8.2: Trigonometric Ratios of special angles:

Since one angle in a right angled triangle is 90⁰ the sum of other two angles has to be 90⁰

The special pair of acute angles are (60⁰ 30⁰ ) and(45⁰ ,45⁰ )

Let us study the properties of ratios in such cases

1. The special pair of (45⁰ ,45⁰):

In the adjacent figure, let A = 45^0 hence C = 45^0. (sum of two angles has to be 90^0) Therefore ABC is equilateral triangle with AB=BC. Let AB =a By Pythagoras theorem AC^2 = AD^2+DC^2 = 2a^2 AC = a By definition sin A = sin 45 = Opposite Side /Hyp. =BC/AC =a/a = 1/ cos A = cos 45 =Adjacent Side /Hyp. = AB/AC =a/a = 1/ tan A = tan 45 =Opposite Side /Adjacent Side =BC/AB = a/a =1

2. The special pair of (60⁰ ,30⁰):

Let us consider an isosceles triangle whose sides are 2a. Let CD be perpendicular to AB Since ABC is an isosceles triangle A = B=C=60^0(all angles are equal) and ACD = 30^0(sum of all angles in triangle ADC = 180^0) By Pythagoras theorem AC^2 = AD^2+DC^2 DC^2 = AC^2-AD^2 = (2a)^2-a^2 = 3a^2 CD = a By definition sin A = sin 60 = O/H= CD/AC =a/2a = /2 cos A = cos 60 = B/H= AD/AC =a/2a = 1/2 tan A = tan 60 = O/A =CD/AD = a/a = By definition sin ACD = sin 30= O/H= AD/AC =a/2a = 1/2 cos ACD = cos 30= A/H= CD/AC =a/2a = /2 tan ACD = tan 30= O/A = AD/CD = a/a =1/

3. The special pair of (0⁰ ,90⁰):

When angle A approaches 90^0 (hypotenuse becomes Opposite Side ) the length of Opposite Side and hypotenuse become same and length of Adjacent Side becomes zero. Thus sin 90 = O/H= 1, cos90= A/H =0 and tan90 = O/A = Opposite Side/0=undefined When angle A approaches 0^0 (hypotenuse becomes Adjacent Side itself) the length of Adjacent Side and hypotenuse become same and Opposite Side becomes zero. Thus sin 0 = O/H= 0 , cos0= A/H =1 and tan 0 = O/A = 0

The ratios for few special angles can be summarized in a table as given below:

Angle =>	0^0	30^0	45^0	60^0	90^0
Ratios	Values for the angles
sin(Angle) =	0	1/2	1/	/2	1
cos(Angle) =	1	/2	1/	1/2	0
tan(Angle) =	0	1/	1		undefined
cosec(Angle) =	undefined	2		2/	1
sec(Angle) =	1	2/		2	undefined
cot(Angle) =	undefined		1	1/	0

For values of = 0, 30^0,45^0,60^0and 90^0 let us draw the graph for sin, cos and tan. First graph has values for both sin and cos represented by blue line and green line respectively. The second graph is for tan. Observations: 1. with the increase in angles value of sin increases from 0 to 1 2. with the increase in angles value of cos decreases from 1 to 0 3. with the increase in angles, value of tan increases from 0 to infinity.	Graph of sin() and cos()	Graph for tan()

Trigonometric table:

The values of sin, cos and tan for different angles (1 to 89^0) are found using a table, part of which is given below	The section of sine table for values from 1 to 10^0 and part of the degrees is given below:

8.2 Problem 1: For any angle prove

1. sin²A+ cos²A =1

2. sec²A-tan²A =1

3. cosec²A-cot²A =1

By definitions

1	sin^2A+ cos^2A = (Opposite Side /hypotenuse)^2+ (Adjacent Side /hypotenuse)^2 = (Opposite Side ^2+ Adjacent Side ^2)/ hypotenuse^2 = (hypoenuse^2)/ hypotenuse^2(by Pythagoras theorem (Opposite Side ^2+ Adjacent Side ^2) = hypotenuse^2) =1
2	sec^2A-tan^2A = (hypotenuse / Adjacent Side )^2-( Opposite Side / Adjacent Side )^2 = (hypotenuse^2- Opposite Side ^2)/ Adjacent Side ^2 =((Opposite Side ^2+ Adjacent Side ^2)- Opposite Side ^2) / Adjacent Side ^2(by Pythagoras theorem (Opposite Side ^2+ Adjacent Side ^2) = hypotenuse^2) = Adjacent Side ^2 / Adjacent Side ^2 =1
3	cosec^2A-cot^2A = (hypotenuse / Opposite Side )^2-( Adjacent Side / Opposite Side )^2 = (hypotenuse^2- Adjacent Side ^2)/ Opposite Side ^2 = (Opposite Side ^2+ Adjacent Side ^2)- Adjacent Side ^2) / Opposite Side ^2(by Pythagoras theorem (Opposite Side ^2+ Adjacent Side ^2) = hypotenuse^2) = Opposite Side ^2/ Opposite Side ^2 =1

Exercise : By substituting A = 30⁰, 45⁰, 60⁰ and corresponding values for sin, cos, sec, tan, cosec, cot observe that equations in Problem 8.2.1 are true.

