On this page you will find all the basic trigonometric formulas that will help you solve many exercises, greatly simplifying the expression itself.
Trigonometric formulas are mathematical equalities for trigonometric functions that are valid for all valid argument values.
The formulas set the ratios between the main trigonometric functions - sine, cosine, tangent, cotangent.
The sine of an angle is the y-coordinate of a point (the ordinate) on the unit circle. The cosine of an angle is the x-coordinate of a point (abscissa).
Tangent and cotangent are, respectively, the ratio of sine to cosine and vice versa.
`sin\\alpha,\cos\\alpha`
`tg \ \alpha=\frac(sin\ \alpha)(cos \ \alpha),` ` \alpha\ne\frac\pi2+\pi n, \ n \in Z`
`ctg \ \alpha=\frac(cos\ \alpha)(sin\ \alpha),` ` \alpha\ne\pi+\pi n, \ n \in Z`
And two that are used less often - secant, cosecant. They denote ratios of 1 to cosine and sine.
`sec \ \alpha=\frac(1)(cos\ \alpha),` ` \alpha\ne\frac\pi2+\pi n,\ n \in Z`
`cosec \ \alpha=\frac(1)(sin \ \alpha),` ` \alpha\ne\pi+\pi n,\ n \in Z`
From the definitions of trigonometric functions, you can see what signs they have in each quarter. The sign of the function depends only on which quadrant the argument is in.
When changing the sign of the argument from "+" to "-", only the cosine function does not change its value. It's called even. Its graph is symmetrical about the y-axis.
The remaining functions (sine, tangent, cotangent) are odd. When the sign of the argument is changed from "+" to "-", their value also changes to negative. Their graphs are symmetrical about the origin.
`sin(-\alpha)=-sin \ \alpha`
`cos(-\alpha)=cos \ \alpha`
`tg(-\alpha)=-tg \ \alpha`
`ctg(-\alpha)=-ctg \ \alpha`
Basic trigonometric identities
Basic trigonometric identities are formulas that establish a relationship between the trigonometric functions of one angle (`sin \ \alpha, \ cos \ \alpha, \ tg \ \alpha, \ ctg \ \alpha`) and which allow you to find the value of each of these functions through any known other.
`sin^2 \alpha+cos^2 \alpha=1`
`tg \ \alpha \cdot ctg \ \alpha=1, \ \alpha\ne\frac(\pi n) 2, \ n \in Z`
`1+tg^2 \alpha=\frac 1(cos^2 \alpha)=sec^2 \alpha,` ` \alpha\ne\frac\pi2+\pi n, \ n \in Z`
`1+ctg^2 \alpha=\frac 1(sin^2 \alpha)=cosec^2 \alpha,` ` \alpha\ne\pi n, \ n \in Z`
Formulas for the sum and difference of angles of trigonometric functions
The formulas for adding and subtracting arguments express the trigonometric functions of the sum or difference of two angles in terms of the trigonometric functions of these angles.
`sin(\alpha+\beta)=` `sin \ \alpha\ cos \ \beta+cos \ \alpha\ sin \ \beta`
`sin(\alpha-\beta)=` `sin \ \alpha\ cos \ \beta-cos \ \alpha\ sin \ \beta`
`cos(\alpha+\beta)=` `cos \ \alpha\ cos \ \beta-sin \ \alpha\ sin \ \beta`
`cos(\alpha-\beta)=` `cos \ \alpha\ cos \ \beta+sin \ \alpha\ sin \ \beta`
`tg(\alpha+\beta)=\frac(tg \ \alpha+tg \ \beta)(1-tg \ \alpha\ tg \ \beta)`
`tg(\alpha-\beta)=\frac(tg \ \alpha-tg \ \beta)(1+tg \ \alpha \ tg \ \beta)`
`ctg(\alpha+\beta)=\frac(ctg \ \alpha \ ctg \ \beta-1)(ctg \ \beta+ctg \ \alpha)`
`ctg(\alpha-\beta)=\frac(ctg \ \alpha\ ctg \ \beta+1)(ctg \ \beta-ctg \ \alpha)`
Double angle formulas
`sin \ 2\alpha=2 \ sin \ \alpha \ cos \ \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=cos^2 \alpha-sin^2 \alpha=` `1-2 \ sin^2 \alpha=2 \ cos^2 \alpha-1=` `\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 \ tg \ \alpha)(1-tg^2 \alpha)=` `\frac(2 \ ctg \ \alpha)(ctg^2 \alpha-1)=` `\frac 2( \ ctg \ \alpha-tg \ \alpha)`
`ctg \ 2\alpha=\frac(ctg^2 \alpha-1)(2 \ ctg \ \alpha)=` `\frac ( \ ctg \ \alpha-tg \ \alpha)2`
Triple Angle Formulas
`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)`
Half Angle Formulas
`sin \ \frac \alpha 2=\pm \sqrt(\frac (1-cos \ \alpha)2)`
`cos \ \frac \alpha 2=\pm \sqrt(\frac (1+cos \ \alpha)2)`
`tg \ \frac \alpha 2=\pm \sqrt(\frac (1-cos \ \alpha)(1+cos \ \alpha))=` `\frac (sin \ \alpha)(1+cos \ \ alpha)=\frac (1-cos \ \alpha)(sin \ \alpha)`
`ctg \ \frac \alpha 2=\pm \sqrt(\frac (1+cos \ \alpha)(1-cos \ \alpha))=` `\frac (sin \ \alpha)(1-cos \ \ alpha)=\frac (1+cos \ \alpha)(sin \ \alpha)`
Half, double, and triple argument formulas express the functions `sin, \cos, \tg, \ctg` of these arguments (`\frac(\alpha)2, \ 2\alpha, \ 3\alpha,… `) in terms of these same functions argument `\alpha`.
