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Sine, cosine and tangent inequalities

Choose arcs, not just points An equation like $\sin x=m$ marks meeting points with a horizontal line. The inequality $\sin x>m$ chooses the arcs above this line; $\sin x

October 21, 2026 Read

Arcsine, arccosine and arctangent

Why is it necessary to restrict? An inverse needs to retrieve a single input from the output. Sine, cosine and tangent repeat values. To define inverses, we choose intervals in which each function is one-to-one and reaches its entire desired image. Restricted function Chosen domain Image Sine $[-\pi/2,\pi/2]$ $[-1,1]$ Cosine $[0,\pi]$ $[-1,1]$ Tangent $(-\pi/2,\pi/2)$ $\mathbb R$ In these intervals, sine and tangent are strictly increasing, and cosine is strictly decreasing. Each output appears only once; therefore, the inverses are well defined.

October 21, 2026 Read

Intervals and classical trigonometric equations

Filter a family To solve $\sin x=1/2$ to $[0,4\pi)$, first obtain the families $x=\pi/6+2k\pi$ and $x=5\pi/6+2k\pi$. Then impose the limits of the range. The solutions are: $$\frac\pi6,\quad\frac{5\pi}6,\quad\frac{13\pi}6,\quad\frac{17\pi}6.$$The inclusion sign of the ends matters. After squaring or multiplying by a variable expression, substitute the answers into the initial equation. A combination of sine and cosine To solve $a\sin x+b\cos x=c$, with $(a,b)\ne(0,0)$, set $R=\sqrt{a^2+b^2}$ and choose $\varphi$ with: $$\cos\varphi=\frac aR,\qquad\sin\varphi=\frac bR.$$This is possible because the squares of these two numbers add up to 1. Using the addition formula:

October 21, 2026 Read

Fundamental equations of sine, cosine and tangent

The equation looks for inputs A trigonometric equation asks for the values of the variable that make an equality true. Before transforming the expression, determine its domain. Divisions by functions that can be zero can eliminate solutions. For $\sin x=m$ or $\cos x=m$, $-1\le m\le1$ is required. For the tangent, any real $m$ is possible, but entries where $\cos x=0$ remain prohibited. Demonstration of sine families A horizontal line cuts the cycle at a maximum of two points. If one of them corresponds to $\alpha$, the other corresponds to $\pi-\alpha$. As full turns repeat the stitch:

October 21, 2026 Read

Transform sums into products and products into sums

Add two known formulas Write the formulas for $\sin(a+b)$ and $\sin(a-b)$. Adding them together, the terms with $\cos a\sin b$ cancel each other out: $$\sin(a+b)+\sin(a-b)=2\sin a\cos b.$$Now do $u=a+b$ and $v=a-b$. Then $a=(u+v)/2$ and $b=(u-v)/2$. This produces the first identity below; the other cases come from adding or subtracting the sine and cosine formulas: $$\sin u+\sin v=2\sin\frac{u+v}{2}\cos\frac{u-v}{2},$$ $$\sin u-\sin v=2\cos\frac{u+v}{2}\sin\frac{u-v}{2},$$ $$\cos u+\cos v=2\cos\frac{u+v}{2}\cos\frac{u-v}{2},$$ $$\cos u-\cos v=-2\sin\frac{u+v}{2}\sin\frac{u-v}{2}.$$These identities are valid for all reals $u,v$. In the latter case, the negative sign is essential.

October 21, 2026 Read

Arc half division and tangent formulas

Isolate the squares We start with the identities $\cos2u=1-2\sin^2u$ and $\cos2u=2\cos^2u-1$. Making $u=x/2$ and isolating: $$\sin^2\frac x2=\frac{1-\cos x}{2},\qquad \cos^2\frac x2=\frac{1+\cos x}{2}.$$Consequently: $$\sin\frac x2=\pm\sqrt{\frac{1-\cos x}{2}},\qquad \cos\frac x2=\pm\sqrt{\frac{1+\cos x}{2}}.$$The signs are not a free choice: they are determined by the image quadrant of $x/2$. If $0

October 21, 2026 Read

Sine, cosine and tangent of double and triple arc

An addition formula applied to itself Replace $b=a$ in addition formulas: $$\sin2a=2\sin a\cos a,$$ $$\cos2a=\cos^2a-\sin^2a.$$The fundamental identity allows two other forms for the cosine: $$\cos2a=1-2\sin^2a=2\cos^2a-1.$$The choice depends on the data available. If we only know the sine, the first alternative expression avoids calculating the cosine and choosing its sign. If tangents exist: $$\tan2a=\frac{2\tan a}{1-\tan^2a}.$$The form with $\tan a$ requires $\cos a\ne0$ and $1-\tan^2a\ne0$; does not cover all cases where $\tan2a$ exists.

