Trigonometry - Double Angle Identities

[Nusc]

Homework Statement

If I have the following relation:

tan(2x) = (B/2) / (A - C)

but tan(2x) = sin(2x) / cos(2x)

How do I obtain an expression for sin(x) and cos(x) in terms of the constants, B,A,C only?

Homework Equations

cos(2x) = 1- 2 sin^2(x)

The Attempt at a Solution

[/B]
I can't just consider B/2 as the opposite and A-C as the adjacent to find the hypotenuse:

sin(x) = O/H = (B/2)/[B^2/4 +(A-C)^2]^1/2
etc.

 

[PeroK]

I'm not sure I understand quite what you are trying to do, but $A, B, C$ cannot be "constants". They must be functions of $x$.

 

[neilparker62]

Couldn't you simply write down:

2 sinx.cosx = B/2 ...Eqn 1 and
1 - 2.(sinx)^2 = A-C....Eqn 2.

Solve simultaneously.

 

[Ray Vickson]

Couldn't you simply write down:

2 sinx.cosx = B/2 ...Eqn 1 and
1 - 2.(sinx)^2 = A-C....Eqn 2.

Solve simultaneously.

No, that won't work: from $\tan(2x) = 4/5$ we cannot conclude that $\sin(2x)=4$ and $\cos(2x) = 5.$
However, the OP could write $R = \pm \sin(2x)/\sqrt{1-\sin(2x)^2}$, where $R = (B/2)/(A-C)$, and choose the "+" sign if $R > 0$ or "-" if $R < 0$. Letting $S = \sin(2x)^2$ gives $R^2 = S/(1-S),$ so $S$ is obtainable. Then $\sin(2x) = \pm \sqrt{S}$, with the sign chosen to match that of $R$.

 

[Nusc]

No, that won't work: from $\tan(2x) = 4/5$ we cannot conclude that $\sin(2x)=4$ and $\cos(2x) = 5.$
However, the OP could write $R = \pm \sin(2x)/\sqrt{1-\sin(2x)^2}$, where $R = (B/2)/(A-C)$, and choose the "+" sign if $R > 0$ or "-" if $R < 0$. Letting $S = \sin(2x)^2$ gives $R^2 = S/(1-S),$ so $S$ is obtainable. Then $\sin(2x) = \pm \sqrt{S}$, with the sign chosen to match that of $R$.

I don't follow. I get as far as $R = \pm \sin(2x)/\sqrt{1-\sin(2x)^2}$
Then how does one determine cos(x) and sin(x), not cos(2x) or sin(2x)?

 

[PeroK]

I don't follow. I get as far as $R = \pm \sin(2x)/\sqrt{1-\sin(2x)^2}$
Then how does one determine cos(x) and sin(x), not cos(2x) or sin(2x)?

$\tan(2x)$ and $\tan(x)$ are related. As are $\cos(2x)$ and $\cos(x)$ and $\sin(2x)$ and $\sin(x)$.

In fact, all these trig functions are related to each other.

 

[neilparker62]

No, that won't work: from $\tan(2x) = 4/5$ we cannot conclude that $\sin(2x)=4$ and $\cos(2x) = 5.$
However, the OP could write $R = \pm \sin(2x)/\sqrt{1-\sin(2x)^2}$, where $R = (B/2)/(A-C)$, and choose the "+" sign if $R > 0$ or "-" if $R < 0$. Letting $S = \sin(2x)^2$ gives $R^2 = S/(1-S),$ so $S$ is obtainable. Then $\sin(2x) = \pm \sqrt{S}$, with the sign chosen to match that of $R$.

Quite right - in the example you give rather solve:

2 sinx.cosx = 4/√41 ...Eqn 1 and
1 - 2.(sinx)^2 = 5/√41....Eqn 2.

Or

2 sinx.cosx = -4/√41 ...Eqn 1 and
1 - 2.(sinx)^2 = -5/√41....Eqn 2.

The signs of cos and sine in the answer do depend on the angle x. In this example if 2x ∈ [0;90) then x ∈ [0;45) in which case sinx and cosx are both +.
But if 2x ∈ [180;270) then x ∈ [90; 135) in which case sinx is + and cosx - .

Sincere apologies for the elementary error - I hope working with a numeric example like this does provide the OP with some insights into the algebraic problem.

