Trigonometry Basics

What is trigonometry?

Trigonometry is the study of the relationship between the side lengths and angles of a triangle. It gets its name from the Greek trigonon (triangle) and metron (measure). In a right triangle, relative to an acute angle θ\theta: the hypotenuse is the side opposite the right angle, the opposite side is opposite θ\theta, and the adjacent side is next to θ\theta.

The six ratios

The three main ratios are sine, cosine, and tangent, defined for an angle θ\theta as:

sin⁡θ=oppositehypotenuse,cos⁡θ=adjacenthypotenuse,tan⁡θ=oppositeadjacent\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}, \quad \cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}, \quad \tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}

Three more ratios are the reciprocals of these:

csc⁡θ=1sin⁡θ,sec⁡θ=1cos⁡θ,cot⁡θ=1tan⁡θ\csc\theta = \dfrac{1}{\sin\theta}, \quad \sec\theta = \dfrac{1}{\cos\theta}, \quad \cot\theta = \dfrac{1}{\tan\theta}

A key fact: the values of these ratios for a given angle don't depend on the size of the triangle, only on the angle itself, since any two right triangles with the same acute angle are similar (their sides are all in the same proportion). This is the foundation trigonometry is built on.

Worked example: finding a building's height

One common application is finding the height of a building without measuring it directly. Stand at a point CC, a distance of 250 m from the building (BCBC), and measure the angle of elevation to the top of the building as 53°. Using the tangent function:

tan⁡53∘=ABBC=AB250\tan 53^\circ = \dfrac{AB}{BC} = \dfrac{AB}{250}

Since tan⁡53∘≈1.327\tan 53^\circ \approx 1.327,

1.327=AB250  ⟹  AB=331.75 m1.327 = \dfrac{AB}{250} \implies AB = 331.75 \text{ m}

The building's height comes out to 331.75 m, without ever measuring it directly.

Standard angle values

It helps to know the ratios for a handful of standard angles: 0°, 30°, 45°, 60°, and 90°. There's a pattern worth noticing: for sine, the numerator (as 0,1,2,3,4\sqrt{0}, \sqrt{1}, \sqrt{2}, \sqrt{3}, \sqrt{4} over 2) increases from 0° to 90°; for cosine, the same five values repeat but in reverse. Tangent is just sine divided by cosine, and cosecant/secant/cotangent are the reciprocals of sine/cosine/tangent, so none of them need to be memorized separately.

0°30°45°60°90°
sin θ0012\frac{1}{2}22\frac{\sqrt{2}}{2}32\frac{\sqrt{3}}{2}11
cos θ1132\frac{\sqrt{3}}{2}22\frac{\sqrt{2}}{2}12\frac{1}{2}00
tan θ0013\frac{1}{\sqrt{3}}113\sqrt{3}not defined
csc θnot defined222\sqrt{2}23\frac{2}{\sqrt{3}}11
sec θ1123\frac{2}{\sqrt{3}}2\sqrt{2}22not defined
cot θnot defined3\sqrt{3}1113\frac{1}{\sqrt{3}}00

Trigonometric identities

An identity is an equation involving trigonometric ratios that holds for every value of θ\theta. The three Pythagorean identities:

  • sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1
  • sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1
  • csc⁡2θ−cot⁡2θ=1\csc^2\theta - \cot^2\theta = 1

The double-angle and triple-angle identities:

  • sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin\theta\cos\theta
  • cos⁡2θ=cos⁡2θ−sin⁡2θ\cos 2\theta = \cos^2\theta - \sin^2\theta
  • tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan 2\theta = \dfrac{2\tan\theta}{1 - \tan^2\theta}
  • sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta = 3\sin\theta - 4\sin^3\theta
  • cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4\cos^3\theta - 3\cos\theta
  • tan⁡3θ=3tan⁡θ−tan⁡3θ1−3tan⁡2θ\tan 3\theta = \dfrac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}

Trigonometry: Solved Examples

Seven worked examples, one per question type: ratios from one given ratio, identities, standard-angle values, co-function identities and product-to-sum.

Finding all ratios from one

Given sec⁡θ=1312\sec\theta = \dfrac{13}{12}, find all other trigonometric ratios.

Solution: sec⁡θ=hypotenuseadjacent\sec\theta = \dfrac{\text{hypotenuse}}{\text{adjacent}}, so take hypotenuse = 13 and adjacent = 12. By the Pythagorean theorem, opposite = 132−122=5\sqrt{13^2 - 12^2} = 5. That gives:

sin⁡θ=513,cos⁡θ=1213,tan⁡θ=512,csc⁡θ=135,cot⁡θ=125\sin\theta = \dfrac{5}{13}, \quad \cos\theta = \dfrac{12}{13}, \quad \tan\theta = \dfrac{5}{12}, \quad \csc\theta = \dfrac{13}{5}, \quad \cot\theta = \dfrac{12}{5}

An identity check

If 3cot⁡A=43\cot A = 4, check whether 1−tan⁡2A1+tan⁡2A\dfrac{1-\tan^2 A}{1+\tan^2 A} equals cos⁡2A−sin⁡2A\cos^2 A - \sin^2 A.

