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

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.

Recent Articles

1. Formation of Greatest and Smallest Numbers | Arranging the Numbers

May 19, 24 03:36 PM

the greatest number is formed by arranging the given digits in descending order and the smallest number by arranging them in ascending order. The position of the digit at the extreme left of a number…

2. Formation of Numbers with the Given Digits |Making Numbers with Digits

May 19, 24 03:19 PM

In formation of numbers with the given digits we may say that a number is an arranged group of digits. Numbers may be formed with or without the repetition of digits.

3. Arranging Numbers | Ascending Order | Descending Order |Compare Digits

May 19, 24 02:23 PM

We know, while arranging numbers from the smallest number to the largest number, then the numbers are arranged in ascending order. Vice-versa while arranging numbers from the largest number to the sma…

4. Comparison of Numbers | Compare Numbers Rules | Examples of Comparison

May 19, 24 01:26 PM

Rule I: We know that a number with more digits is always greater than the number with less number of digits. Rule II: When the two numbers have the same number of digits, we start comparing the digits…