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| ≤ \(\sqrt{a^{2} + b^{2}}\).

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 = \(\frac{π}{2}\), - \(\frac{π}{6}\), - \(\frac{3π}{2}\), \(\frac{11π}{6}\).

● Trigonometric Equations







11 and 12 Grade Math

From a cos θ + b sin θ = c to HOME PAGE


New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.



Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



Share this page: What’s this?