We will learn how to solve different types of problems on compound angles using formula.
We will see step-by-step how to deal with the trigonometrical ratios of compound angles in different questions.
1. An angle θ is divided into two parts so that the ratio of the tangents of the parts is k; if the difference between the parts be ф, prove that, sin ф = (k - 1)/(k + 1) sin θ .
Solution:
Let, α and β be the two parts of the angle θ.
Therefore, θ = α + β.
By question, θ = α - β. (assuming a >β)
and tan α/tan β = k
⇒ sin α cos β/sin β cos α = k/1
⇒ (sin α cos β + cos α sin β)/(sin α cos β - cos α sin β) = (k + 1)/(k - 1), [by componendo and dividendo]
⇒ sin (α + β)/sin (α - β) = (k + 1)/(k - 1)
⇒ (k + 1) sin Ø = (k - 1) sin θ, [Since we know that α + β = θ; α + β = ф]
⇒ sin ф = (k - 1)/(k + 1) sin θ. Proved.
2. If x + y = z and
tan x = k tan y, then prove that sin (x - y) = [(k - 1)/(k + 1)] sin z
Solution:
Given tan x = k tan y
⇒ sin x/cos x = k ∙ sin y/cos y
⇒ sin x cos y/cos x sin y = k/1
Applying componendo and dividend, we get
sin x cos y + cos x sin y/ sin x cos y - cos x sin y = k + 1/k - 1
⇒ sin (x + y)/sin (x – y) = k + 1/k - 1
⇒ sin z/sin (x – y) = k + 1/k - 1, [Since x + y = z given]
⇒ sin (x – y) = [k + 1/k – 1] sin z Proved.
3. If A + B + C = π and cos A = cos B cos C, show that, tan B tan C = 2
Solution:
A + B + C = π
Therefore, B + C = π - A
⇒ cos (B + C) = cos (π - A)
⇒ cos B cos C - sin B sin C = - cos A
⇒ cos B cos C + cos B cos C = sin B sin C,[Since we know, cos A = cos B cos C]
⇒ 2 cos B cos C = sin B sin C
⇒ tan B tan C = 2 Proved.
Note: In different problems on compound angles we need to use the formula as required.
4. Prove that cot 2x + tan x = csc 2x
Solution:
L.H.S. = cot 2x + tan x
= cos 2x/sin 2x + sin x/cos x
= cos 2x cos x + sin 2x sin x/sin 2x cos x
= cos (2x - x)/sin 2x cos x
= cos x/sin 2x cos x
= 1/sin 2x
= csc 2x = R.H.S. Proved.
5. If sin (A + B) + sin (B + C) + cos (C - A) = -3/2 show that,
sin A + cos B + sin C = 0; cos A + sin B + cos C = 0.
Solution:
Since, sin (A + B) + sin (B + C) + cos (C - A) = -3/2
Therefore, 2 (sin A cos B + cos A sin B + sin B cos C + cos B sin C + cos C cos A + sin C sin A) = -3
⇒ 2 (sin A cos B + cos A sin B + sin B cos C + cos B sin C + cos C cos A + sin C sin A) = - (1 + 1 + 1)
⇒ 2 (sin A cos B + cos A sin B + sin B cos C + cos B sin C + cos C cos A + sin C sin A) = - [(sin^2 A + cos^2 A) + (sin^2 B + cos^2 B) + (sin^2 C + cos^2 C)]
⇒ (sin^2 A + cos^2 B + sin^2 C + 2 sin A sin C + 2 sin A cos B + 2 cos B sin C) + (cos^2 A + sin^2 B + cos^2 C + 2 cos A sin B + 2 sin B cos C + 2 cos A cos C) = 0
⇒ (sin A + sin B + sin C)^2 + (cos A + sin B + cos C)^2
Now the sum of squares of two real quantities is zero if each quantity is separately zero.
Therefore, sin A + cos B + Sin C = 0
and cos A + sin B + cos C = 0. Proved.
11 and 12 Grade Math
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