# Trigonometry - Double Angle Identities

• Nusc
Oh, yes, ##D##. I think that was a better choice, but it doesn't matter.Using ##D = \tan(2x) = \frac{\sin(2x)}{\cos(2x)}##, you can express ##\cos(2x)## in terms of ##D##, and similarly for ##\sin(2x)##, and then solve for ##\sin(x)## and ##\cos(x)##.I'm not sure what you are trying to do with ##A## and ##B##. You might need to post a clear statement of the problem.Are you able to get an expression from sin(2x)/cos(
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.

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##.

Couldn't you simply write down:

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

Solve simultaneously.

neilparker62 said:
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##.

PeroK
Ray Vickson said:
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)?

Nusc said:
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.

Ray Vickson said:
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.

PeroK said:
##\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

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

neilparker62 said:
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).

Nusc said:
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##.

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

Nusc said:
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
Nusc said:
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##.

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

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
$$\frac{\pm\sqrt{1-\cos^2(2x)}}{\cos(2x)}=D$$.

Nusc said:
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.

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.

Exactly - see post #9 above!

## What are the double angle identities in trigonometry?

The double angle identities in trigonometry are formulas that express a trigonometric function of an angle in terms of two times that angle.

## What is the double angle identity for cosine?

The double angle identity for cosine is cos(2x) = cos^2(x) - sin^2(x).

## What is the double angle identity for sine?

The double angle identity for sine is sin(2x) = 2sin(x)cos(x).

## What is the double angle identity for tangent?

The double angle identity for tangent is tan(2x) = 2tan(x) / 1 - tan^2(x).

## How are double angle identities useful in trigonometry?

Double angle identities are useful in trigonometry because they allow us to simplify complex trigonometric expressions and solve equations involving trigonometric functions.

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