8.2: Trigonometric Ratios of standard 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 = 450 hence C = 450. (sum of  two angles has to be 900) Therefore ABC is equilateral triangle with AB=BC. Let AB =a By Pythagoras theorem   AC2 = AD2+DC2 = 2a2  AC = a   By definition sin A = sin 45 = Perp./Hyp. =BC/AC =a/a = 1/ cos A = cos 45 =Base/Hyp.  = AB/AC =a/a = 1/ tan A = tan 45 =Perp./Base =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=600(all angles are equal) and ACD = 300(sum of all angles in triangle ADC = 1800) By Pythagoras theorem  AC2 = AD2+DC2   DC2 = AC2-AD2  = (2a)2-a2  = 3a2 CD = a By definition sin A = sin 60 =  P/H = CD/AC =a/2a = /2 cos A = cos 60 = B/H= AD/AC =a/2a = 1/2 tan A = tan 60 = P/B =CD/AD = a/a = By definition sin ACD = sin 30=  P/H = AD/AC =a/2a = 1/2 cos ACD = cos 30= B/H = CD/AC =a/2a = /2 tan ACD  = tan 30= P/B = AD/CD = a/a =1/

 

3. The special pair of (0⁰ ,90⁰):

When angle A approaches 90⁰ (hypotenuse becomes perpendicular) the length of perpendicular and hypotenuse become same and length of base becomes zero.

Thus sin 90 = 1 and cos 90 =0 and tan90 = undefined

When angle A approaches 0⁰ (hypotenuse becomes base itself) the length of base and hypotenuse become same and perpendicular becomes zero.

Thus sin 0 = 0 and cos 0 =1 and tan 0 = 0

 

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

 

Angle =>	00	300	450	600	900
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, 300,450,600 and 900 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 890) are found using a table, part of which is given below	The section of sine table for values from 1 to 100 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	sin2A+ cos2A = (perpendicular/hypotenuse)2+ (base/hypotenuse)2 = (perpendicular2+ base2)/ hypotenuse2 = (hypoenuse2)/ hypotenuse2 (by Pythagoras theorem (perpendicular2+ base2) = hypotenuse2) =1
2	sec2A-tan2A = (hypotenuse / base)2-( perpendicular / base)2 = (hypotenuse 2 - perpendicular 2)/ base2 = (perpendicular 2+ base2 )- perpendicular 2) / base2 (by Pythagoras theorem (perpendicular2+ base2) = hypotenuse2) =  base2  / base2 =1
3	cosec2A-cot2A = (hypotenuse / perpendicular)2-( base / perpendicular)2 = (hypotenuse 2 - base2)/ perpendicular 2 = (perpendicular 2+ base2) - base 2) / perpendicular 2 (by Pythagoras theorem (perpendicular2+ base2) = hypotenuse2) = perpendicular 2 / perpendicular 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) = sinAcosB+cosAsinB and cos(A+B) = 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) BC2/ AB2+AC2 /AB2 = (BC2+AC2)/AB2 =1(By Pythagoras theorem) Since A and B are acute angles of the triangle A+B = 900 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 = 60⁰ and B  = 30⁰ by using the values in the table for  standard 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:

LHS=2sin Acos A –cos A-2sinA+1

= cos A(2sinA-1) –(2sinA-1)

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

Thus it is given that

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

 (2sinA-1) =0 or (cos A-1)=0

1. If (2sinA-1) =0 then 2sina A =1

I.e. sin A =1/2

Since sin 30 =1/2,A=30

2. If (cos A-1)=0 then cos A =1

Since cos A =1,A=0

Thus A=30 or A=0 satisfy the given equation

 

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 40⁰How 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 =900(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 300,450,600