Originally Posted by

**ticbol** :

:

sinA in terms of tanA:

tanA = sinA/cosA ----identity

Multiply both sides by cosA,

cosA*tanA = sinA

Express the cosA in terms of sinA,

[sqrt(1 -sin^2(A)]tanA = sinA ------------(i)

There, only sinA and tanA are involved.

Solve for sinA in terms of tanA and you're done.

Here is one way of doing that,

Square both sides of (i),

(1 -sin^2(A)[tan^2(A)] = sin^2(A)

Expand,

tan^2(A) -[sin^2(A)][tan^2(A)] = sin^2(A)

Collect the sinA terms,

tan^2(A) = sin^2(A) +[sin^2(A)][tan^2(A)]

Factor out the sin^2(A),

tan^2(A) = [sin^2(A)][1 +tan^2(A)]

Isolate the sin^2(A),

[tan^2(A)] /[1 +tan^2(A)] = sin^2(A)

So, take the square roots of both sides,

sinA = (tanA) /[sqrt(1 +tan^2(A))] -------------answer.

This is OK in the first and third quadrants, but in the second and third

sin and tan have different signs; so in full this should be:

Code:

sinA=(tanA) /[sqrt(1 +tan^2(A))] for 0<=A<90 or 270<pi<=360
=-(tanA) /[sqrt(1 +tan^2(A))] for 90<A<270

also as tan is undefined at 90, 270 there is strictly no expression for sin in

terms of tan at these points.

RonL

(Note: It's a bit late to comment on it as Jameson has locked this thread and

I will not unlock something another has locked, but Soroban's post also suffers

from this problem)