8.4: Trigonometric Identities:
8.4.1
Fundamental identities:
We have learnt in Section
8.1 the ratios for sin, cos and tan of angles.
sin_{} 
Opposite Side /Hypotenuse 
PQ/OP; Cosec_{} =1/sin_{}; OP/PQ 

cos_{} 
Adjacent Side /Hypotenuse 
OQ/OP; sec_{} = 1/cos_{}; OP/OQ 

tan_{} 
Opposite Side /Adjacent
Side = sin_{}/ cos_{} 
PQ/OQ;
cot_{} = 1/tan_{};OQ/PQ 

By Pythagoras theorem we know that PQ^{2 }+
OQ^{2 }= OP^{2 }ŕ(1) _{} PQ^{2}/OP^{2} + OQ^{2}/OP^{2 }= 1(By dividing both sides of equation (1) by OP^{2})^{} _{}(PQ/OP)^{2 }+ (OQ/OP)^{2 }= 1 _{}(sin_{})^{2 }+ (cos_{})^{2 }= 1 _{}sin^{2}_{} + cos^{2}_{} = 1
(I) By dividing both sides of equation (1) by OQ^{2}
we get PQ^{2}/OQ^{2 }+
1 = OP^{2}/OQ^{2} _{}(PQ/OQ)^{2 }+ 1 = (OP/OQ)^{2} _{}1 + (tan_{})^{2 }= (sec_{})^{2}
_{}tan^{2}_{} + 1 = sec^{2}_{} (II) By dividing both sides of equation (1) by PQ^{2}
we get 1 +OQ^{2}/PQ^{2 }=
OP^{2}/PQ^{2} _{}1 + (OQ/PQ)^{2 }= (OP/PQ)^{2} _{}1 + (cot_{})^{2 }= (cosec_{})^{2} _{}1 + cot^{2}_{} = cosec^{2}_{} (III) The equations (I), (II) and (III) are called ‘Fundamental identities’. From
the first fundamental identity we can also arrive at the following:
Since sin_{}and cos_{} are positive when _{} is acute, sin_{} = +_{}(1cos^{2}_{}) cos_{} = +_{}(1sin^{2}_{}) From the other two fundamental identities we can
arrive at the following:

In summary we have the following
relationships between various standard trigonometric ratios.
Note: All these relationships can be derived by using just sin^{2}_{}+cos^{2}_{}=1.
The Fundamental identities are useful for simplifying various
trigonometric expressions.
8.4 Problem 1: If _{} (1+x^{2})*sin_{} = x prove that sin^{2}_{}/ cos^{2}_{} + cos^{2}_{}/ sin^{2}_{} = x^{2} + 1/x^{2}
Solution:
It is given that
_{} (1+x^{2})*sin_{} = x
_{} sin_{} = x/_{} (1+x^{2})
_{} sin^{2}_{} = x^{2}/(1+x^{2})
(By squaring both sides)(1)
But sin^{2}_{}+cos^{2}_{}=1 (Fundamental identity)
_{} cos^{2}_{} = 1  sin^{2}_{} (By transposition)
= 1  x^{2}/(1+x^{2})
(By
substitution)
= (1+x^{2 } x^{2})/(1+x^{2})
= 1/(1+x^{2}) (2)
From (1) and (2)
sin^{2}_{}/cos^{2}_{} =
{x^{2}/(1+x^{2})}/{1/(1+x^{2})} = x^{2} (3)
Similarly, cos^{2}_{}/sin^{2}_{} = 1/x^{2}
(4)
From (3) and (4)
sin^{2}_{}/cos^{2}_{} + cos^{2}_{}/sin^{2}_{} = x^{2} + 1/x^{2}
8.4 Problem 2: prove that sin^{6}_{}+cos^{6}_{}=13*sin^{2}_{}.cos^{2}_{}
Solution:
Let x = sin^{2}_{} and y = cos^{2}_{}
Since sin^{2}_{}+cos^{2}_{}=1. It follows that x+y
= 1
Note LHS of the given
equation is of the form x^{3}+y^{3}
We also know the
identity x^{3}+y^{3 }= (x+y)^{3}3xy(x+y) = 13xy(_{}x+y =1)
= 1 – 3*sin^{2}_{}.cos^{2}_{}( By substituting values for x and y)
8.4 Problem 3: Prove that
tanA/(secA1)+tanA/(secA+1) = 2cosecA
Solution:
LHS LHS =
tanA{(secA+1)+(secA1)}/(sec^{2}A1) (By taking out tanA as common
factor and having (secA+1)*(secA1) as common denominator)
= 2tanA.secA/tan^{2}A
(_{}sec^{2}_{}1 = tan^{2}_{})
= 2secA/tanA (canceling
of tanA)
= 2secA*cosA/sinA (_{}tanA = sinA/cosA)
= 2/sinA (_{}cosA = 1/secA)
= 2cosecA
8.4.2 Trigonometric ratios of complimentary angles:
In a right angled triangle, if _{} is one angle then the
other angle has to be 90^{0}_{}(sum of all angles in a triangle is 180^{0}).
In the adjacent figure, _{}QOP = _{}^{ }hence
_{}QPO = 90^{0}_{} If we
consider _{}QOP then sin_{} = PQ/OP ŕ(1) cos_{} = OQ/OP ŕ(2) tan_{} = PQ/OQ ŕ(3) If we
consider _{}QPO then cos(90^{0}_{}) = PQ/OP ŕ (4) sin(90^{0}_{}) = OQ/OP ŕ(5) cot(90^{0}_{}) = PQ/OQ ŕ(6) By comparing (1), (2) and (3)
with (4), (5) and (6) respectively and then by comparing their inverses, we
note that


8.4 Problem 4: Evaluate 3sin62^{0}/cos28^{0 } sec42^{0}/cosec48^{0}
Solution:
Note that 28 = 9062
and 48 = 9042
cos(28) = cos(9062) =
sin62
cosec(48) =
cosec(9042) = sec(42)
3sin62^{0}/cos28^{0 } sec42^{0}/cosec48^{0}
= 3sin62^{0}/sin62^{0 } sec42^{0}/sec42^{0}
= 31 = 2
8.4 Problem 5: If sec4A=Cosec(A20^{0}),
Where 4A is acute angle, Find the value of A.
Solution:
Since we are familiar
with sin and cos more than Sec, let us convert the problem as follows by taking
reciprocals
1/ sec4A = 1/
Cosec(A20^{0})
Ie, cos4A= sin(A20^{0})
sin(904A)= sin(A20^{0})
( since 4A is acute, it is less than 90^{0}. Hence we can use cos_{} = sin(90^{0}_{})
_{}904A= A20^{0}
Ie, 90+20= A+4A
110= 5A
_{}A= 22^{0}
8.4. Summary of
learning
No 
Points
studied 
1 
sin^{2}_{}+cos^{2}_{}=1,tan^{2}_{} + 1 = sec^{2},1
+ cot^{2}_{} = cosec^{2}_{} 
2 
Trigonometric ratios of
complimentary angles 