We will learn how to solve various types of problems using compound angle formulae. While solving the problems we need to keep in mind all the formulas of trigonometric ratios of compound angles and use the formula according to the question.
1. If ABCD is a cyclic quadrilateral, then show that cos A + cos B + cos C + cos D = 0.
Solution:
Since ABCD is a cyclic quadrilateral,
A + C = π ⇒ C = π - A
B + D = π ⇒ D = π - B
Therefore, cos A + cos B + cos C + cos D
= cos A + cos B + cos (π - A) + cos (π - B)
= cos A + cos B - cos A - cos B, [Since, cos (π - A) = - cos A and cos (π - B) = - cos B]
= 0
2. Show that, cos^2A + cos^2 (120°
- A) + cos^2 (120° + A) = 3/2
Solution:
L. H. S. = cos^2 A + (cos 120° cos A + sin 120° sin A)^2 + (cos 120° cos A - sin 120° sin A)^2
= cos^2 A + 2(cos^2 120° cos^2 α + sin^2 120° sin^2 α), [Since, (a + b)^2 + (a - b)^2 = 2(a^2 + b^2)]
= cos^2 A + 2[(-1/2)^2 cos^2 A + (√3/2)^2 sin^2 A], [Since, cos 120° = cos (2 ∙ 90° - 60°) = - cos 60°= -1/2 and sin 120°
= sin (2 ∙ 90° - 60°) = sin 60° = √3/2]
= cos^2 A + 2[1/4 cos^2 A + 3/4 sin^2 A]
= 3/2(cos^2 A + sin^2 A)
= 3/2 Proved.
3. If A, B, and C are angles of a triangle, then prove that tan A/2 = cot (B + C)/2
Solution:
Since A, B, and C are angles of a triangle, A + B + C = π
⇒ B + C = π - A
⇒ (B + C)/2 = π/2 - A/2
Therefore, cot (B + C)/2 = cot (π/2 - A/2) = tan A/2 Proved.
Proof the problems using compound angle formulae.
4. If tan x - tan y = m
and cot y - cot x = n, prove
that,
1/m + 1/n
= cot (x - y).
Solution:
We have, m = tan x - tan y
⇒
m = sin x / cos x - sin y/cos y = (sin x cos y - cos x sin y)/cos x cos y
Therefore, 1/m = cos x cos y/sin (x - y) (1)
Again, n
= cot y - cot x = cos y/sin y - cos x/sin x = (sin x cos y - cos x sin
y)/sin y sin x
⇒ n = sin (x - y)/sin y sin x
Therefore, 1/n = sin y sin x/sin (x - y) (2)
Now, (1) + (2) gives,
1/m + 1/n = (cos x cos y + sin y sin x)/sin (x - y) = cos (x - y)/sin (x - y)
⇒ 1/m + 1/n = cot (x - y). Proved.
5. If tan β = sin α cos α/(2 + cos^2 α) prove that 3 tan (α - β) = 2 tan α.
Solution:
We have, tan (α - β) = (tan α - tan β)/1 + tan α tan β
⇒ tan (α - β) = [(sin α/cos α) - sin α cos α/(2 + cos^2 α)]/[1 + (sin α / cos α) ∙ sin α cos α/(2 + cos^2 α)], [Since, tan β = sin α cos α/(2 + cos^2 α)]
= (2 sin α + sin α cos^2 α - sin αcos^2 α)/(2 cos α + cos^3 α + sin^2 α cos α)
= 2 sin α/cos α (2 + cos^2 α + sin^2 α)
= 2 sin α/3 cos α
⇒ 3 tan (α - β) = 2 tan α Proved.
11 and 12 Grade Math
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