Double angle formulas are used to express sines, cosines, tangents, cotangents of an angle with a value of 2 α, using trigonometric functions of the angle α. This article will introduce all double angle formulas with proofs. Examples of application of formulas will be considered. In the final part, the formulas for triple and quadruple angles will be shown.
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List of double angle formulas
To convert double angle formulas, you should remember that angles in trigonometry have the form n α notation, where n is a natural number, the value of the expression is written without parentheses. Thus, the notation sin n α is considered to have the same meaning as sin (n α) . When denoting sin n α, we have a similar notation (sin α) n. The use of notation is applicable to all trigonometric functions with powers n.
Below are the double angle formulas:
sin 2 α = 2 · sin α · cos α cos 2 α = cos 2 α - sin 2 α , cos 2 α = 1 - 2 · sin 2 α , cos 2 α = 2 · cos 2 α - 1 t g 2 α = 2 t g α 1 - t g 2 α c t g 2 α - c t g 2 α - 1 2 c t g α
Note that these formulas sin and cos are applicable with any value of the angle α. The double angle tangent formula is valid for any value of α, where t g 2 α makes sense, that is, α ≠ π 4 + π 2 · z, z is any integer. The double angle cotangent exists for any α, where c t g 2 α is defined at α ≠ π 2 z.
The cosine of a double angle has the triple notation of a double angle. All of them are applicable.
Proof of double angle formulas
The proof of the formulas starts from the addition formulas. Let's apply the formulas for the sine of the sum:
sin (α + β) = sin α · cos β + cos α · sin β and the cosine of the sum cos (α + β) = cos α · cos β - sin α · sin β. Let's assume that β = α, then we get that
sin (α + α) = sin α · cos α + cos α · sin α = 2 · sin α · cos α and cos (α + α) = cos α · cos α - sin α · sin α = cos 2 α - sin 2 α
Thus, the formulas for the sine and cosine of the double angle sin 2 α = 2 · sin α · cos α and cos 2 α = cos 2 α - sin 2 α are proven.
The remaining formulas cos 2 α = 1 - 2 · sin 2 α and cos 2 α = 2 · cos 2 α - 1 lead to the form cos 2 α = cos 2 α = cos 2 α - sin 2 α, when replacing 1 with the sum of squares by the main identity sin 2 α + cos 2 α = 1 . We get that sin 2 α + cos 2 α = 1. So 1 - 2 sin 2 α = sin 2 α + cos 2 α - 2 sin 2 α = cos 2 α - sin 2 α and 2 cos 2 α - 1 = 2 cos 2 α - (sin 2 α + cos 2 α) = cos 2 α - sin 2 α.
To prove the formulas for the double angle of tangent and cotangent, we apply the equalities t g 2 α = sin 2 α cos 2 α and c t g 2 α = cos 2 α sin 2 α. After the transformation, we obtain that t g 2 α = sin 2 α cos 2 α = 2 · sin α · cos α cos 2 α - sin 2 α and c t g 2 α = cos 2 α sin 2 α = cos 2 α - sin 2 α 2 · sin α · cos α . Divide the expression by cos 2 α, where cos 2 α ≠ 0 with any value of α when t g α is defined. We divide another expression by sin 2 α, where sin 2 α ≠ 0 with any values of α, when c t g 2 α makes sense. To prove the double angle formula for tangent and cotangent, we substitute and get:
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Double angle formulas make it possible to express the trigonometric functions (sine, cosine, tangent, cotangent) of the angle `2\alpha` through these same functions of the angle `\alpha`.
The list below is the basic double angle formulas most commonly used in trigonometry. For cosine there are three of them, they are all equivalent and equally important.
`sin \ 2\alpha=` `2 \ sin \ \alpha \ cos \ \alpha`
`cos \ 2\alpha=cos^2 \alpha-sin^2 \alpha`, ` cos \ 2\alpha=1-2 \ sin^2 \alpha`, `cos \ 2\alpha=2 \ cos^2 \alpha-1`
`tg \ 2\alpha=\frac(2 \ tg \ \alpha)(1-tg^2 \alpha)`
`ctg \ 2\alpha=\frac(ctg^2 \alpha-1)(2 \ ctg \ \alpha)`
The following identities express all trigonometric functions of the angle ` 2\alpha` through the functions tangent and cotangent of the angle `\alpha`.