Their output can be obtained from the previous group (addition and subtraction of arguments). For example, double angle identities are easily obtained by replacing `\beta` with `\alpha`.
Reduction Formulas
Formulas of squares (cubes, etc.) of trigonometric functions allow you to go from 2,3, ... degrees to trigonometric functions of the first degree, but multiple angles (`\alpha, \ 3\alpha, \ ...` or `2\alpha, \ 4\alpha, \...`).
`sin^2 \alpha=\frac(1-cos \ 2\alpha)2,` ` (sin^2 \frac \alpha 2=\frac(1-cos \ \alpha)2)`
`cos^2 \alpha=\frac(1+cos \ 2\alpha)2,` ` (cos^2 \frac \alpha 2=\frac(1+cos \ \alpha)2)`
`sin^3 \alpha=\frac(3sin \ \alpha-sin \ 3\alpha)4`
`cos^3 \alpha=\frac(3cos \ \alpha+cos \ 3\alpha)4`
`sin^4 \alpha=\frac(3-4cos \ 2\alpha+cos \ 4\alpha)8`
`cos^4 \alpha=\frac(3+4cos \ 2\alpha+cos \ 4\alpha)8`
Formulas for the sum and difference of trigonometric functions
Formulas are transformations of the sum and difference of trigonometric functions of different arguments into a product.
`sin \ \alpha+sin \ \beta=` `2 \ sin \frac(\alpha+\beta)2 \ cos \frac(\alpha-\beta)2`
`sin \ \alpha-sin \ \beta=` `2 \ cos \frac(\alpha+\beta)2 \ sin \frac(\alpha-\beta)2`
`cos \ \alpha+cos \ \beta=` `2 \ cos \frac(\alpha+\beta)2 \ cos \frac(\alpha-\beta)2`
`cos \ \alpha-cos \ \beta=` `-2 \ sin \frac(\alpha+\beta)2 \ sin \frac(\alpha-\beta)2=` `2 \ sin \frac(\alpha+\ beta)2\sin\frac(\beta-\alpha)2`
`tg \ \alpha \pm tg \ \beta=\frac(sin(\alpha \pm \beta))(cos \ \alpha \ cos \ \beta)`
`ctg \ \alpha \pm ctg \ \beta=\frac(sin(\beta \pm \alpha))(sin \ \alpha \ sin \ \beta)`
`tg \ \alpha \pm ctg \ \beta=` `\pm \frac(cos(\alpha \mp \beta))(cos \ \alpha \ sin \ \beta)`
Here, the addition and subtraction of functions of one argument are converted into a product.
`cos \ \alpha+sin \ \alpha=\sqrt(2) \ cos (\frac(\pi)4-\alpha)`
`cos \ \alpha-sin \ \alpha=\sqrt(2) \sin (\frac(\pi)4-\alpha)`
`tg \ \alpha+ctg \ \alpha=2 \cosec \2\alpha;` `tg \ \alpha-ctg \ \alpha=-2 \ctg \2\alpha`
The following formulas convert the sum and difference of a unit and a trigonometric function to a product.
