Write 25 - 9 sin^2 x as a product using the difference of squares.
A(5 - 3 sin x)(5 + 3 sin x)
B(5 - 9 sin x)(5 + 9 sin x)
C(25 - 3 sin x)(1 + 3 sin x)
D(5 - 3 sin x)(5 - 3 sin x)
Answer & Solution
Correct answer: A. (5 - 3 sin x)(5 + 3 sin x)
1. Read the expression as a^2 - b^2 with a^2 = 25 and b^2 = 9 sin^2 x.
2. Then a = 5 because 5^2 = 25.
3. And b = 3 sin x because (3 sin x)^2 = 9 sin^2 x.
4. The difference of squares formula a^2 - b^2 = (a - b)(a + b) now gives (5 - 3 sin x)(5 + 3 sin x).
5. Expanding that product returns 25 - 9 sin^2 x, so the factorisation checks out.
6. Using 9 sin x inside the brackets squares to 81 sin^2 x, which is nine times too large.
7. Repeating the same bracket would give 25 - 30 sin x + 9 sin^2 x, which has a middle term and the wrong sign on the square.
_Source: OpenStax Algebra and Trigonometry (CC BY 4.0), Ch 9 "Trigonometric Identities and Equations", section 9.1 Solving Trigonometric Equations with Identities_
Keep practising Trigonometric Identities and Equations
Browse all Trigonometric Identities and Equations questions for B.Sc Mathematics, or explore more Mathematics topics.
Related questions
If a ladder's base is a feet from the wall and its length is 4a feet, how high up the wallSafety rules place a ladder's base 1 foot from the wall for every 4 feet of ladder length.Compared with y = sin t, the graph of y = sin(2t) is which transformation?Solving sin(n t) = c across one turn requires going around the unit circle how many times?Solve cos(2t) = 1/2 on the interval from 0 to 2 pi.Solve 3 cos t + 3 = 2 sin^2 t exactly on the interval from 0 to 2 pi.Solve cos t cos(2t) + sin t sin(2t) = sqrt(3)/2 on the interval from 0 to 2 pi.Solve 2 sin^2 t + sin t = 0 exactly on the interval from 0 to 2 pi.