Trigonometric equations of the form a cos theta plus b sin theta equals c (i.e. a cos θ + b sin θ = c) where a, b, c are constants (a, b, c ∈ R) and |c| ≤ \(\sqrt{a^{2} + b^{2}}\).
To solve this type of questions, we first reduce them in the form cos θ = cos α or sin θ = sin α.
We use the following ways to solve the equations of the form a cos θ + b sin θ = c.
(i) First write the equation a cos θ + b sin θ = c.
(ii) Let a = r cos ∝ and b = r sin ∝ where, r > 0 and - \(\frac{π}{2}\) ≤ ∝ ≤ \(\frac{π}{2}\).
Now, a\(^{2}\) + b\(^{2}\) = r\(^{2}\) cos\(^{2}\) ∝ + r\(^{2}\) sin\(^{2}\) ∝ = r\(^{2}\) (cos\(^{2}\) ∝ + sin\(^{2}\) ∝) = r\(^{2}\)
or, r = \(\sqrt{a^{2} + b^{2}}\)
and tan ∝ = \(\frac{r sin ∝}{r cos ∝}\) = \(\frac{b}{a}\) i.e. ∝ = tan\(^{-1}\) (\(\frac{b}{a}\)).
(iii) Using the substitution in step (ii), the equation
reduce to r cos (θ - ∝) = c
⇒ cos (θ - ∝) = \(\frac{c}{r}\) = cos β
Now, putting the value of a and b in a cos θ + b sin θ = c we get,
r cos ∝ cos θ + r sin ∝ sin θ = c
⇒ r cos (θ - ∝) = c
⇒ cos (θ - ∝) = \(\frac{c}{r}\) = cos β (say)
(iv) Solve the equation obtained in step (iii) by using the formula of cos θ = cos ∝.
cos (θ - ∝) = cos β
Therefore, θ - ∝ = 2nπ ± β
⇒ θ = 2nπ ± β + ∝ where n ∈ Z
and cos β = \(\frac{c}{r}\) = \(\frac{c}{\sqrt{a^{2} + b^{2}}}\)
Note: If |c| > \(\sqrt{a^{2} + b^{2}}\), the given equation has no solution.
From the above discussion we observe that a cos θ + b sin θ = c can be solved when |cos β| ≤ 1
⇒ |\(\frac{c}{\sqrt{a^{2} + b^{2}}}\)| ≤ 1
⇒ |c| ≤ \(\sqrt{a^{2} + b^{2}}\)
1. Solve the trigonometric equation √3 cos θ + sin θ = √2.
Solution:
√3 cos θ + sin θ = √2
This trigonometric equation is of the form a cos θ + b sin θ = c where a = √3, b = 1 and c = √2.
Let a = r cos ∝ and b = r sin ∝ i.e., √3 = r cos ∝ and 1 = r sin ∝.
Then r = \(\sqrt{a^{2} + b^{2}}\) = \(\sqrt{(√3)^{2} + 1^{2}}\) = 2
and tan ∝ = \(\frac{1}{√3}\) ⇒ ∝ = \(\frac{π}{6}\)
Substituting a = √3 = r cos ∝ and b = 1 = r sin ∝ in the given equation √3 cos θ + sin θ = √2 we get,
r cos ∝ cos θ + r sin ∝ sin θ = √2
⇒ r cos (θ - ∝) = √2
⇒ 2 cos (θ - \(\frac{π}{6}\)) = √2
⇒ cos (θ - \(\frac{π}{6}\)) = \(\frac{√2}{2}\)
⇒ cos (θ - \(\frac{π}{6}\)) = \(\frac{1}{√2}\)
⇒ cos (θ - \(\frac{π}{6}\)) = cos \(\frac{π}{4}\)
⇒ (θ - \(\frac{π}{6}\))= 2nπ ± \(\frac{π}{4}\), where n = 0, ± 1, ± 2,…………
⇒ θ = 2nπ ± \(\frac{π}{4}\) + \(\frac{π}{6}\), where n = 0, ± 1, ± 2,…………
⇒ θ = 2nπ + \(\frac{π}{4}\) + \(\frac{π}{6}\) or θ = 2nπ - \(\frac{π}{4}\) + \(\frac{π}{6}\), where n = 0, ± 1, ± 2,…………
⇒ θ = 2nπ + \(\frac{5π}{12}\) or θ = 2nπ - \(\frac{π}{12}\), where n = 0, ± 1, ± 2,…………
2. Solve √3 cos θ + sin θ = 1 (-2π < θ < 2π)
Solution:
√3 cos θ + sin θ = 1
This trigonometric equation is of the form a cos θ + b sin θ = c where a = √3, b = 1 and c = 1.
Let a = r cos ∝ and b = r sin ∝ i.e., √3 = r cos ∝ and 1 = r sin ∝.
Then r = \(\sqrt{a^{2} + b^{2}}\) = \(\sqrt{(√3)^{2} + 1^{2}}\) = 2
and tan ∝ = \(\frac{1}{√3}\) ⇒ ∝ = \(\frac{π}{6}\)
Substituting a = √3 = r cos ∝ and b = 1 = r sin ∝ in the given equation √3 cos θ + sin θ = √2 we get,
r cos ∝ cos θ + r sin ∝ sin θ = 1
⇒ r cos (θ - ∝) = 1
⇒ 2 cos (θ - \(\frac{π}{6}\)) = 1
⇒ cos (θ - \(\frac{π}{6}\)) = \(\frac{1}{2}\)
⇒ cos (θ - \(\frac{π}{6}\)) = cos \(\frac{π}{3}\)
⇒ (θ - \(\frac{π}{6}\))= 2nπ ± \(\frac{π}{3}\), where n = 0, ± 1, ± 2, …………
⇒ θ = 2nπ ± \(\frac{π}{3}\) + \(\frac{π}{6}\), where n = 0, ± 1, ± 2, …………
⇒ Either, θ = 2nπ + \(\frac{π}{3}\) + \(\frac{π}{6}\) (4n + 1)\(\frac{π}{2}\) ………..(1) or, θ = 2nπ - \(\frac{π}{3}\) + \(\frac{π}{6}\) = 2nπ - \(\frac{π}{6}\) ………..(2) Where 0, ± 1, ± 2, …………
Now, putting n = 0 in equation (1) we get, θ = \(\frac{π}{2}\),
Putting n = 1 in equation (1) we get, θ = \(\frac{5π}{2}\),
Putting n = -1 in equation (1) we get, θ = - \(\frac{3π}{2}\),
and putting n = 0 in equation (2) we get, θ = - \(\frac{π}{6}\)
Putting n = 1 in equation (2) we get, θ = \(\frac{11π}{6}\)
Putting n = -1 in equation (2) we get, θ = - \(\frac{13π}{6}\)
Therefore,the required solution of the trigonometric equation √3 cos θ + sin θ = 1 in -2π < θ < 2π are θ = \(\frac{π}{2}\), - \(\frac{π}{6}\), - \(\frac{3π}{2}\), \(\frac{11π}{6}\).
11 and 12 Grade Math
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