Square of Identities Involving Squares of Sines and Cosines

We will learn how to solve identities involving square of sines and cosines of multiples or submultiples of the angles involved.

We use the following ways to solve the identities involving square of sines and cosines.

(i) Express the first two squares of L.H.S. in terms of cos 2A (or cos A).

(ii) Either retain the third term unchanged or make a change using the formula sin\(^{2}\) A+ cos\(^{2}\) A = 1.

(iii) Keeping the numericais (if any) apart, express the sum of two cosines in the form of product.

(iv) Then use the condition A + B + C = π (or A + B + C = \(\frac{π}{2}\))and take one sine or cosine term common.

(v) Finally, express the sum or difference of two sines (or cosines) in the brackets as product.

1. If A + B + C = π, prove that,

cos\(^{2}\) A + cos\(^{2}\) B - cos\(^{2}\) C = 1 - 2 sin A sin B cos C.

Solution:

L.H.S. =  cos\(^{2}\) A + cos\(^{2}\) B - cos\(^{2}\) C

= cos\(^{2}\) A + (1 - sin\(^{2}\) B) - cos\(^{2}\) C

= 1 + [cos\(^{2}\) A - sin\(^{2}\) B] - cos\(^{2}\) C

= 1 + cos (A + B) cos (A - B) - cos\(^{2}\) C

= 1 + cos (π - C) cos (A - B) - cos\(^{2}\) C, [Since A + B + C = π ⇒ A + B = π - C]

= 1 - cos C cos (A - B) - cos\(^{2}\) C

= 1 - cos C [cos (A - B) + cos C]

= 1 - cos C [cos (A - B) + cos {π - (A + B)}], [Since A + B + C = π ⇒ C = π - (A + B)]

= 1 - cos C [cos (A - B) - cos (A + B)]

= 1 - cos C [2 sin A sin B]

= 1 - 2 sin A sin B cos C = R.H.S.                    Proved.


2. If A + B + C = π, prove that,

sin\(^{2}\) \(\frac{A}{2}\) + sin\(^{2}\) \(\frac{A}{2}\) + sin\(^{2}\) \(\frac{A}{2}\) = 1 - 2 sin \(\frac{A}{2}\) - sin \(\frac{B}{2}\) sin \(\frac{C}{2}\)

Solution:

L.H.S. = sin\(^{2}\) \(\frac{A}{2}\) + sin\(^{2}\) \(\frac{B}{2}\) + sin\(^{2}\) \(\frac{C}{2}\)

= \(\frac{1}{2}\)(1 - cos A) + \(\frac{1}{2}\)(1 - cos B) + sin\(^{2}\) \(\frac{C}{2}\), [Since, 2 sin\(^{2}\) \(\frac{A}{2}\) = 1 - cos A                   

⇒ sin\(^{2}\) \(\frac{A}{2}\) = \(\frac{1}{2}\)(1 - cos A)

Similarly, sin\(^{2}\)  \(\frac{B}{2}\) = \(\frac{1}{2}\)( 1 - cos B)]

= 1 - \(\frac{1}{2}\)(cos A + cos B) + sin\(^{2}\) \(\frac{C}{2}\)

= 1 -  \(\frac{1}{2}\) ∙ 2 cos \(\frac{A + B}{2}\)  ∙ cos \(\frac{A - B}{2}\) + sin\(^{2}\) \(\frac{C}{2}\)

=1 - sin \(\frac{C}{2}\)  cos \(\frac{A - B}{2}\)  + sin 2 \(\frac{C}{2}\)

[A + B + C = π ⇒ \(\frac{A + B}{2}\) = \(\frac{π}{2}\)  - \(\frac{C}{2}\).

 Therefore, cos \(\frac{A + B}{2}\) = cos (\(\frac{π}{2}\)  - \(\frac{C}{2}\)) = sin \(\frac{C}{2}\)]

= 1 - sin \(\frac{C}{2}\)[cos \(\frac{A - B}{2}\) - sin \(\frac{C}{2}\)]

= 1 - sin \(\frac{C}{2}\)[cos \(\frac{A - B}{2}\) - cos \(\frac{A + B}{2}\)]   [Since, sin \(\frac{C}{2}\) = cos \(\frac{A + B}{2}\)]

= 1 - sin \(\frac{C}{2}\)[2 sin \(\frac{A}{2}\) ∙ sin \(\frac{B}{2}\)]

= 1 - 2 sin \(\frac{A}{2}\) sin \(\frac{B}{2}\) sin \(\frac{C}{2}\) = R.H.S.                    Proved.

