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Maths Inverse Trigonometry

Whats the rules for inverse trigonometry? Like when you have sin-1 or something, to get the angles and stuff? Much appreciated if someone could explain how to do it that would be great!
Reply 1
Avatar for smsgd
smsgd
1
All you would need to know is SOHCAHTOA

so, the sin of an angle = opposite/Hypotenuse
cos of an angle = adjacent/Hypotenuse
tan of an angle = opposite/Adjacent

sin^-1(opposite/Hypotenuse ) = the angle
cos^-1(adjacent/Hypotenuse ) = the angle
tan^-1(opposite/Adjacent ) = the angle

Not sure if this is what you're after, but hope it helps! (If you need me to elaborate or you needed something else, just pm me or reply on the thread)
I'm not 100% sure, but I think you use sin-1 when you have something like:

Two sides and need to find an angle, so you get something like sinx = a/b and to get x=........., you do the inverse of sin, which is sin-1 and get x= sin-1(a/b).

I'm not sure if this was what you were asking for, so sorry if I couldn't help.
They are the inverse functions of sine, cosine and tangent.

At GCSE:
sin⁡−1\sin^{-1} takes a ratio of edges and returns an angle. The ratio can be between -1 and 1 and the function returns values between -90 and 90(its range is restricted to make it a function). It does the opposite of the sine function, but only returns one solution. At GCSE it's probably best to use the sine graph to find the other solutions.

cos⁡−1\cos^{-1} takes a ratio of edges and returns an angle. The ratio can be between -1 and 1 and the function returns values between 0 and 180(its range is restricted to make it a function). It does the opposite of the cosine function, but only returns one solution. At GCSE it's probably best to use the cosine graph to find the other solutions.

tan⁡−1\tan^{-1} takes a ratio of edges and returns an angle. The ratio can be between −∞-\infty and ∞\infty and the function returns values between -90 and 90(its range is restricted to make it a function). It does the opposite of the tangent function, but only returns one solution. At GCSE it's probably best to use the tangent graph to find the other solutions.
Reply 4
Avatar for Jasminee_03
Jasminee_03
OP
Original post by smsgd
All you would need to know is SOHCAHTOA

so, the sin of an angle = opposite/Hypotenuse
cos of an angle = adjacent/Hypotenuse
tan of an angle = opposite/Adjacent

sin^-1(opposite/Hypotenuse ) = the angle
cos^-1(adjacent/Hypotenuse ) = the angle
tan^-1(opposite/Adjacent ) = the angle

Not sure if this is what you're after, but hope it helps! (If you need me to elaborate or you needed something else, just pm me or reply on the thread)


Thats perfect, thank you! I wasn't sure if you just used sohcahtoa or there was like another set rule :3
Reply 5
Avatar for smsgd
smsgd
1
Original post by Jasminee_03
Thats perfect, thank you! I wasn't sure if you just used sohcahtoa or there was like another set rule :3


No problem - just remembered that you need to know the sin and cosine rules as well (for the trig parts of the test), thought i'd mention it just in case you needed it

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