# Trigonometric Ratios , Definition , Formula , Solved Examples

Consider the system of rectangular co-ordinate axes dividing the plane into four quadrants.

A line OP makes angle θ with the positive x-axis. The angle θ is said to be positive if measured in counter clockwise direction from the positive x-axis and is negative if measured in clockwise direction.

The positive values of the trigonometric ratios in the various quadrants are shown, the signs of the other ratios may be derived.

Note that ∠xoy = π/2, ∠xox’ = π , ∠xoy’ = 3π/2.

PiQi is positive if above the x-axis, negative if below the x-axis,  OPi is always taken as positive.

OQi is positive if along positive x-axis, negative if in the opposite direction.

$\large sin\angle P_i O Q_i = \frac{P_i Q_i}{O P_i}$

$\large cos\angle P_i O Q_i = \frac{O Q_i}{O P_i}$

$\large tan\angle P_i O Q_i = \frac{P_i Q_i}{O Q_i}$

(where  i  = 1, 2, 3, 4 )

Thus depending on signs of OQi and PiQi the various trigonometrical ratios will have different signs.

 α equals sinα cosα tanα cotα secα cosecα –θ $– sinθ$ cosθ $–tanθ$ $– cotθ$ secθ $–cosecθ$ 90° – θ cosθ sinθ cotθ tanθ cosecθ secθ 90° + θ cosθ $– sinθ$ $–cotθ$ $– tanθ$ $–cosecθ$ secθ 180°– θ sinθ $– cosθ$ $– tanθ$ $– cotθ$ $– secθ$ cosecθ 180°+ θ $– sinθ$ $– cosθ$ tanθ cotθ $– secθ$ $–cosecθ$ 360°– θ $– sinθ$ cosθ $– tanθ$ $– cotθ$ secθ $–cosecθ$ 360°+ θ sinθ cosθ tanθ cotθ secθ cosecθ

Notes:

Angle θ and 90°– θ are complementary angles, θ and 180°– θ are supplementary angles

sin(nπ + (–1)nθ) = sinθ , n∈I .

cos(2nπ ± θ) = cosθ , n∈I .

tan(nπ + θ) = tanθ , n∈I .

i.e. sine of general angle of the form nπ + (–1)nθ will have same sign as that of sine of angle θ and so on.

The same is true for the respective reciprocal functions also.

Notes:

Angle θ and 90°– θ are complementary angles, θ and 180°– θ
are supplementary angles

sin(nπ + (–1)nθ) = sinθ , n∈I .

cos(2nπ ± θ) = cosθ , n∈I .

tan(nπ + θ) = tanθ , n∈I .

i.e. sine of general angle of the form nπ + (–1)nθ will have same sign as that of sine of angle θ and so on. The same is true for the respective reciprocal functions also.

Illustration : Find the general value of θ satisfying both Sin θ = -1/2 and tan θ = 1/√3

Solution: Let us first find out θ lying between 0 and 360°.

Since , Sin θ = -1/2

=> θ = 210° or 330°

and , tan θ = 1/√3

=> θ = 30° or 210°

Hence θ = 210° or 7π/6 is the value of θ satisfying both the equations.

∴The general value of

$\large \theta = (2 n \pi + \frac{7\pi}{6})$ ; n ∈ I

Illustration : Are the set of angles α and β given by $\alpha = (2n+\frac{1}{2})\pi \pm A$ and $\beta = m\pi + (-1)^m (\frac{\pi}{2}-A)$  same where n, m ∈ I.

Solution: Let m = 2k i.e. m is even where k ∈ I

Now , $\beta = 2 k\pi + \frac{\pi}{2} -A = (2k + \frac{1}{2})\pi – A$ …(i)

If m = 2k + 1 i.e. m is odd, then

$\beta = (2 k + 1)\pi – (\frac{\pi}{2} -A ) = (2k + \frac{1}{2})\pi + A$ …(ii)

From (i) and (ii) follows that β can be expressed as

$\large \beta = (2k + \frac{1}{2})\pi \pm A$ , k ∈ I Which is same as α .

Exercise :

(i) Find the value of tan1° tan2°tan3° ….. tan89°

(ii) Find the value of : tan π/3 + 2 tan 2π/3 + 4 tan 4π/3 + 8 tan 8π/3

(iii) Show that Sin² 6° + Sin² 12° + Sin² 18° + ……..+ Sin² 90° = 8

(iv) If 4nα = π show that tan α tan 2α tan 3α …… tan (2n − 1)α = 1