cosA/(1+sinA) = 0,5?

Question

I had to prove this 2 years ago in maths and it wasn't difficult for me, but now I have to solve a physics problem, where the answer is 1/2 and my solution is cosA/(1+sinA).

Answer 1

Multiply top and bottom of the left side by (1-sinA)

cosA(1-sinA) / (1-sin^2 A)
=cosA(1-sinA) / cos^2 A
=(1-sinA) / cosA

These two are equivalent but that's about it.

Answer 2

Sheen is correct.
Here is a somewhat simpler solution:

Substitute sqrt(1-sin^2(A)) for cos A. Note that (1-sin^2(A)) is the same as (1-sinA)*(1+sinA).

Square both sides.

At this point you have:

[(1+sinA)*(1+sinA)]/
[(1+sinA)(1-sinA)] = 4

Cancel out the common 1+sinA terms to get:

1 + sinA = 4*(1-sinA)

Solve for sinA and finish.

Answer 3

The above is an equation not identity

becuase put A = 0

LHS = 1/1 = 1 not RHS

to solve the equation

cos A = .5 + .5 sin A
or cos A - 1/2 sin A = 1/2
let 1 = r sin t

and 1/2 = r cos t
r^2 = 1+ 1/4 = 5/4
tan t = 1/2

we get LHS = r sin(t-A) = 1/2

sin (t-A) = 1/(2r)

above can be solved as

t-A = npi+(-1)^n/2r

by putting the proper values

Answer 4

cosA/(1+sinA) = 0,5

Apply submultiple angle formulae and sin^2 + cos^2 = 1

[(CosA/2)^2 - (SinA/2)^2] / [(CosA/2)^2 + (SinA/2)^2 + 2 SinA/2 CosA/2] = 0.5
[(CosA/2 - SinA/2)(CosA/2 + SinA/2)] / [(CosA/2 + SinA/2)(CosA/2 + SinA/2)] = 0.5
(CosA/2 - SinA/2) / (CosA/2 + SinA/2) = 0.5

Addition-substraction rule
If a/b = c/d, then (a+b)/(a-b) = (c+d)/(c-d)

(2 CosA/2) / (-2 SinA/2) = 1.5/(-0.5)
Tan(A/2) = 1/3
A = 2 (arctan(1/3)) = 36.87 deg

Answer 5

Thus cosA = 0.5(1+sinA) or 2cosA – sinA =1 or cos(A+B) =1/R , where R= sqrt (2*2 +1) = sqrt5, and B= atan(1/2); thence A= -+acos(1/R) –B +2kpi; quite possible! but I guess you mean another formula:
(CosA)^2 / (1+cos(2A)) = 0.5 or (sinA)^2 / (1-cos(2A)) = 0.5;

Answer 6

It sounds not right. There is no such an identity.

Can you post the physics problem? I might have time to help you solve the physics problem directly.