Question

simplify. sin(arcsinx+arccosx)

step by step

Answer 1

first use angle sum identity
\[\sin(a+b) = \sin a \cos b + \sin b \cos a\]
then note that
\[\sin x = \sqrt{1-\cos^{2}x}\]
\[\cos x = \sqrt{1-\sin^{2}x}\]

Answer 2

okay but then that is supposed to be arcsin?

Answer 3

im confused

Answer 4

ok so you have
\[\sin (\sin^{-1} x) \cos(\cos^{-1} x) + \sin(\cos^{-1} x) \cos(\sin^{-1} x)\]
\[=x*x + \sin(\cos^{-1} x) \cos(\sin^{-1} x)\]

Answer 5

now change "sin" into "cos" so you get situation where its
\[\cos(\cos^{-1} x)\]

Answer 6

isn't sin(sin^-1x)= 1

Answer 7

and same thing for cos

Answer 8

=x not 1

Answer 9

oh ok so then its x(x)+ cos(sin^-1x) sin(cos^-1x)where do u go after that?

Answer 10

\[\cos(sin^{-1}x) = \sqrt{1-\sin^{2}(sin^{-1}x)}\]
due to pythagorean identity

Answer 11

i don't know can u just write out it all in one message please, I'm lost in it all

Answer 12

so you are not following my helpful steps at all huh...
ok its too much work to rewrite everything
answer is 1

Answer 13

well see I'm trying to do test corrections, and i don't know how to show that lol

Answer 14

great site for checking answers....wont help you on a test though :)

http://www.wolframalpha.com/input/?i=sin%28arcsin%28x%29%2Barccos%28x%29%29

Answer 15

i see well if you look at my posts again ive given you everything you need to show how to do it...see if you can piece it together

Answer 16

ok thank you

Answer 17

ok i have more questions

Answer 18

ok:)

Answer 19

okay thanks for coming back!

Answer 20

now i understand how u got to where you had the sin(cos^-1)cos(sin^-1) but how do u change it to cos cos thing

Answer 21

using pythagorean identity
sin^2 + cos^2 = 1

Answer 22

now change "sin" into "cos" so you get situation where its
cos(cos−1x)
this? how do you use this?

Answer 23

so then you have x^2 +radical 1-cos^2(cos^-1) *radical 1-sin^2(sin^-1)?

Answer 24

now you got it , then change cos(cos^-1 x) = x ... same for sin

Answer 25

why does that equal x though?

Answer 26

thats the part I'm not getting

Answer 27

oh ok...because they are inverse functions they cancel each other out leaving the input "x"
same reason
\[\sqrt{x^{2}} = x\]
the square and square root are inverse functions

Answer 28

okay so then you would have x(x) +radical 1-x *radical 1-x?

Answer 29

almost they were squared
\[\cos^{2} (\cos^{-1} x) = x^{2}\]

Answer 30

okay i was wondering about that, so then you would have \[\sqrt{1-x ^{2}}\] \[\sqrt{1-x ^{2}}\]

Answer 31

correct

Answer 32

those times each other
which would then be \[\sqrt{1}x \sqrt{1}x\]

Answer 33

hmm u lost me there?

Answer 34

nevermind

Answer 35

what is
\[\sqrt{2} * \sqrt{2}\]

Answer 36

i totally just figured it out

Answer 37

i got the correct answer that you said earlier also, which is 1

Answer 38

cool :)