NOTE!!!Don't overthink Tangent problems! You don't always have to use the whole formal equation stated above. Isn't Tanget= (Sin / Cosine)??
So woudn't it only make sense that if Tanx= (Sinx / Cosx) , then Tan2x = (Sin2x / Cos2x)!
If you already have the sine and cosine values in the problem, why do extra work by plugging into that long Tanget formula??
If you have already found Sin2x and Cos2x, finding Tan2x is a no brainer!

Sin(2a)Proof: The proof of the double angle formula for sine proceeds as follows.

external image anglesum14.gif

Cos(2a)
Why are there three identities for cos(2a)? Cos(2a)'s different identities are easily explained by the Pythagorean identities. The Pythagorean identity you need to know is sin2(a) + cos2(a) = 1.

cos(2a) = 2cos2(a) -1 because you can plug 1 - cos2(a) in for sin2(a). cos(2a) = cos2(a) - sin2(a) cos(2a) = cos2(a) - (1- cos2(a))
cos(2a) = cos2(a) - 1+cos2(a)
cos(2a) = 2cos2(a) -1

cos(2a) = 1- 2sin2(a) because you can plug 1-sin2(a) --in for cos2(a). cos(2a) = cos2(a) - sin2(a) cos(2a) = (1- sin2(a)) - sin2a
cos(2a) = 1- sin2(a) - sin2(a)
cos(2a) = 1 - 2sin2(a)

Proving the Three Identities for Cos(2A): Identity #1:
cos2x = cos2x - sin2x
cosxcosx -sinxsinx =cos2x - sin2x *use sum & difference formulas*
cos2x - sin2x = cos2x - sinx

Use the double-angle identity to find the exact value for cos 2 x given that sin x = .
Because sin x is positive, angle x must be in the first or second quadrant. The sign of cos 2 x will depend on the size of angle x. If 0° < x < 45° or 135° < x < 180°, then 2 x will be in the first or fourth quadrant and cos2 x will be positive. On the other hand, if 45° < x < 90° or 90° < x < 135”, then 2 x will be in the second or third quadrant and cos 2 x will be negative.

Tan(2a)
There is a much easier way to use the identity of tan(2a) as long as you remember that tan(a)=sin(a)/cos(a). Keeping this in mind, you can substitute the following for tan's Double Angle Identity: sin(2a)
cos(2a) Alternate Proof: Let positive angles A and B be given, whose sum is less than 90 degrees. Construct segment PU with length 1. Construct triangle TPU so that angle TPU is equal to angle A, and angle TUP is equal to the complement of A. Construct the circumscribed rectangle PQRS so that angle QPT is equal to angle B, angle QPU is equal to the sum of angles A and B, T is on segment QR and U is on segment RS. Note that angle RTU is also equal to angle B.

: The following identities are true for all values for which they are defined:

external image anglesum15.gif

♦ Proof: To find the power-reducing formula for the sine, we start with the cosine double angle formula and replace the cosine squared term using the Pythagorean identity. The resulting equation can be solved for the sine squared term. The proofs of the power-reducing formulas for the other five functions are similar.♦ Videos

Double Angle theorem and the unit circle

When using the double angle theorem, it is often helpful to draw triangles. If the triangle is one of the special right triangles found on the unit circle then the corresponding sin and cos sides can be used. If the triangle is not a special right triangle then sin and cos must be calculated using the opposite over hypotenuse and adjacent over hypotenuse rules.
Example:

The cosine would be 12/13 and the sine would be 5/13.

If the triangle is a special right triangle on the unit circle it would be:

Sine:
√3/2
Cosine:
½
Note: tangent is still opposite over adjacent.

_

Different double angle identity problems might come up concerning the unit circle. You may the following type of question:

Find the exact values of sin2a, cos2a, and tan2a given that ‘a’ terminates in the given quadrant.

GIVEN: sin(a)= 11/61 tan(a)<0

ASSESS IT!: Sine is positive and Tangent is negative. This means that we are in Quadrant 2
Now we draw out our triangle with the sine values and use Pythagorean triples to find the last side of the triangle is -60. (This is negative because aren’t we going left of zero!?) Draw out the triangle if you need a visual.

SOLVE:
We can figure out according to a drawing of the triangle that
cos(a) = - 60 / 61
sin(a) = 11 / 61

Tan2a = -1320 / 3721 x 3721 / 3479 = -1320 / 3479
(Isn’t Tangent Sine over Cosine so put the sine answer over the cosine answer and multiply by the reciprocal!)

Double Angle Theorem and Identities

## Table of Contents

NOTE!!!Don't overthink Tangent problems! You don't always have to use the whole formal equation stated above. Isn't Tanget= (Sin / Cosine)??

So woudn't it only make sense that if

Tanx= (Sinx / Cosx), thenTan2x = (Sin2x / Cos2x)!If you already have the sine and cosine values in the problem, why do extra work by plugging into that long Tanget formula??

If you have already found Sin2x and Cos2x, finding Tan2x is a no brainer!

