Trigonometrical Ratios of Standard Angles

We shall study trigonometrical ratios of some standard angles for other angles, and we can also use trigonometrical table.

Trigonometrical Ratio of 0° and 90° :

Let consider an acute angle ∠ABC = 𝜭 and let P be any point on its terminal side BC. Draw perpendicular PQ from P on BA.

In ⃤ BPQ, we have

$\sin \theta =\frac{PQ}{BP},\; \cos \theta = \frac{BQ}{BP}, \; \tan\theta = \frac{PQ}{BQ}$

In ⃤ AMP, If 𝜃 becomes smaller and smaller line segment PQ also becames smaller and smaller; and Finally when 𝜃 becomes 0°; the point P will parallel with Q.

PQ = 0 and BP = BQ

$\small \therefore \:\sin0\degree = \frac{PQ}{BP} = \frac{0}{BP} = 0, \: \cos0\degree = \frac{BQ}{BP}=\frac{BP}{BP} = 1, \:and\: \tan\theta = \frac{PQ}{BP} = \frac{0}{BP} = 0$

Thus, we have

$\sin0\degree = 0, \cos0\dgree = 1, \tan0\degree=\infty$

Similarly as above as and on using reciprocal relation,

$\csc0\degree = \frac{1}{\sin0\degree}=\infty, \sec0\degree=\frac{1}{\cos0\degree}=1, \cot0\degree=\frac{1}{\tan0\degree}=\infty$

From ⃤ BPQ, it is obvious that as 𝜃 increase, line segment BQ becomes smaller and smaller and finally when 𝜭 became 90° the point Q will parallel with B.
Consequently, we have

PQ=0 and BP = PQ

$\therefore \sin90\degree=\frac{PQ}{BP}=\frac{PQ}{PQ}=1$

$\;\;\; \cos90\degree=\frac{BQ}{BP}=\frac{0}{BP}=0$

Thus, we have

Sin90° = 1, cos90° = 0, tan90° = ∞, cot90° = 0, sec90° = ∞ ,and csc90° = 0

Trigonometrical ratio of 30° and 60° :

Let consider an equilateral triangle PQR with each side of length 2a. since each angle of an equilateral triangle is of 60° . therefore, each angle of ⃤ PQR is of 60°. Let PS be perpendicular from P and QR. Since the triangle is equilateral.

Therefore, PS is the bisector of ∠P and ∠S the mid-point of QR.

∴ QS = SR = a and ∠QPS = 30°

Thus, in ⃤ PQS, ∠S is a right angle, hypotenuse PQ = 2a and QS = a.

Applying pythagoros theorem, in ⃤ PQS, we obtain

PQ² = PS² + QS²

(2a)² = PS² + a²

PS² = 4a² - a²

PS = √3a

Trigonometrical ratios of 30° :

In ◿ PSQ, we obtain Base = PS = √3a,

Perpendicular = QS = a,

Hypotenuse = PQ = 2a, and

∠ SPQ = 30°

$\therefore \;\;\;\sin30\degree = \frac{QS}{PQ} = \frac{a}{2a} = \frac{1}{2} \;;$

$\Rightarrow\:\;\;\cos30\degree = \frac{PS}{PQ} = \frac{\sqrt{3}a}{2a} = \frac{\sqrt{3}}{2} \;;$

$\Rightarrow\:\;\;\tan30\degree = \frac{QS}{PS} = \frac{a}{\sqrt{3}a} = \frac{1}{\sqrt{3}} \;;$

$\Rightarrow\:\;\;\csc30\degree = \frac{1}{\sin30\degree} = 2 \;;$

$\Rightarrow\:\;\;\sec30\degree = \frac{1}{\cos30\degree} = \frac{2}{\sqrt{3}} \;;$

$\Rightarrow\:\;\;\cot30\degree = \frac{1}{\tan30\degree} =\sqrt{3} \;;$

Trigonometrical ratios of 60° :

In◿PSQ, we have

Base = QS = a

Perpendicular = PS √3a

Hypotenuse = PS = 2a, and

∠PQS = 60°

$\therefore \;\;\;\sin60\degree = \frac{PS}{PQ} = \frac{\sqrt{3}a}{2a} = \frac{\sqrt{3}}{2} \;;$

$\Rightarrow\:\;\;\cos60\degree = \frac{QS}{PQ} = \frac{a}{2a} = \frac{1}{2} \;;$

$\Rightarrow\:\;\;\tan60\degree = \frac{PS}{QS} = \frac{\sqrt{3}a}{a} = \sqrt{3} \;;$

$\Rightarrow\:\;\;\csc60\degree = \frac{1}{\sin60\degree} = \frac{2}{\sqrt{3}} \;;$