Solution: cot⁡A=43\cot A = \dfrac{4}{3}, so tan⁡A=34\tan A = \dfrac{3}{4}: a 3-4-5 triangle gives sin⁡A=35\sin A = \dfrac{3}{5}, cos⁡A=45\cos A = \dfrac{4}{5}.

1−tan⁡2A1+tan⁡2A=1−9161+916=725,cos⁡2A−sin⁡2A=1625−925=725\dfrac{1-\tan^2 A}{1+\tan^2 A} = \dfrac{1 - \frac{9}{16}}{1 + \frac{9}{16}} = \dfrac{7}{25}, \qquad \cos^2 A - \sin^2 A = \dfrac{16}{25} - \dfrac{9}{25} = \dfrac{7}{25}

Both sides equal 725\dfrac{7}{25}, so the identity holds here.

Evaluating a standard-angle expression

Evaluate cos⁡45∘sec⁡30∘+csc⁡60∘\dfrac{\cos 45^\circ}{\sec 30^\circ + \csc 60^\circ}.

Solution: cos⁡45∘=12\cos 45^\circ = \dfrac{1}{\sqrt{2}}, and since cos⁡30∘=sin⁡60∘=32\cos 30^\circ = \sin 60^\circ = \dfrac{\sqrt{3}}{2}, both sec⁡30∘\sec 30^\circ and csc⁡60∘\csc 60^\circ equal 23\dfrac{2}{\sqrt{3}}:

1223+23=1243=342=68\dfrac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}} + \frac{2}{\sqrt{3}}} = \dfrac{\frac{1}{\sqrt{2}}}{\frac{4}{\sqrt{3}}} = \dfrac{\sqrt{3}}{4\sqrt{2}} = \dfrac{\sqrt{6}}{8}

Solving for two unknown angles

If tan⁡(A+B)=3\tan(A+B) = \sqrt{3} and tan⁡(A−B)=13\tan(A-B) = \dfrac{1}{\sqrt{3}}, with 0∘<A+B≤90∘0^\circ < A+B \le 90^\circ and A>BA > B, find AA and BB.

Solution: tan⁡(A+B)=tan⁡60∘⇒A+B=60∘\tan(A+B) = \tan 60^\circ \Rightarrow A+B = 60^\circ, and tan⁡(A−B)=tan⁡30∘⇒A−B=30∘\tan(A-B) = \tan 30^\circ \Rightarrow A-B = 30^\circ. Adding the two equations gives 2A=90∘2A = 90^\circ, so A=45∘A = 45^\circ and B=15∘B = 15^\circ.

Co-function identities

Every trig ratio has a complementary partner: sin⁡(90∘−θ)=cos⁡θ\sin(90^\circ-\theta)=\cos\theta, tan⁡(90∘−θ)=cot⁡θ\tan(90^\circ-\theta)=\cot\theta, and csc⁡(90∘−θ)=sec⁡θ\csc(90^\circ-\theta)=\sec\theta. Rewriting one ratio in terms of its complement often makes an expression collapse.

Evaluate sin⁡18∘cos⁡72∘\dfrac{\sin 18^\circ}{\cos 72^\circ}.

Solution: Since 18∘=90∘−72∘18^\circ = 90^\circ - 72^\circ, sin⁡18∘=cos⁡72∘\sin 18^\circ = \cos 72^\circ, so the ratio is cos⁡72∘cos⁡72∘=1\dfrac{\cos 72^\circ}{\cos 72^\circ} = 1.

Proving an identity

Prove that (csc⁡θ−cot⁡θ)2=1−cos⁡θ1+cos⁡θ\left(\csc\theta - \cot\theta\right)^2 = \dfrac{1-\cos\theta}{1+\cos\theta}.

Solution: Write both terms on the left over a common denominator, then simplify using the difference of squares:

(1sin⁡θ−cos⁡θsin⁡θ)2=(1−cos⁡θ)2sin⁡2θ=(1−cos⁡θ)2(1−cos⁡θ)(1+cos⁡θ)=1−cos⁡θ1+cos⁡θ\left(\dfrac{1}{\sin\theta} - \dfrac{\cos\theta}{\sin\theta}\right)^2 = \dfrac{(1-\cos\theta)^2}{\sin^2\theta} = \dfrac{(1-\cos\theta)^2}{(1-\cos\theta)(1+\cos\theta)} = \dfrac{1-\cos\theta}{1+\cos\theta}

Worked example: a product-to-sum calculation

Calculate sin⁡65∘sin⁡25∘\sin 65^\circ \sin 25^\circ.

Solution: Since 25∘=90∘−65∘25^\circ = 90^\circ - 65^\circ, sin⁡25∘=cos⁡65∘\sin 25^\circ = \cos 65^\circ, turning the product into sin⁡65∘cos⁡65∘\sin 65^\circ \cos 65^\circ. Using the double-angle identity 2sin⁡θcos⁡θ=sin⁡2θ2\sin\theta\cos\theta = \sin 2\theta:

sin⁡65∘cos⁡65∘=12sin⁡130∘=12sin⁡50∘\sin 65^\circ \cos 65^\circ = \dfrac{1}{2}\sin 130^\circ = \dfrac{1}{2}\sin 50^\circ

Continue learning

Every trigonometric ratio here comes directly from the right-triangle relationship the Pythagorean theorem describes.