Trigonometrical ratios of an acute angle
Let $\theta$ be an angle of a right angled triangle. Then, the six trigonometric functions are defined as;
\[\sin\theta=\frac{PQ}{OP}=\frac{y}{r}=\frac{p}{h}=\frac{\text{perpendicular}}{\text{hypotenuse}}\]
\[\cos\theta=\frac{OQ}{OP}=\frac{x}{r}=\frac{b}{h}=\frac{\text{base}}{\text{hypotenuse}}\]
\[\tan\theta=\frac{PQ}{OQ}=\frac{y}{x}=\frac{p}{b}=\frac{\text{perpendicular}}{\text{base}}\]
\[\cot\theta=\frac{OQ}{PQ}=\frac{x}{y}=\frac{b}{p}=\frac{\text{base}}{\text{perpendicular}}\]
\[\sec\theta=\frac{OP}{OQ}=\frac{r}{x}=\frac{h}{b}=\frac{\text{hypotenuse}}{\text{base}}\]
\[\operatorname{cosec}\theta=\frac{OP}{OQ}=\frac{r}{x}=\frac{h}{b}=\frac{\text{hypotenuse}}{\text{perpendicular}}\]
Some basic properties
For trigonometric angles of any size, positive or negative, the following properties follow directly from the definitions:
1. Arms or side length
For a given angle of reference, each of the six trigonometric ratios has the same value whatever be the size of the arms (or triangle).
2. Elementary identities
The trigonometric ratios; sine, cosine and tangent of an angle are the reciprocals of cosecant, secant and contangent of the angle i.e. \[\sin\theta=\frac{1}{\operatorname{cosec}\theta},\;\cos\theta=\frac{1}{\sec\theta}\;\text{and}\;\tan\theta=\frac{1}{\cot\theta}\]
3. Limit of values
Since hypotenuse is the greatest side in a right angled triangle, \[\sin\theta=\frac{p}{h}=\frac{y}{r}\;\text{and}\;\cos\theta=\frac{b}{h}=\frac{x}{r}\] are both less than $1$. \[\therefore -1≤\sin\theta≤1 \;\text{and}\; -1≤\cos\theta≤1\]
4. Quadrant rule of signs
In the first quadrant, the given angle is positive and acute. So, x-coordinate, y-coordinate and the distance $r$ of the given point $P(x,y)$ from the origin are all positive. Hence, the six trigonometric ratios are all positive.
In the second quadrant, the given angle is positive and obtuse, so $y>0$, $r>0$ but $x<0$. Thus, \[\sin\theta=\frac{y}{r}>0,\;\cos\theta=\frac{x}{r}<0\;\text{and}\;\tan\theta=\frac{y}{x}<0\]
Similarly, signs of the various trigonometric functions can be found in the four quadrants which is shown in the following table.
| I | II | III | IV | |
| $\sin\theta$ | + | + | – | – |
| $\cos\theta$ | + | – | – | + |
| $\tan\theta$ | + | – | + | – |
Pythagorean Identities
Pythagorean theorem states that “the sum of the squares on the two sides forming the right angle is equal to the square on the hypotenuse.” By using this theorem, we can find an unknown side whenever two other sides are given in any right angled triangle.
From Pythagorean theorem, we have, \[h^2=p^2+b^2\] Dividing both sides ny $h^2$, we get, \[1=\left(\frac{p}{h}\right)^2+\left(\frac{b}{h}\right)^2\] \[1=\sin^2\theta+\cos^2\theta\] \[\therefore \sin^2\theta+\cos^2\theta=1\text{ __(1)}\]
Similarly, we can find that \[\sec^2\theta=1+\tan^2\theta\text{ __(2)}\] \[\operatorname{cosec}^2\theta =1+\cot^2\theta\text{ __(3)}\]
From $\text{(1)}$, $\text{(2)}$ and $\text{(3)}$, we can easily obtain the following identities: \[\sin^2\theta=1-\cos^2\theta\] \[\cos^2\theta=1-\sin^2\theta\] \[\cos\theta=\sqrt{1-\sin^2\theta}\] \[\sec^2\theta-\tan^2\theta=1\] \[\operatorname{cosec}^2\theta-\cot^2\theta=1, \;\;\text{etc.}\]
Addition and Subtraction Formulae