`sin \ 2\alpha=` `\frac (2 \ tg \ \alpha)(1+tg^2 \alpha)=\frac (2 \ ctg \ \alpha)(1+ctg^2 \alpha)=` `\frac 2(tg \ \alpha+ctg \ \alpha)`
`cos\2\alpha=` `\frac(1-tg^2\alpha)(1+tg^2\alpha)=\frac(ctg^2\alpha-1)(ctg^2\alpha+1) =` `\frac(ctg \ \alpha-tg \ \alpha)(ctg \ \alpha+tg \ \alpha)`
`tg \ 2\alpha=` `\frac(2 \ctg \ \alpha)(ctg^2 \alpha-1)=` `\frac 2(\ctg \ \alpha-tg \ \alpha)`
`ctg \ 2\alpha=\frac ( \ ctg \ \alpha-tg \ \alpha)2`
The formulas for cosine and sine of a double angle work for any angle `\alpha`. The formulas for the tangent of a double angle are valid for those `\alpha` for which `tg\2\alpha` is defined, that is, for ` \alpha\ne\frac\pi4+\frac\pi2 n, \n \in Z`. Similarly, for the cotangent they occur for those `\alpha` for which `ctg \2\alpha` is defined, that is, for ` \alpha\ne\frac\pi2 n, \n \in Z`.
Proof of double angle formulas
All double angle formulas are derived from the formulas for the sum and difference of angles of trigonometric functions.
Let's take two formulas for the sum of the angles of sine and cosine:
`sin(\alpha+\beta)=` `sin \ \alpha\ cos \ \beta+cos \ \alpha\ sin \ \beta` and `cos(\alpha+\beta)=` `cos \ \alpha\ cos \ \beta-sin \ \alpha\ sin \ \beta`. Take `\beta=\alpha`, then `sin(\alpha+\alpha)=` `sin \ \alpha\ cos \ \alpha+cos \ \alpha\ sin \ \alpha=2 \ sin \ \alpha \cos \ \alpha`, similar to `cos(\alpha+\alpha)=` `cos \ \alpha\ cos \ \alpha-sin \ \alpha\ sin \ \alpha=cos^2 \alpha-sin^2 \alpha`, which and proves double angle formulas for sine and cosine.
Two other equalities for the cosine ` cos \ 2\alpha=1-2 \ sin^2 \alpha ` and ` cos \ 2\alpha=2 \ cos^2 \alpha-1` are reduced to what has already been proven if we replace 1 in them to `sin^2 \alpha+cos^2 \alpha=1`. So `1-2 \sin^2 \alpha=` `sin^2 \alpha+cos^2 \alpha-2 \sin^2 \alpha=` `cos^2 \alpha-sin^2 \alpha` and ` 2 \cos^2 \alpha-1=` `2 \cos^2 \alpha-(sin^2 \alpha+cos^2 \alpha)=` `cos^2 \alpha-sin^2 \alpha`.
To prove the formulas for the tangent of a double angle and cotangent, we will use the definition of these functions. Let's write `tg \ 2\alpha` and `ctg \ 2\alpha` as `tg \ 2\alpha=\frac (sin \ 2\alpha)(cos \ 2\alpha)` and `ctg \ 2\alpha= \frac (cos\2\alpha)(sin\2\alpha)`. Applying the already proven double angle formulas for sine and cosine, we obtain `tg\2\alpha=\frac (sin\2\alpha)(cos\2\alpha)=\frac (2\sin\\alpha\cos\\alpha )(cos^2 \alpha-sin^2 \alpha)` and `ctg \ 2\alpha=\frac (cos \ 2\alpha)(sin \ 2\alpha)=` `\frac (cos^2 \alpha -sin^2 \alpha)(2 \ sin \ \alpha \cos \ \alpha)`.
In the case of tangent, we divide the numerator and denominator of the final fraction by `cos^2 \alpha`, for the cotangent, in turn, by `sin^2 \alpha`.
`tg \ 2\alpha=\frac (sin \ 2\alpha)(cos \ 2\alpha)=\frac (2 \ sin \ \alpha \ cos \ \alpha)(cos^2 \alpha-sin^2 \ alpha)=` \frac (\frac(2 \ sin \ \alpha \ cos \ \alpha)(cos^2 \alpha))(\frac(cos^2 \alpha-sin^2 \alpha)(cos^ 2 \alpha))=` `\frac (2 \cdot \frac( sin \alpha )(cos \alpha))(1-\frac(sin^2 \alpha)(cos^2 \alpha))=\frac (2\tg\\alpha)(1-tg^2 \alpha)`.
`ctg \ 2\alpha=\frac (cos \ 2\alpha)(sin \ 2\alpha)=` `\frac (cos^2 \alpha-sin^2 \alpha)(2 \ sin \ \alpha \ cos \ \alpha)=` `\frac (\frac(cos^2 \alpha-sin^2 \alpha)(sin^2 \alpha))(\frac(2 \ sin \ \alpha \cos \ \alpha)( sin^2 \alpha))=` `\frac (\frac(cos^2 \alpha)(sin^2 \alpha)-1)(2 \cdot \frac(cos \alpha)( sin \alpha ))= \frac(ctg^2 \alpha-1)(2 \ctg \ \alpha)`.