$\Rightarrow\:\;\;\sec60\degree = \frac{1}{\cos60\degree} = 2 \;;$

$\Rightarrow\:\;\;\cot60\degree = \frac{1}{\tan60\degree} =\frac{1}{\sqrt{3}} \;;$

Trigonometrical ratios of 45° :

Consider a right angle triangle ◿ PQR with right angle at R such that ∠P = 45°. Then,

∠P + ∠Q +∠R = 180°

⇒ 45° + 90° + 45° = 180°

⇒ ∠Q = 45°

∴ ∠P = ∠R

⇒ PR = RQ

Let PR² = RQ² = a . Applying pythagoros theorem, in ⊿PQR ,

we obtain

PQ² = PR² + RQ² PQ² = a² + a² PQ² = 2a² PQ = √2a Thus, in PQR, we have P = 45, Base = PR = a, Perpendicular = RQ = a, and Hypotenuse = PQ = √2a

PQ² = a² + a²

PQ² = 2a²

PQ = √2a

Thus, in PQR, we have

∠P = 45°, Base = PR = a, Perpendicular = RQ = a, and Hypotenuse = PQ = √2a

$\therefore \;\;\;\sin45\degree = \frac{RS}{PQ} = \frac{a}{\sqrt{2}a} = \frac{1}{\sqrt{2}} \;;$

$\Rightarrow\:\;\;\cos45\degree = \frac{PR}{PQ} = \frac{a}{\sqrt{2}a} = \frac{1}{\sqrt{2}} \;;$

$\Rightarrow\:\;\;\tan45\degree = \frac{RQ}{RQ} = \frac{a}{a} = 1 \;;$

$\Rightarrow\:\;\;\csc45\degree = \frac{1}{\sin45\degree} = \sqrt{2} \;;$

$\Rightarrow\:\;\;\sec45\degree = \frac{1}{\cos45\degree} = \sqrt{2} \;;$

$\Rightarrow\:\;\;\cot60\degree = \frac{1}{\tan60\degree} =\frac{1}{\sqrt{3}} \;;$

Trigonometric Table For Standard Angle :

***Trigonomeric Table***
𝜭	0°	30°	45°	60°	90°
sin	0	1/2	1/√2	√3/2	1
cos	1	√3/2	1/√2	1/2	0
tan	0	1/√3	1	√3	∞
csc	∞	2	2	√2/3	1
sec	1	2/√3	√2	2	∞
cot	∞	√3	1	1/√3	0

Trigonometric Ratios of Complementary Angles :

In this section, we will obtain the trigonometric ratios of complementary angles in terms of the trigonometric ratios of the given angles.

Complementary Angle ⇒

When sum of two angles are in 90° then it is called Complementary Angle.

from the above definition, If 𝜭 and (90 - 𝜭 ) are complementary angles for an acute angle 𝜭.

Theorem :

If 𝜽 is an acute angle, than prove that

sin (90 - 𝜽) = cos 𝜽,

cos (90 - 𝜽) = sin 𝜽,

tan (90 - 𝜽) = cot 𝜽,

cot (90 - 𝜽) = tan 𝜽,

sec (90 - 𝜽) = csc 𝜽, and

csc (90 - 𝜽) = sec 𝜽

Proof :

consider a right triangle ⊿ OAB, right angled at B as shown in fig Let ∠BOA = 𝜽, then ∠OAB = (90 - 𝜽).

For the reference angle 𝜽, we have

$\csc\theta = \frac{OA}{AB},\;\cot\theta = \frac{OB}{AB},\;\sec\theta=\frac{OA}{OB}.$

For the reference angle (90° - 𝜽), we have

Base = AB, Perpendicular = OB, and Hypotenuse = OA

$\therefore\;\; \sin\:(90\degree-\theta) = \frac{OB}{OA},$

$\Rightarrow \;\; \cos\:(90\degree-\theta) = \frac{AB}{OA}, \;$

$\Rightarrow \;\; \tan\:(90\degree-\theta) = \frac{OB}{AB}, \;$

$\Rightarrow \;\; \csc\:(90\degree-\theta) = \frac{OA}{OB}, \;$

$\Rightarrow \;\; \sec\:(90\degree-\theta) = \frac{OA}{AB}, \;$

By solving Both, we get

$\Rightarrow \;\; \sin\:(90\degree-\theta) = \frac{OA}{OB} = \cos\theta$

$\Rightarrow \;\; \cos\:(90\degree-\theta) = \frac{AB}{OA} = \sin\theta$

$\Rightarrow \;\; \tan\:(90\degree-\theta) = \frac{OB}{A} = \cot\theta$

$\Rightarrow \;\; \csc\:(90\degree-\theta) = \frac{OA}{OB} = \sec\theta$

$\Rightarrow \;\; \sec\:(90\degree-\theta) = \frac{OA}{AB} = \csc\theta$

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