 

[Nusc]

$\tan(2x)$ and $\tan(x)$ are related. As are $\cos(2x)$ and $\cos(x)$ and $\sin(2x)$ and $\sin(x)$.

In fact, all these trig functions are related to each other.

I solved algebraically for sin(2x) using the above relation and obtained the right hand side of the expression I wrote in the original post but that was for sin(x) not sin(2x)

sin(2x) = (B/2)/[B^2/4 +(A-C)^2]^1/2

 

[neilparker62]

Rather get an expression for cos(2x) since if we let cos(2x) = k, then $$ cos(x) = \sqrt { \frac{1+k} {2}} $$ and $$ sin(x) = \sqrt { \frac{1-k} {2}} $$

 

[Nusc]

Rather get an expression for cos(2x) since if we let cos(2x) = k, then $$ cos(x) = \sqrt { \frac{1+k} {2}} $$ and $$ sin(x) = \sqrt { \frac{1-k} {2}} $$

Are you able to get an expression from sin(2x)/cos(2x) strictly in terms of cos(2x)?
I didn't think you could so I went with sin(2x).

 

[PeroK]

I solved algebraically for sin(2x) using the above relation and obtained the right hand side of the expression I wrote in the original post but that was for sin(x) not sin(2x)

sin(2x) = (B/2)/[B^2/4 +(A-C)^2]^1/2

I'm still not sure what you are trying to do. I would let:

$\tan(2x) = \frac{B/2}{A-C} = D$, then express $\cos(x), \sin(x)$ in terms of $D$.

 

[Nusc]

I need to solve for x strictly in terms of the constants. What you describe is valid for sin(2x) and cos(2x),

tan(2x) = O/AO = B/2
A = A-C
H = sqrt( (B/2)^2 + (A-C)^2 )

What I want are exprsssions for sinx and cosx

 

[neilparker62]

Are you able to get an expression from sin(2x)/cos(2x) strictly in terms of cos(2x)?
I didn't think you could so I went with sin(2x).

Yes $$ cos(2x) = 2cos^2x - 1 $$ and $$ cos(2x) = 1 - 2sin^2x $$. You just need to re-arrange as I have done above to get your expressions for sin(x) and cos(x).

 

[PeroK]

I need to solve for x strictly in terms of the constants. What you describe is valid for sin(2x) and cos(2x),

tan(2x) = O/AO = B/2
A = A-C
H = sqrt( (B/2)^2 + (A-C)^2 )

What I want are exprsssions for sinx and cosx

Well, you have to start thinking of the trig functions as something more general than the ratio of sides of a triangle. Although, for $x < \frac{\pi}{4}$ you could use a geometric approach. That would be interesting.

Can you see that there must be some relationship between $\sin(x)$ and $\tan(2x)$? You should be able to derive this using common trig identities. Remember that $\sin^2(x) + \cos^2(x) = 1$.

I would use the intermediate value $D$ as above. In fact, this is what Ray showed you in post #4, where he used $R$ instead of $D$.

 

[vela]

Try using the double-angle identity for tangent to eliminate the $2x$ from the start.

 

[eys_physics]

I would suggest the following. Put as proposed by PeroK,
$$D=\frac{B}{2(A-C)}(B/2) $$.
Then, use
$$\sin(2x)=\pm \sqrt{1-\cos^2(2x)}$$.
Solve the equation
$$\frac{\pm\sqrt{1-\cos^2(2x)}}{\cos(2x)}=D$$.

 

[neilparker62]

I need to solve for x strictly in terms of the constants. What you describe is valid for sin(2x) and cos(2x),

tan(2x) = O/AO = B/2
A = A-C
H = sqrt( (B/2)^2 + (A-C)^2 )

What I want are exprsssions for sinx and cosx

Not sure if I'm perhaps missing something here - very possible in light of previous error - but just to repeat what the OP said: he's not primarily interested in sin(2x) or cos(2x) . He wants sin(x) and cos(x) in terms of A,B and C ? And he's already demonstrated above how to get sin(2x) and/or cos(2x) as needed en route.

 

[eys_physics]

But, if you, e.g, have an expression for $\cos(2x)$ it is easy to get $\cos x$ in terms of A, B and C.

 

[neilparker62]

Exactly - see post #9 above!