

we can use another right triangle here. \(7^2+24^2=25^2\)

if arccos of our angle is -7/25

Answer 6

\(\\\dfrac{5}{13}*\dfrac{-7}{25}-\dfrac{12}{13}*\dfrac{24}{25}=\dfrac{-35}{325}-\dfrac{288}{325}=\dfrac{-35-288}{325}=-\dfrac{323}{325} \)

Answer 7

Ohhh, I started with the triangles first, and got lost. Thank you so much for walking me through that!

Answer 8

no problem, pls buy me lunch or dinner or linner as payment

Answer 9

p.s. that was a joke, now I'm off to run some errands and get food, feel free to tag me in any of your future questions using " @bibby "

Answer 10

Hahaha, I would if I wasn't a poor college student now. :p thanks @bibby