Problems on elimination of unknown angles using trigonometric identities.
1. If x = tan θ + sin θ and y = tan θ - sin θ, prove that x^{2} – y^{2} = 4\(\sqrt{xy}\).
Solution:
Given that
x = tan θ + sin θ ……………………. (i)
and
y = tan θ - sin θ ……………………. (ii)
Adding (i) and (ii), we get
x + y = 2 tan θ ……………………. (iii)
⟹ tan θ = \(\frac{x + y}{2}\) ……………………. (iv)
Subtracting (ii) from (i), we get,
x - y = 2 sin θ ……………………. (v)
Now, dividing (iii) by (v) we get,
\(\frac{x + y}{x - y}\) = \(\frac{2 tan θ}{2 sin θ}\)
= \(\frac{tan θ}{sin θ}\)
= \(\frac{\frac{sin θ}{cos θ}}{sin θ}\)
= \(\frac{sin θ}{cos θ}\) ∙ \(\frac{1}{sin θ}\)
= \(\frac{1}{cos θ}\)
= sec θ
Therefore, sec θ = \(\frac{x + y}{x - y}\) ……………………. (vi)
We know that the Pythagorean identity, sec\(^{2}\) θ - tan\(^{2}\) θ = 1.
Now from (iv) and (vi) we get,
\((\frac{x + y}{x - y})^{2}\) - \((\frac{x + y}{2})^{2}\) = 1
Taking common (x + y)\(^{2}\) we get,
⟹ (x + y)\(^{2}\) ∙ {\(\frac{1}{(x - y)^{2}} - \frac{1}{4}\)} = 1
⟹ (x + y)\(^{2}\) ∙ \(\frac{4 – (x – y)^{2}}{4(x – y)^{2}}\)= 1
⟹ (x + y)\(^{2}\) ∙ {4 – (x – y)\(^{2}\)} = 4(x – y)\(^{2}\)
⟹ 4(x + y)\(^{2}\) - (x + y)\(^{2}\) ∙ (x – y)\(^{2}\) = 4(x – y)\(^{2}\)
⟹ 4(x + y)\(^{2}\) - 4(x – y)\(^{2}\) = (x + y)\(^{2}\) ∙ (x – y)\(^{2}\)
⟹ 4(x\(^{2}\) + y\(^{2}\) + 2xy - x\(^{2}\) - y\(^{2}\) + 2xy) = \((x^{2} + y^{2})^{2}\)
⟹ 4 ∙ 4xy = \((x^{2} + y^{2})^{2}\)
⟹ 16xy = \((x^{2} + y^{2})^{2}\)
⟹ 4\(\sqrt{xy}\) = \(x^{2} + y^{2}\)
Therefore, \(x^{2} + y^{2}\) = 4\(\sqrt{xy}\). (Proved)
2. If a = r cos θ ∙ sin β, b = r cos θ ∙ cos β and c = r sin θ then prove that a\(^{2}\) + b\(^{2}\) + c\(^{2}\) = r\(^{2}\).
Solution:
a\(^{2}\) + b\(^{2}\) + c\(^{2}\) = r\(^{2}\) cos\(^{2}\) θ ∙ sin\(^{2}\) β + r\(^{2}\) cos\(^{2}\) θ ∙ cos\(^{2}\) β + r\(^{2}\) sin\(^{2}\) θ
= r\(^{2}\) cos\(^{2}\) θ(sin\(^{2}\) β + cos\(^{2}\) β) + r\(^{2}\) sin\(^{2}\) θ
= r\(^{2}\) cos\(^{2}\) θ ∙ (1) + r\(^{2}\) sin\(^{2}\) θ, [since We know that the Pythagorean identity, sin\(^{2}\) θ + cos\(^{2}\) θ = 1.]
= r\(^{2}\) cos\(^{2}\) θ + r\(^{2}\) sin\(^{2}\) θ
= r\(^{2}\) (cos\(^{2}\) θ + sin\(^{2}\) θ)
= r\(^{2}\) ∙ (1), [since, sin\(^{2}\) θ + cos\(^{2}\) θ = 1]
= r\(^{2}\)
Therefore, a\(^{2}\) + b\(^{2}\) + c\(^{2}\) = r\(^{2}\). (proved)
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