tough question, thanks for your help
tan[sin-1(4/5)+cos-1 (1)]
can someone show me how to do it, I know how to using a calculator. thanks
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tough question, thanks for your help
tan[sin-1(4/5)+cos-1 (1)]
can someone show me how to do it, I know how to using a calculator. thanks
Here's one way to do it.Quote:
Originally Posted by mcdaking84
See the diagram below.
this problem is a lot simpler than it seems. at first it seems like you will need the double angle formula, but in fact you don't. remember that cos^-1(1) = 0
so we want tan(sin^-1(4/5))
let sin^-1(4/5) = x
so we want tanx
now if sin^-1(4/5) = x
=> sinx = 4/5 .............we can find tanx using several identities, or just draw the triangle, see below.
we can find the that the missing side of the triangle is 3 using pythagoras' theorem, or we can simply recognize that this is the famous 3-4-5 triangle.
so tanx is simply opposite/adjacent
=> tan[sin^-1(4/5)+cos^-1 (1)] = tan(sin^-1(4/5)) = tanx = 4/3
Hello, mcdaking84!
Quote:
tan[arcsin(4/5) + arccos(1)]
Let .A = arcsin(4/5) . . . then: .sin(A) = 4/5
So A is an acute angle in a right triangle with opp = 4, hyp = 5.
. . Using Pythagorus, we find that: adj = 3.
Hence: .tan(A) = 4/3
Let B = arccos(1) . . . then B = 0
Hence: .tan(B) = 0
. . . . . . . . . . . . . . . . . . . . . . . . . . .tan(A) + tan(B)
We use the identity: .tan(A + B) .= .---------------------
. . . . . . . . . . . . . . . . . . . . . . . . . .1 + tan(A)·tan(B)
. . . . . . . . . . . . . . . . . . . . . . . 4/3 + 0. . . . . 4
Then we have: .tan(A + B) .= .-------------- .= .---
. . . . . . . . . . . . . . . . . . . . . .1 + (4/3)(0) - - .3