Sin(2a)The proof of the double angle formula for sine proceeds as follows.Proof:Cos(2a)Cos(2a)'s different identities are easily explained by the Pythagorean identities. The Pythagorean identity you need to know is sin2(a) + cos2(a) = 1.Why are there three identities for cos(2a)?

cos(2a) = 2cos2(a) -1because you can plug1 - cos2(a)in forsin2(a).cos(2a) = cos2(a) - sin2(a)cos(2a) = cos2(a) - (1- cos2(a))

because you can plugcos(2a) = cos2(a) - 1+cos2(a)

cos(2a) = 2cos2(a) -1

cos(2a) = 1- 2sin2(a)

1-sin2(a)--in forcos2(a).cos(2a) = cos2(a) - sin2(a)cos(2a) = (1- sin2(a)) - sin2acos(2a) = 1- sin2(a) - sin2(a)

cos(2a) = 1 - 2sin2(a)

Proving the Three Identities for Cos(2A):Identity #1:cos2x = cos2x - sin2x

cosxcosx -sinxsinx =cos2x - sin2x

*use sum & difference formulas*cos2x - sin2x = cos2x - sinx

Identity #2:cos2x = 2cos2x -1

cosxcosx - sinxsinx = 2cos2x -1

*use sum and difference formulas*cos2x - sin2x = 2cos2x -1

cos2x - (1- cos2x) = 2cos2x -1

*use pythagorean identities*cos2x -1+cos2x = 2cos2x -1

2cos2x -1 = 2cosx -1

Identity #3:cos2x = 1- 2sin2x

cos(x+x) =1- 2sin2x

cosxcosx - sinxsinx =1- 2sin2x

*use sum and difference formulas*cos2x - sin2x = 1- 2sin2x

(1-sin2x) - sin2x =1- 2sin2x

*use pythagorean identities*1- 2sin2x = 1- 2sin2x

Example for Cos2A:

xgiven that sinx= .Because sin

xis positive, anglexmust be in the first or second quadrant. The sign of cos 2xwill depend on the size of anglex. If 0° <x< 45° or 135° <x< 180°, then 2xwill be in the first or fourth quadrant and cos2xwill be positive. On the other hand, if 45° <x< 90° or 90° <x< 135”, then 2xwill be in the second or third quadrant and cos 2xwill be negative.Tan(2a)There is a much easier way to use the identity of tan(2a) as long as you remember that tan(a)=sin(a)/cos(a). Keeping this in mind, you can substitute the following for tan's Double Angle Identity:

sin(2a)cos(2a)

Let positive anglesAlternate Proof:AandBbe given, whose sum is less than 90 degrees. Construct segmentPUwith length 1. Construct triangleTPUso that angleTPUis equal to angleA, and angleTUPis equal to the complement ofA. Construct the circumscribed rectanglePQRSso that angleQPTis equal to angleB, angleQPUis equal to the sum of anglesAandB,Tis on segmentQRandUis on segmentRS. Note that angleRTUis also equal to angleB.By the Triangle Ratios Theorem, we have:

## Power-Reducing Theorem

: The following identities are true for all values for which they are defined:To find the power-reducing formula for the sine, we start with the cosine double angle formula and replace the cosine squared term using the Pythagorean identity. The resulting equation can be solved for the sine squared term. The proofs of the power-reducing formulas for the other five functions are similar.♦Proof:Videos## Double Angle theorem and the unit circle

When using the double angle theorem, it is often helpful to draw triangles. If the triangle is one of the special right triangles found on the unit circle then the corresponding sin and cos sides can be used. If the triangle is not a special right triangle then sin and cos must be calculated using the opposite over hypotenuse and adjacent over hypotenuse rules.Example:

The cosine would be 12/13 and the sine would be 5/13.

If the triangle is a special right triangle on the unit circle it would be:

Sine:

√3/2

Cosine:

½

Note: tangent is still opposite over adjacent.

_

Different double angle identity problems might come up concerning the unit circle. You may the following type of question:

Find the exact values of

sin2a, cos2a, and tan2agiven that ‘a’ terminates in the given quadrant.GIVEN: sin(a)= 11/61 tan(a)<0

ASSESS IT!: Sine is positive and Tangent is negative. This means that we are in Quadrant 2

Now we draw out our triangle with the sine values and use Pythagorean triples to find the last side of the triangle is -60. (This is negative because aren’t we going left of zero!?) Draw out the triangle if you need a visual.

SOLVE:

We can figure out according to a drawing of the triangle that

cos(a) = - 60 / 61

sin(a) = 11 / 61

SO...

Cos2a = 1 – 2 sin²a

Cos2a = 1 - 2 (11/ 61)²

Cos2a = 3479 / 3721And so..

Sin2a = 2 sin (a) cos (a)

Sin2a = 2 (11 / 61) (-60 / 61)

Sin2a = -1320 / 3721Tan2a = -1320 / 3721 x 3721 / 3479 = -1320 / 3479

(Isn’t Tangent Sine over Cosine so put the sine answer over the cosine answer and multiply by the reciprocal!)

And the answer shall set you free!!!