8.2 Problem 2: If A and B are two acute angles in a right angled triangle prove that

sin(A+B) =1= sinAcosB+cosAsinB and cos(A+B) =0= cosAcosB-sinAsinB

Solution:

1. sin(A+B) = sinAcosB+cosAsinB 2. cos(A+B) = cosAcosB-sinAsinB By definition sinAcosB+CosAsinB = (BC/AB)*(BC/AB) + (AC/AB)*(AC/AB) BC^2/ AB^2+AC^2 /AB^2 = (BC^2+AC^2)/AB^2=1(By Pythagoras theorem) Since A and B are acute angles of the triangle, A+B = 90^0 Thus sin(A+B) = sin 90 = 1 This proves the first statement Similarly the other statement can be proved.

Exercise : Verify the statements in 8.2 Problem 2 when (A,B) = (60⁰ ,30⁰ ), (30⁰ ,60⁰ ), (0⁰ ,90⁰ ) , , ,using the values in the table for special angles.

8.2 Problem 3: If A = 30⁰ then prove that

cos 2A = cos²A - sin²A = (1-tan²A)/(1+ tan²A)

Solution:

Since A = 30⁰, 2A = 60⁰

cos 2A =cos 60 = 1/2 -----à(1)

cos² A = (cosA)²= (cos30)²= (/2)² =3/4

sin² A= (sin30)²= (1/2)² =1/4

cos²A - sin²A = 3/4 -1/4 = 1/2 -----à(2)

tan²A = (tan 30)²= (1/)² =1/3

(1-tan²A)/(1+ tan²A) = (1-1/3)/(1+1/3)

= (2/3)/(4/3) = 2/4 = 1/2 ------à(3)

From (1), (2) and (3) we conclude that cos 2A = cos²A - sin²A = (1-tan²A)/(1+ tan²A)

8.2 Problem 4: Find the magnitude of angle A if

2sin Acos A –cos A-2sinA+1=0

Solution:

2sin Acos A –cos A-2sinA+1 =0

cos A(2sinA-1) –(2sinA-1)=0

(2sinA-1)(cos A-1)=0

(2sinA-1) =0 Or (cos A-1)=0

I.e. sin A =1/2 Or Cos A =1

A=30 Or A=0 ( sin 30⁰ =1/2, Cos 0⁰ =1)

Verification:

Let A =30 then

2sin Acos A –cos A-2sinA+1 = 2sin30cos30 –cos30 -2sin30+1

= 2*(1/2)* (/2) – (/2) -2*(1/2) +1

= 1*(/2) - (/2) -1+1

= (/2) - (/2) +0

=0

This proves that A=30 is one of the solution. Similarly verify that A=0 also satisfies the relationship.

8.2 Problem 5: Solve sin² 60+ cos² (3x-9) =1

Solution:

The given equation can be rewritten as cos² (3x-9) =1- sin² 60

Since sin 60= (/2)

sin² 60 = 3/4

Substituting this value in the given equation we get

cos² (3x-9) =1-3/4 =1/4 = (1/2)²

cos(3x-9) =1/2

Since 1/2 =cos 60

3x-9 =60

i.e. 3x =60+9=69

Therefore x =23

(Note: we can also use the property : sin²A+ cos²A =1 )

Verification:

By substituting x=23 in cos² (3x-9)

We get cos² (3x-9) = cos² (69-9) = cos² (60) = (cos60)² = (1/2)²= 1/4

sin² 60+cos² (3x-9)=(/2)²+1/4=3/4+1/4 = 4/4 =1 which is RHS of the given equation.

8.2 Problem 5: A point outside a circle of radius 2cm is to be chosen such that the angle between two tangents from this point to the circle is 40⁰. How far away from the centre of the circle should this point be, if sin 20 = 0.342

Hint: Draw a rough diagram as in the adjacent figure. 1. Join OA and OP 2. Note OP bisects APB and OAP =90^0(Refer section 6.14 Theorems) 3. Use value for sin20 (=0.342) from sine table to get the length of PO.

8.2 Summary of learning

No	Points studied
1	Values of sin, cos, tan and others for standard angles of 30^0,45^0,60^0