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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
By definition sin A = sin 45
= Perp./Hyp. =BC/AC =a/ cos A = cos 45 =Base/Hyp. = AB/AC =a/ tan A = tan 45 =Perp./Base =BC/AB = a/a =1 |
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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 and By Pythagoras theorem
AC2 = AD2+DC2 By definition sin A = sin 60 = P/H =
CD/AC = cos A = cos 60 = B/H= AD/AC =a/2a = 1/2 tan A = tan 60 = P/B =CD/AD = By definition sin ACD = sin 30= P/H =
AD/AC =a/2a = 1/2 cos ACD = cos 30= B/H = CD/AC = tan ACD = tan 30= P/B = AD/CD = a/ |
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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/ |
|
1 |
cos(Angle) = |
1 |
|
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 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( |
Graph for tan( |
|
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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. |
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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 3. Use value for sin20 (=0.342) from sine table to get
the length of |
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8.2 Summary of learning
No |
Points studied |
1 |
Values of sin, cos, tan and
others for standard angles of 300,450,600 |