October 21, 2026 Read

Arc addition and subtraction formulas

Rotate a vector Consider the horizontal and vertical unit vectors. After a rotation of $a$, they start to point, respectively, to $(\cos a,\sin a)$ and $(-\sin a,\cos a)$. The second is perpendicular to the first, in the positive direction. The point corresponding to $b$ is the sum of $\cos b$ times the horizontal vector with $\sin b$ times the vertical. Rotation preserves this decomposition. After turning $a$: $$\begin{aligned} (\cos(a+b),\sin(a+b)) &=\cos b(\cos a,\sin a)\\ &\quad+\sin b(-\sin a,\cos a). \end{aligned}$$Demonstrate and assemble the formulas Comparing the coordinates, we obtain the sum formulas. Replacing $b$ with $-b$ and using parity gives the difference:

October 20, 2026 Read

Complements, reduction identities, and parity

Reduce the table further In a right-angled triangle, the acute angles are complementary. The side opposite to one is adjacent to the other. Like this: $$\sin\left(\frac\pi2-x\right)=\cos x,\qquad \cos\left(\frac\pi2-x\right)=\sin x.$$For $\pi/4\le x\le\pi/2$, the complement $\pi/2-x$ belongs to $[0,\pi/4]$. In this way, values in the second half of the first quadrant can be calculated from those in the first half, exchanging sine for cosine. Also: $$\tan\left(\frac\pi2-x\right)=\cot x,$$ $$\sec\left(\frac\pi2-x\right)=\csc x,$$in their respective domains. The geometric interpretation by exchanging the legs demonstrates the relationships for acute angles; reading the cycle coordinates extends the identities to the other allowed values.

October 20, 2026 Read

Reduction of the second, third and fourth quadrants

A point and its reflections Choose $0<\alpha<\pi/2$, whose image is $(\cos\alpha,\sin\alpha)$. A reflection on the vertical axis changes the sign of the first coordinate; a reflection on the horizontal axis exchanges that of the second; symmetry about the origin exchanges both. These three operations generate the points of $\pi-\alpha$, $\pi+\alpha$ and $2\pi-\alpha$. Logo: Angle Sine Cosine Tangent $\pi-\alpha$ $\sin\alpha$ $-\cos\alpha$ $-\tan\alpha$ $\pi+\alpha$ $-\sin\alpha$ $-\cos\alpha$ $\tan\alpha$ $2\pi-\alpha$ $-\sin\alpha$ $\cos\alpha$ $-\tan\alpha$ Demonstration for the second quadrant The reflection of $P=(u,v)$ on the vertical axis is $P'=(-u,v)$. The height remains the same, but the horizontal position becomes negative. Since $P'$ corresponds to angle $\pi-\alpha$:

October 20, 2026 Read

Identities and three demonstration paths

An equality between functions A trigonometric identity is an equality valid for all entries of a declared domain. It is necessary to check both the values and the points at which the expressions exist. For example, $\sec^2x-\tan^2x=1$ is valid when $\cos x\ne0$. The right side also exists in the excluded points, but this does not extend the domain of the expression to the left. First way: transform one of the members To demonstrate $\sin x\tan x+\cos x=\sec x$, with $\cos x\ne0$, we start from the left:

October 20, 2026 Read

Fundamental relationships demonstrated

Pythagoras in the circle The point $P=(\cos x,\sin x)$ is one unit away from the origin. The projections on the axes form a right-angled triangle with sides $|\cos x|$ and $|\sin x|$. Applying Pythagoras, we obtain: $$\sin^2x+\cos^2x=1.$$On the axes, equality is verified directly: one coordinate is zero and the other has modulus 1. Therefore, the identity holds for every real $x$. Bring relationships together The definitions and previous identity provide:

October 20, 2026 Read