 

3. If A + B + C = π, prove that,

cos\(^{2}\)  \(\frac{A}{2}\)  + cos\(^{2}\) \(\frac{B}{2}\) - cos\(^{2}\) \(\frac{C}{2}\)  = 2 cos \(\frac{A}{2}\) cos \(\frac{B}{2}\)  sin \(\frac{C}{2}\)

Solution:

L.H.S. = cos\(^{2}\)  \(\frac{A}{2}\)  + cos\(^{2}\) \(\frac{B}{2}\) - cos\(^{2}\) \(\frac{C}{2}\)

= \(\frac{1}{2}\)(1 + cos A) + \(\frac{1}{2}\)(1 + cos B) - cos\(^{2}\) \(\frac{C}{2}\), [Since, 2 cos\(^{2}\) \(\frac{A}{2}\)  = 1 + cos A  ⇒ cos\(^{2}\) \(\frac{A}{2}\) = \(\frac{1}{2}\)(1 + cos A)

Similarly, cos\(^{2}\) \(\frac{B}{2}\) = \(\frac{1}{2}\)(1 + cos B)]

=  1 + \(\frac{1}{2}\)(cos A + cos B) - cos\(^{2}\) \(\frac{C}{2}\)

= 1 + \(\frac{1}{2}\) ∙ 2 cos \(\frac{A + B}{2}\) cos \(\frac{A - B}{2}\) - 1 + sin\(^{2}\)  \(\frac{C}{2}\)

= cos \(\frac{A + B}{2}\) cos \(\frac{A - B}{2}\) + sin\(^{2}\)  \(\frac{C}{2}\)

= sin C/2 cos \(\frac{A - B}{2}\) + sin\(^{2}\)  \(\frac{C}{2}\)

[Since, A + B + C = π ⇒ \(\frac{A + B}{2}\)  = \(\frac{π}{2}\) - \(\frac{C}{2}\).

Therefore, cos (\(\frac{A + B}{2}\)) = cos (\(\frac{π}{2}\) - \(\frac{C}{2}\)) = sin \(\frac{C}{2}\)]

= sin \(\frac{C}{2}\) [cos \(\frac{A - B}{2}\) + sin \(\frac{C}{2}\)]

= sin \(\frac{C}{2}\) [cos \(\frac{A - B}{2}\) + cos \(\frac{A + B}{2}\)], [Since, sin \(\frac{C}{2}\) = cos \(\frac{A - B}{2}\)]

= sin \(\frac{C}{2}\) [2 cos \(\frac{A}{2}\) cos \(\frac{B}{2}\)]

= 2 cos \(\frac{A}{2}\) cos \(\frac{B}{2}\) sin \(\frac{C}{2}\) = R.H.S.          Proved.

 Conditional Trigonometric Identities






11 and 12 Grade Math

From Square of Identities Involving Squares of Sines and Cosines to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.

Share this page: What’s this?

Recent Articles

  1. Addition and Subtraction of Fractions | Solved Examples | Worksheet

    Jul 18, 24 03:08 PM

    Addition and subtraction of fractions are discussed here with examples. To add or subtract two or more fractions, proceed as under: (i) Convert the mixed fractions (if any.) or natural numbers

    Read More

  2. Worksheet on Simplification | Simplify Expressions | BODMAS Questions

    Jul 18, 24 01:19 AM

    In worksheet on simplification, the questions are based in order to simplify expressions involving more than one bracket by using the steps of removal of brackets. This exercise sheet

    Read More

  3. Fractions in Descending Order |Arranging Fractions an Descending Order

    Jul 18, 24 01:15 AM

    We will discuss here how to arrange the fractions in descending order. Solved examples for arranging in descending order: 1. Arrange the following fractions 5/6, 7/10, 11/20 in descending order. First…

    Read More

  4. Fractions in Ascending Order | Arranging Fractions | Worksheet |Answer

    Jul 18, 24 01:02 AM

    Comparison Fractions
    We will discuss here how to arrange the fractions in ascending order. Solved examples for arranging in ascending order: 1. Arrange the following fractions 5/6, 8/9, 2/3 in ascending order. First we fi…

    Read More

  5. Worksheet on Comparison of Like Fractions | Greater & Smaller Fraction

    Jul 18, 24 12:45 AM

    Worksheet on Comparison of Like Fractions
    In worksheet on comparison of like fractions, all grade students can practice the questions on comparison of like fractions. This exercise sheet on comparison of like fractions can be practiced

    Read More