\[\sin(A+B)=\sin A\cos B+\cos A\sin B\] \[\sin(A-B)=\sin A\cos B-\cos A\sin B\] \[\cos(A+B)=\cos A\cos B-\sin A\sin B\] \[\cos(A-B)=\cos A\cos B+\sin A\sin B\]
\[\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\] \[\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}\] \[\cot(A+B)=\frac{\cot A\cot B-1}{\cot B+\cot A}\] \[\cot(A-B)=\frac{\cot A\cot B+1}{\cot B-\cot A}\]
Some Trigonometric Formulae Deduced from the Addition and Subtraction Formulae
1. Put $A=0°$, then \[\sin(-B)=-\sin B\] \[\cos(-B)=\cos B\] \[\tan(-B)=-\tan B\]
2. Put $A=90°$, then \[\sin(90°-B)=\cos B\] \[\cos(90°-B)=\sin B\] \[\tan(90°-B)=\cot B\] Also, \[\sin(90°+B)=\cos B\] \[\cos(90°+B)=-\sin B\] \[\tan(90°+B)=-\cot B\]
3. Put $A=180°$, then \[\sin(180°-B)=\sin B\] \[\cos(180°-B)=-\cos B\] \[\tan(180°-B)=-\tan B\] Also, \[\sin(180°+B)=-\sin B\] \[\cos(180°+B)=\cos B\] \[\tan(180°+B)=-\tan B\]
Multiple and Submultiple Angle Formulae
1. Double Angle Formulae
Put $B=A$ in the addition formulae, we get, \[\sin 2A=2\sin A\cos A\] \[\cos 2A=\cos^2A-\sin^2A\] \[=2\cos^2A-1\] \[=1-2\sin^2A\]
\[\tan 2A=\frac{2\tan A}{1-\tan^2A}\] \[\sin 2A=\frac{2\tan A}{1+\tan^2A}\] \[\cos 2A=\frac{1-\tan^2 A}{1+\tan^2A}\] \[\text{Also,}\;\;1-\cos 2A=2\sin^2A\] \[\text{and,}\;\;1+\cos 2A=2\cos^2A\]
2. Half Angle Formulae
Replace $A$ by $\frac{A}{2}$ in Double Angle Formulae, we get, \[\sin A=2\sin\frac{A}{2}\cos\frac{A}{2}\] \[\cos A=\cos^2\frac{A}{2}-\sin^2\frac{A}{2}\] \[=2\cos^2\frac{A}{2}-1\] \[=1-2\sin^2\frac{A}{2}\]
\[\tan A=\frac{2\tan\frac{A}{2}}{1-\tan^2\frac{A}{2}}\] \[\sin A=\frac{2\tan\frac{A}{2}}{1+\tan^2\frac{A}{2}}\] \[\cos A=\frac{1-\tan^2\frac{A}{2}}{1+\tan^2\frac{A}{2}}\] \[\text{Also,}\;\;1-\cos A=2\sin^2\frac{A}{2}\] \[\text{and,}\;\;1+\cos A=2\cos^2\frac{A}{2}\]
3. Triple Angle Formulae
Put $B=2A$ in the addition formulae, we get, \[\sin 3A=3\sin A-4\sin^3A\] \[\cos 3A=4\cos^3A-3\cos A\] \[\tan 3A=\frac{3\tan A-\tan^3A}{1-3\tan^2A}\]
Transformation Formulae
By using the addition and subtraction formulae, we can deduce the following formulae:
\[\sin(A+B)+\sin(A-B)=2\sin A\cos B\] \[\sin(A+B)-\sin(A-B)=2\cos A\sin B\] \[\cos(A+B)+\cos(A-B)=2\cos A\cos B\] \[\cos(A-B)-\cos(A+B)=2\sin A\sin B\]
If we replace $A+B$ and $A-B$ by $C$ and $D$ respectively, then, we have the following formulae:
\[\sin C+\sin D=2\sin\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)\] \[\sin C-\sin D=2\cos\left(\frac{C+D}{2}\right)\sin\left(\frac{C-D}{2}\right)\] \[\cos C+\cos D=2\cos\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)\] \[\cos C-\cos D=2\sin\left(\frac{C+D}{2}\right)\sin\left(\frac{D-C}{2}\right)\]
Conditional Identities
Trigonometrical identities involving certain conditions are known as conditional identities. Generally, the condition will be $A+B+C=π$ where $A$, $B$ and $C$ are the angles of a triangle. Now, let us see the following results:
If $A+B+C=π$, then $A+B=π-C$. \[\sin(A+B)=\sin(π-C)=\sin C\] \[\cos(A+B)=\cos(π-C)=-\cos C\] \[\tan(A+B)=\tan(π-C)=-\tan C\]
Again, if $A+B+C=\frac{π}{2}$, then $\frac{A+B}{2}=\frac{π}{2}-\frac{C}{2}$. \[\sin\left(\frac{A+B}{2}\right)=\sin\left(\frac{π}{2}-\frac{C}{2}\right)=\cos\frac{C}{2}\] \[\cos\left(\frac{A+B}{2}\right)=\cos\left(\frac{π}{2}-\frac{C}{2}\right)=\sin\frac{C}{2}\] \[\tan\left(\frac{A+B}{2}\right)=\tan\left(\frac{π}{2}-\frac{C}{2}\right)=\cot\frac{C}{2}\;\;\text{etc.}\]
Next: Trigonometric Equation