We also suggest watching the video to better consolidate the theoretical material:
Examples of using formulas to solve problems
Double angle formulas are most often used to convert trigonometric expressions. Let's look at some of the cases and how they can be applied in practice when solving specific problems.
Example 1. Check the validity of the double angle identities for `\alpha=30^\circ`.
Solution. Our formulas use two angles `\alpha` and `2\alpha`. The value of the first angle is specified in the condition, the second accordingly will be `2\alpha=60^\circ`. We also know the numerical values for all trigonometric functions of these angles. Let's write them down:
`sin 30^\circ=\frac 1 2`, `cos 30^\circ=\frac (\sqrt 3)2`, `tg 30^\circ=\frac (\sqrt 3)3`, `ctg 30 ^\circ=\sqrt 3` and
`sin 60^\circ=\frac (\sqrt 3)2`, `cos 60^\circ=\frac 1 2`, `tg 60^\circ=\sqrt 3`, `ctg 60^\circ=\ frac (\sqrt 3)3`.
Then we will have
`sin 60^\circ=2 sin 30^\circ cos 30^\circ=` `2 \cdot \frac 1 2 \cdot \frac (\sqrt 3)2=\frac (\sqrt 3)2`,
`cos 60^\circ=cos^2 30^\circ-sin^2 30^\circ=` `(\frac (\sqrt 3)2)^2 \cdot (\frac 1 2)^2=\frac 1 2`,
`tg 60^\circ=\frac(2 tg 30^\circ)(1-tg^2 30^\circ)=` `\frac(2 \cdot \frac (\sqrt 3)3)(1-( \frac (\sqrt 3)3)^2)=\sqrt 3`,
`ctg 60^\circ=\frac(ctg^2 30^\circ-1)(2 \ctg 30^\circ)=` `\frac((\sqrt 3)^2-1)(2 \cdot \ sqrt 3)=\frac (\sqrt 3)3`.
Which proves the validity of the equalities for the angle specified in the condition.
Example 2. Express `sin \frac (2\alpha)3` in terms of trigonometric functions of the angle `\frac (\alpha)6`.
Solution. Let's write the sine angle as follows: ` \frac (2\alpha)3=4 \cdot \frac (\alpha)6`. Then, by applying the double angle formula twice, we can solve our problem.
First, we will use the equality of the sine of the double angle: ` sin\frac (2\alpha)3=2 \cdot sin\frac (\alpha)3 \cdot cos\frac (\alpha)3 `, now we will again apply our formulas for sine and cosine respectively. As a result we get:
` sin\frac (2\alpha)3=2 \cdot sin\frac (\alpha)3 \cdot cos\frac (\alpha)3=` `2 \cdot (2 \cdot sin\frac (\alpha)6 \cdot cos\frac (\alpha)6) \cdot (cos^2\frac (\alpha)6-sin^2\frac (\alpha)6)=` `4 \cdot sin\frac (\alpha)6 \cdot cos^3 \frac (\alpha)6-4 \cdot sin^3\frac (\alpha)6 \cdot cos \frac (\alpha)6`.
Answer. ` sin\frac (2\alpha)3=` `4 \cdot sin\frac (\alpha)6 \cdot cos^3 \frac (\alpha)6-4 \cdot sin^3\frac (\alpha)6 \cdot cos \frac (\alpha)6`.
Triple angle formulas
These formulas, similar to the previous ones, make it possible to express the functions of the angle ` 3\alpha` through these same functions of the angle `\alpha`.
`sin \ 3\alpha=3 \ sin \ \alpha-4sin^3 \alpha`
`cos \ 3\alpha=4cos^3 \alpha-3 \ cos \ \alpha`
`tg \ 3\alpha=\frac(3 \ tg \ \alpha-tg^3 \alpha)(1-3 \ tg^2 \alpha)`
`ctg \ 3\alpha=\frac(ctg^3 \alpha-3 \ ctg \ \alpha)(3 \ ctg^2 \alpha-1)`
They can be proven using sum equalities and angle differences, as well as the double angle formulas that are well known to us.
`sin \ 3\alpha= sin (2\alpha+ \alpha)=` `sin 2\alpha cos \alpha+cos 2\alpha sin \alpha=` `2 sin \alpha cos \alpha cos \alpha+(cos^2 \alpha-sin^2 \alpha) sin \alpha=` `3 sin \alpha cos^2 \alpha-sin^3 \alpha`.