`1+cos \ \alpha=2 \ cos^2 \frac(\alpha)2`
`1-cos \ \alpha=2 \ sin^2 \frac(\alpha)2`
`1+sin \ \alpha=2 \ cos^2 (\frac (\pi) 4-\frac(\alpha)2)`
`1-sin \ \alpha=2 \ sin^2 (\frac (\pi) 4-\frac(\alpha)2)`
`1 \pm tg \ \alpha=\frac(sin(\frac(\pi)4 \pm \alpha))(cos \frac(\pi)4 \ cos \ \alpha)=` `\frac(\sqrt (2) sin(\frac(\pi)4 \pm \alpha))(cos \ \alpha)`
`1 \pm tg \ \alpha \ tg \ \beta=\frac(cos(\alpha \mp \beta))(cos \ \alpha \ cos \ \beta);` ` \ctg \ \alpha \ ctg \ \ beta \pm 1=\frac(cos(\alpha \mp \beta))(sin \ \alpha \ sin \ \beta)`
Function conversion formulas
Formulas for converting the product of trigonometric functions with `\alpha` and `\beta` arguments into the sum (difference) of these arguments.
`sin \ \alpha \ sin \ \beta =` `\frac(cos(\alpha - \beta)-cos(\alpha + \beta))(2)`
`sin\alpha \ cos\beta =` `\frac(sin(\alpha - \beta)+sin(\alpha + \beta))(2)`
`cos \ \alpha \ cos \ \beta =` `\frac(cos(\alpha - \beta)+cos(\alpha + \beta))(2)`
`tg \ \alpha \ tg \ \beta =` `\frac(cos(\alpha - \beta)-cos(\alpha + \beta))(cos(\alpha - \beta)+cos(\alpha + \ beta)) =` `\frac(tg \ \alpha + tg \ \beta)(ctg \ \alpha + ctg \ \beta)`
`ctg \ \alpha \ ctg \ \beta =` `\frac(cos(\alpha - \beta)+cos(\alpha + \beta))(cos(\alpha - \beta)-cos(\alpha + \ beta)) =` `\frac(ctg \ \alpha + ctg \ \beta)(tg \ \alpha + tg \ \beta)`
`tg \ \alpha \ ctg \ \beta =` `\frac(sin(\alpha - \beta)+sin(\alpha + \beta))(sin(\alpha + \beta)-sin(\alpha - \ beta))`
Universal trigonometric substitution
These formulas express trigonometric functions in terms of the tangent of a half angle.
`sin \ \alpha= \frac(2tg\frac(\alpha)(2))(1 + tg^(2)\frac(\alpha)(2)),` ` \alpha\ne \pi +2\ pi n, n \in Z`
`cos \ \alpha= \frac(1 - tg^(2)\frac(\alpha)(2))(1 + tg^(2)\frac(\alpha)(2)),` ` \alpha \ ne \pi +2\pi n, n \in Z`
`tg \ \alpha= \frac(2tg\frac(\alpha)(2))(1 - tg^(2)\frac(\alpha)(2)),` ` \alpha \ne \pi +2\ pi n, n \in Z,` ` \alpha \ne \frac(\pi)(2)+ \pi n, n \in Z`
`ctg \ \alpha = \frac(1 - tg^(2)\frac(\alpha)(2))(2tg\frac(\alpha)(2)),` ` \alpha \ne \pi n, n \in Z,` `\alpha \ne \pi + 2\pi n, n \in Z`
Cast formulas
Reduction formulas can be obtained using such properties of trigonometric functions as periodicity, symmetry, the shift property by a given angle. They allow arbitrary angle functions to be converted to functions whose angle is between 0 and 90 degrees.
