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

• General solution of the equation sin x = ½
• General solution of the equation cos x = 1/√2
• General solution of the equation tan x = √3
• General Solution of the Equation sin θ = 0
• General Solution of the Equation cos θ = 0
• General Solution of the Equation tan θ = 0
• General Solution of the Equation sin θ = sin ∝
  
• General Solution of the Equation sin θ = 1
• General Solution of the Equation sin θ = -1
• General Solution of the Equation cos θ = cos ∝
• General Solution of the Equation cos θ = 1
• General Solution of the Equation cos θ = -1
• General Solution of the Equation tan θ = tan ∝
• General Solution of a cos θ + b sin θ = c
• Trigonometric Equation Formula
• Trigonometric Equation using Formula
• General solution of Trigonometric Equation
• Problems on Trigonometric Equation

11 and 12 Grade Math

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