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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- Apr 17th 2007, 09:30 PMmcdaking84Anyone know how to solve this trig problem?
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 - Apr 17th 2007, 10:19 PMJhevon
Here's one way to do it.

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 - Apr 17th 2007, 10:26 PMSoroban
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