For angle (`\frac (\pi)2 \pm \alpha`) or (`90^\circ \pm \alpha`):
`sin(\frac (\pi)2 - \alpha)=cos \ \alpha;` ` sin(\frac (\pi)2 + \alpha)=cos \ \alpha`
`cos(\frac (\pi)2 - \alpha)=sin \ \alpha;` ` cos(\frac (\pi)2 + \alpha)=-sin \ \alpha`
`tg(\frac (\pi)2 - \alpha)=ctg \ \alpha;` ` tg(\frac (\pi)2 + \alpha)=-ctg \ \alpha`
`ctg(\frac (\pi)2 - \alpha)=tg \ \alpha;` ` ctg(\frac (\pi)2 + \alpha)=-tg \ \alpha`
For angle (`\pi \pm \alpha`) or (`180^\circ \pm \alpha`):
`sin(\pi - \alpha)=sin \ \alpha;` ` sin(\pi + \alpha)=-sin \ \alpha`
`cos(\pi - \alpha)=-cos \ \alpha;` ` cos(\pi + \alpha)=-cos \ \alpha`
`tg(\pi - \alpha)=-tg \ \alpha;` ` tg(\pi + \alpha)=tg \ \alpha`
`ctg(\pi - \alpha)=-ctg \ \alpha;` ` ctg(\pi + \alpha)=ctg \ \alpha`
For angle (`\frac (3\pi)2 \pm \alpha`) or (`270^\circ \pm \alpha`):
`sin(\frac (3\pi)2 - \alpha)=-cos \ \alpha;` ` sin(\frac (3\pi)2 + \alpha)=-cos \ \alpha`
`cos(\frac (3\pi)2 - \alpha)=-sin \ \alpha;` ` cos(\frac (3\pi)2 + \alpha)=sin \ \alpha`
`tg(\frac (3\pi)2 - \alpha)=ctg \ \alpha;` ` tg(\frac (3\pi)2 + \alpha)=-ctg \ \alpha`
`ctg(\frac (3\pi)2 - \alpha)=tg \ \alpha;` ` ctg(\frac (3\pi)2 + \alpha)=-tg \ \alpha`
For angle (`2\pi \pm \alpha`) or (`360^\circ \pm \alpha`):
`sin(2\pi - \alpha)=-sin \ \alpha;` ` sin(2\pi + \alpha)=sin \ \alpha`
`cos(2\pi - \alpha)=cos \ \alpha;` ` cos(2\pi + \alpha)=cos \ \alpha`
`tg(2\pi - \alpha)=-tg \ \alpha;` ` tg(2\pi + \alpha)=tg \ \alpha`
`ctg(2\pi - \alpha)=-ctg \ \alpha;` ` ctg(2\pi + \alpha)=ctg \ \alpha`
Expression of some trigonometric functions in terms of others
`sin \ \alpha=\pm \sqrt(1-cos^2 \alpha)=` `\frac(tg \ \alpha)(\pm \sqrt(1+tg^2 \alpha))=\frac 1( \pm \sqrt(1+ctg^2 \alpha))`
`cos \ \alpha=\pm \sqrt(1-sin^2 \alpha)=` `\frac 1(\pm \sqrt(1+tg^2 \alpha))=\frac (ctg \ \alpha)( \pm \sqrt(1+ctg^2 \alpha))`
`tg \ \alpha=\frac (sin \ \alpha)(\pm \sqrt(1-sin^2 \alpha))=` `\frac (\pm \sqrt(1-cos^2 \alpha))( cos \ \alpha)=\frac 1(ctg \ \alpha)`
`ctg \ \alpha=\frac (\pm \sqrt(1-sin^2 \alpha))(sin \ \alpha)=` `\frac (cos \ \alpha)(\pm \sqrt(1-cos^ 2 \alpha))=\frac 1(tg \ \alpha)`
Trigonometry literally translates as "measurement of triangles". It begins to be studied at school, and continues in more detail at universities. Therefore, the basic formulas for trigonometry are needed, starting from the 10th grade, as well as for passing the exam. They denote connections between functions, and since there are many of these connections, there are quite a few formulas themselves. Remembering them all is not easy, and it is not necessary - if necessary, they can all be deduced.
Trigonometric formulas are used in integral calculus, as well as in trigonometric simplifications, calculations, and transformations.
Exercise.
Find the value of x at .
Decision.
To find the value of the function argument, at which it is equal to some value, means to determine for which arguments the value of the sine will be exactly the same as indicated in the condition.
In this case, we need to find out at what values the value of the sine will be equal to 1/2. This can be done in several ways.
For example, use , by which to determine at what values of x the sine function will be equal to 1/2.
Another way is to use . Let me remind you that the values of the sines lie on the Oy axis.
The most common way is to use , especially when it comes to such standard values for this function as 1/2.
In all cases, one should not forget about one of the most important properties of the sine - its period.
Let's find the value 1/2 for the sine in the table and see what arguments correspond to it. The arguments we are interested in are Pi / 6 and 5Pi / 6.
Write down all the roots that satisfy the given equation. To do this, we write down the unknown argument x of interest to us and one of the values of the argument obtained from the table, that is, Pi / 6. Let's write down for it, taking into account the sine period, all the values of the argument:
Let's take the second value and follow the same steps as in the previous case:
The complete solution to the original equation will be: 
and 
q can take the value of any integer.
