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a cos θ + b sin θ = c

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| ≤ a2+b2.

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}.

 Trigonometric Equations







11 and 12 Grade Math

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