In the resulting formula, replace `sin \ 3\alpha=3 sin \alpha cos^2 \alpha-sin^3 \alpha` `cos^2\alpha` with `1-sin^2\alpha` and get `sin \ 3 \alpha=3 \sin \ \alpha-4sin^3 \alpha`.
Also for the cosine of a triple angle:
`cos \ 3\alpha= cos (2\alpha+ \alpha)=` `cos 2\alpha cos \alpha-sin 2\alpha sin \alpha=` `(cos^2 \alpha-sin^2 \alpha) cos \alpha-2 sin \alpha cos \alpha sin \alpha+=` `cos^3 \alpha-3 sin^2 \alpha cos \alpha`.
Replacing `cos \ 3\alpha=cos^3 \alpha-3 sin^2 \alpha cos \alpha` `sin^2\alpha` in the final equality with `1-cos^2\alpha`, we get `cos \ 3 \alpha=4cos^3 \alpha-3 \cos \ \alpha`.
Using the proven identities for sine and cosine, we can prove for tangent and cotangent:
`tg \ 3\alpha=\frac (sin \ 3\alpha)(cos \ 3\alpha)=` `\frac (3 sin \alpha cos^2 \alpha-sin^3 \alpha)(cos^3 \ alpha-3 sin^2 \alpha cos \alpha)=` `\frac (\frac(3 sin \alpha cos^2 \alpha-sin^3 \alpha)(cos^3 \alpha))(\frac(cos ^3 \alpha-3 sin^2 \alpha cos \alpha)(cos^3 \alpha))=` `\frac (3 \cdot \frac( sin \alpha )(cos \alpha)-\frac( sin^ 3 \alpha )(cos^3 \alpha))(1-3\frac(sin^2 \alpha)(cos^2 \alpha))=` `\frac(3 \ tg \ \alpha-tg^3 \ alpha)(1-3tg^2 \alpha)`;
`ctg \ 3\alpha=\frac (cos \ 3\alpha)(sin \ 3\alpha)=` `\frac (cos^3 \alpha-3 sin^2 \alpha cos \alpha)(3 sin \alpha cos^2 \alpha-sin^3 \alpha)=` `\frac (\frac(cos^3 \alpha-3 sin^2 \alpha cos \alpha)(sin^3 \alpha))(\frac(3 sin \alpha cos^2 \alpha-sin^3 \alpha)(sin^3 \alpha))=` `\frac (\frac( cos^3 \alpha )(sin^3 \alpha)-3 \cdot \ frac(cos \alpha)( sin \alpha ))(3\frac(cos^2 \alpha)(sin^2 \alpha)-1)=` `ctg \ 3\alpha=\frac(ctg^3 \alpha -3\ctg\\alpha)(3\ctg^2\alpha-1)`.
To prove the formulas for the angle ` 4\alpha`, you can represent it as ` 2 \cdot 2\alpha` and try the double angle formulas twice.
To derive similar equalities for the angle ` 5\alpha`, you can write it as ` 3\alpha + 2\alpha` and apply the identities of the sum and difference of angles and double and triple angles.
All formulas for other multiple angles are derived similarly, but they are rarely needed in practice.
In trigonometry, many formulas are easier to derive than to memorize. Cosine of double angle is a wonderful formula! It allows you to obtain formulas for reducing degrees and formulas for half angles.
So, we need the cosine of the double angle and the trigonometric unit:
They are even similar: in the double angle cosine formula it is the difference between the squares of the cosine and sine, and in the trigonometric unit it is their sum. If we express the cosine from the trigonometric unit:
and substitute it into the cosine of the double angle, we get:
This is another double angle cosine formula:
This formula is the key to obtaining the reduction formula:
So, the formula for reducing the degree of sine is:
If in it the alpha angle is replaced by a half angle alpha in half, and the double angle two alpha is replaced by an alpha angle, then we obtain the half angle formula for sine:
Now we can express the sine from the trigonometric unit:
Let's substitute this expression into the double angle cosine formula:
We got another formula for the cosine of a double angle:
This formula is the key to finding the formula for reducing the power of cosine and the half angle for cosine.
Thus, the formula for reducing the degree of cosine is:
If we replace α with α/2, and 2α with α, we obtain the formula for the half argument for the cosine:
Since tangent is the ratio of sine to cosine, the formula for tangent is:
Cotangent is the ratio of cosine to sine. Therefore, the formula for cotangent is:
Of course, in the process of simplifying trigonometric expressions, there is no point in deriving the formula for half an angle or reducing a degree every time. It is much easier to put a sheet of paper with formulas in front of you. And simplification will move faster, and visual memory will turn on memorization.
But it’s still worth deriving these formulas several times. Then you will be absolutely sure that during the exam, when it is not possible to use a cheat sheet, you will easily get them if the need arises.