The sine values are in the range [-1; 1], i.e. -1 ≤ sin α ≤ 1. Therefore, if |a| > 1, then the equation sin x = a has no roots. For example, the equation sin x = 2 has no roots.
Let's turn to some tasks.
Solve the equation sin x = 1/2.
Decision.
Note that sin x is the ordinate of the point of the unit circle, which is obtained as a result of the rotation of the point Р (1; 0) by the angle x around the origin.
An ordinate equal to ½ is present at two points of the circle M 1 and M 2.
Since 1/2 \u003d sin π / 6, then the point M 1 is obtained from the point P (1; 0) by turning through the angle x 1 \u003d π / 6, as well as through the angles x \u003d π / 6 + 2πk, where k \u003d +/-1, +/-2, …
The point M 2 is obtained from the point P (1; 0) as a result of turning through the angle x 2 = 5π/6, as well as through the angles x = 5π/6 + 2πk, where k = +/-1, +/-2, ... , i.e. at angles x = π – π/6 + 2πk, where k = +/-1, +/-2, ….
So, all the roots of the equation sin x = 1/2 can be found by the formulas x = π/6 + 2πk, x = π - π/6 + 2πk, where k € Z.
These formulas can be combined into one: x \u003d (-1) n π / 6 + πn, where n € Z (1).
Indeed, if n is an even number, i.e. n = 2k, then from formula (1) we obtain х = π/6 + 2πk, and if n is an odd number, i.e. n = 2k + 1, then from formula (1) we obtain х = π – π/6 + 2πk.
Answer. x \u003d (-1) n π / 6 + πn, where n € Z.
Solve the equation sin x = -1/2.
Decision.
The ordinate -1/2 have two points of the unit circle M 1 and M 2, where x 1 = -π/6, x 2 = -5π/6. Therefore, all the roots of the equation sin x = -1/2 can be found by the formulas x = -π/6 + 2πk, x = -5π/6 + 2πk, k ∈ Z.
We can combine these formulas into one: x \u003d (-1) n (-π / 6) + πn, n € Z (2).
Indeed, if n = 2k, then by formula (2) we obtain x = -π/6 + 2πk, and if n = 2k – 1, then by formula (2) we find x = -5π/6 + 2πk.
Answer. x \u003d (-1) n (-π / 6) + πn, n € Z.
Thus, each of the equations sin x = 1/2 and sin x = -1/2 has an infinite number of roots.
On the segment -π/2 ≤ x ≤ π/2, each of these equations has only one root:
x 1 \u003d π / 6 - the root of the equation sin x \u003d 1/2 and x 1 \u003d -π / 6 - the root of the equation sin x \u003d -1/2.
The number π/6 is called the arcsine of the number 1/2 and is written: arcsin 1/2 = π/6; the number -π/6 is called the arcsine of the number -1/2 and they write: arcsin (-1/2) = -π/6.
In general, the equation sin x \u003d a, where -1 ≤ a ≤ 1, on the segment -π / 2 ≤ x ≤ π / 2 has only one root. If a ≥ 0, then the root is enclosed in the interval; if a< 0, то в промежутке [-π/2; 0). Этот корень называют арксинусом числа а и обозначают arcsin а.
Thus, the arcsine of the number a € [–1; 1] such a number is called a € [–π/2; π/2], whose sine is a.
arcsin a = α if sin α = a and -π/2 ≤ x ≤ π/2 (3).
For example, arcsin √2/2 = π/4, since sin π/4 = √2/2 and – π/2 ≤ π/4 ≤ π/2;
arcsin (-√3/2) = -π/3, since sin (-π/3) = -√3/2 and – π/2 ≤ 
– π/3 ≤ π/2.
Similarly to how it was done when solving problems 1 and 2, it can be shown that the roots of the equation sin x = a, where |a| ≤ 1 are expressed by the formula
x \u003d (-1) n arcsin a + πn, n € Z (4).
We can also prove that for any a € [-1; 1] the formula arcsin (-a) = -arcsin a is valid.
From formula (4) it follows that the roots of the equation
sin x \u003d a for a \u003d 0, a \u003d 1, a \u003d -1 can be found using simpler formulas:
sin x \u003d 0 x \u003d πn, n € Z (5)
sin x \u003d 1 x \u003d π / 2 + 2πn, n € Z (6)
sin x \u003d -1 x \u003d -π / 2 + 2πn, n € Z (7)
site, with full or partial copying of the material, a link to the source is required.