CONDITIONAL IDENTITIES TRIGONOMETRIC EXAMPLES

Trigonometric identities are true for all admissible values of the angle involved. There are some trigonometric identities which satisfy the given additional conditions. Such identities are called conditional trigonometric identities.

Here we are going to see some examples to show how to solve conditional trigonometric identities problems.

Example 1 :

If A + B + C  =  180°, prove that 

tan(A/2)tan(B/2) + tan(B/2)tan(C/2) + tan(C/2)tan(A/2) = 1

Solution :

A + B + C  =  180

A + B  =  180 - C

A/2 + B/2  =  90 - C/2

tan(A/2 + B/2)  =  tan[90 - (C/2)]

tan(A/2 + B/2)  =  cot(C/2) -----(1)

Formula :  

tan(A + B)  =  tan A + tan B / (1 - tan A tan B)

Use the above formula on the left side of (1). 

(1)----->

[tan(A/2) + tan(B/2)] / [(1 - tan(A/2)tan(B/2)]  =  1/tan(C/2)

Take reciprocal on both sides

 [1 - tan(A/2)tan(B/2)] / [tan(A/2) + tan(B/2)]  = tan (C/2)

1 - tan(A/2)tan(B/2)  =  tan(C/2)[tan(A/2) + tan(B/2)]

1 - tan(A/2)tan(B/2) = tan(A/2)tan(C/2)+tan(B/2)tan(C/2)

1  =  tan(A/2)tan(C/2)+tan(B/2)tan(C/2)+tan(A/2)tan(B/2)

Example 2 :

If A + B + C  =  180°, prove that 

sinA + sinB + sinC  =  4cos A/2 cos B/2 cos C/2

Solution :

sinA + sinB + sinC :

=  2sin(A + B)/2cos(A - B)/2 + sinC

=  2sin(180 - C)/2 cos (A - B)/2 + sin C

=  2 sin C/2 cos (A - B)/2 + 2 sin C/2 cos C/2

=  2 sin C/2 [cos (A - B)/2 + sin C/2]

=  2 sin C/2 [cos (A-B)/2 + sin (180-(A+B))/2]

=  2 sin C/2 [cos (A-B)/2 + sin (90-(A+B)/2)]

=  2 sin C/2 [cos (A-B)/2 + cos (A+B)/2]

=  2 sin C/2 [2 cos A/2 cos B/2]

=  4 cos A/2 cos B/2 sin C/2

Example 3 :

If A + B + C  =  180°, prove that 

sin(B + C - A) + sin(C + A − B) + sin(A + B - C) = 4sinAsinB sinC.

Solution :

A + B + C  =  180

A + B  =  180 - C

B + C  =  180 - A

C + A  =  180 - B

sin(B + C - A) + sin(C + A - B) + sin(A + B - C) :

=  sin(180 - A - A) + sin(180 - B - B) + sin(180 - C - C)

=  sin(180 - 2A) + sin(180 - 2B) + sin(180 - 2C)

=  sin 2A + sin 2B + sin 2C

=  2sin(A + B)cos(A - B) + sin2C

=  2sin(180 - C)cos(A - B) + 2sinCcosC

=  2sinCcos(A - B) + 2sinCcosC

=  2sinC[cos(A - B) + cosC]

=  2sinC[cos(A - B) + cos(180 - (A + B)]

=  2sinC[cos(A - B) - cos(A + B)]

=  2sinC[-2sinAsin(-B)]

=  4sinCsinAsinB

=  4sinAsinBsinC

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. SAT Math Resources (Videos, Concepts, Worksheets and More)

    Jul 26, 24 11:27 AM

    SAT Math Resources (Videos, Concepts, Worksheets and More)

    Read More

  2. SAT Math Videos (Part -22)

    Jul 26, 24 11:21 AM

    satmath22.png
    SAT Math Videos (Part -22)

    Read More

  3. Problems on Angles (Part - 3)

    Jul 26, 24 12:39 AM

    problemsonangles11.png
    Problems on Angles (Part - 3)

    Read More