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 PQ2 + OQ2 = OP2   -----ŕ(1)  PQ2/OP2 + OQ2/OP2 = 1(By dividing both sides of equation (1) by OP2) (PQ/OP)2 + (OQ/OP)2 = 1 (sin)2 + (cos)2 = 1    sin2 + cos2 = 1      ----------(I) By dividing both sides of equation (1) by OQ2 we get PQ2/OQ2 + 1 = OP2/OQ2 (PQ/OQ)2 + 1 = (OP/OQ)2 1 + (tan)2 = (sec)2                                                  tan2 + 1 = sec2      ----------(II) By dividing both sides of equation (1) by PQ2 we get 1 +OQ2/PQ2 = OP2/PQ2  1 + (OQ/PQ)2 = (OP/PQ)2 1 + (cot)2 = (cosec)2                                     1 + cot2 = cosec2    ---------(III) The equations (I), (II) and (III) are called ‘Fundamental identities’. From the first fundamental identity we can also arrive at the following: sin2= 1-cos2  sin = (1-cos2) cos2= 1-sin2  cos = (1- sin2) Since sinand  cos are positive when  is acute, sin = +(1-cos2) cos = +(1-sin2) From the other two fundamental identities we can arrive at the following: tan = +(sec2-1), sec = +(1+tan2), cot = +(cosec2-1), cosec = +(1+cot2) tan = sin/ cos = sin/ +(1-sin2)

In summary we have the following relationships between various standard trigonometric ratios.

Note: All these relationships can be derived by using just sin2+cos2=1.

The Fundamental identities are useful for simplifying various trigonometric expressions.

8.4 Problem 1: If  (1+x2)*sin = x prove that sin2/ cos2 + cos2/ sin2 = x2 + 1/x2

Solution:

It is given that

(1+x2)*sin = x

sin = x/ (1+x2)

sin2 = x2/(1+x2) (By squaring both sides)--------(1)

But sin2+cos2=1 (Fundamental identity)

cos2 = 1 - sin2 (By transposition)

= 1 - x2/(1+x2) (By substitution)

= (1+x2 - x2)/(1+x2)

= 1/(1+x2)                            ----------(2)

From (1) and (2)

sin2/cos2 =

{x2/(1+x2)}/{1/(1+x2)} = x2  -----------(3)

Similarly, cos2/sin2 = 1/x2           -----------(4)

From (3) and (4)

sin2/cos2 + cos2/sin2 = x2 + 1/x2

8.4 Problem 2: prove that sin6+cos6=1-3*sin2.cos2

Solution:

Let x = sin2 and y = cos2

Since sin2+cos2=1. It follows that x+y = 1

Note LHS of the given equation is of the form x3+y3

We also know the identity x3+y3 = (x+y)3-3xy(x+y) = 1-3xy(x+y =1)

= 1 – 3*sin2.cos2( By substituting values for x and y)

8.4 Problem 3: Prove that tanA/(secA-1)+tanA/(secA+1) = 2cosecA

Solution:

LHS LHS = tanA{(secA+1)+(secA-1)}/(sec2A-1) (By taking out tanA as common factor and having (secA+1)*(secA-1) as common denominator)

= 2tanA.secA/tan2A (sec2-1 = tan2)

= 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 900-(sum of all angles in a triangle is 1800).

In the adjacent figure, QOP =  hence QPO = 900-

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(900-) = PQ/OP    --ŕ (4)

sin(900-) = OQ/OP    ---ŕ(5)

cot(900-) = 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

 1 sin = cos(900-) 2 cos = sin(900-) 3 tan= cot(900-) 4 cosec = sec(900-) 5 sec = cosec(900-) 6 cot= tan(900-)

8.4 Problem 4: Evaluate 3sin620/cos280 - sec420/cosec480

Solution:

Note that 28 = 90-62 and 48 = 90-42

cos(28) = cos(90-62) = sin62

cosec(48) = cosec(90-42) = sec(42)

3sin620/cos280 - sec420/cosec480

= 3sin620/sin620 - sec420/sec420

= 3-1 = 2

8.4 Problem 5: If sec4A=Cosec(A-200), 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(A-200)

Ie,  cos4A= sin(A-200)

sin(90-4A)= sin(A-200) ( since 4A is acute, it is less than 900. Hence we can use cos = sin(900-)

90-4A= A-200

Ie, 90+20= A+4A

110= 5A

A= 220

8.4. Summary of learning

 No Points studied 1 sin2+cos2=1,tan2 + 1 = sec2,1 + cot2 = cosec2 2 Trigonometric ratios of complimentary angles