Name: Trigonometric Equation to Convert Cosine or Sine
Uploaded: 2019-07-19
Duration: 4 min 11 s
Description: Watch the video to learn how to solve trigonometric equations involving both sine and cosine simultaneously to find the general solution of the given equation.

Question

Show that the solutions of

\cos p θ = \sin q θ

form two series each of which is an A.P. Find also the common difference of each A.P.

(p \neq \pm q)

Accepted Answer

Given trigonometric equation: $\cos p θ = \sin q θ$

$\Rightarrow \cos p θ = \cos (\frac{π}{2} - q θ) [∵ \cos (\frac{π}{2} - θ) = \sin θ]$

$\Rightarrow p θ = 2 n π \pm (\frac{π}{2} - q θ), n \in Z [∵ \cos α = \cos β \Rightarrow α = 2 n π \pm β, n \in Z]$

$\Rightarrow θ = \frac{1}{(p \pm q)} (4 n \pm 1) \frac{π}{2}, n \in Z$

Case $(1)$ : $θ = \frac{1}{(p \pm q)} (4 n + 1) \frac{π}{2}, n \in Z$

Now, if $n = 0 \Rightarrow θ = \frac{π}{2 (p \pm q)}$

If $n = 1 \Rightarrow θ = \frac{5 π}{2 (p \pm q)}$

If $n = 2 \Rightarrow θ = \frac{9 π}{2 (p \pm q)}$ and so on.

The sequence formed as $\frac{π}{2 (p \pm q)}, \frac{5 π}{2 (p \pm q)}, \frac{9 π}{2 (p \pm q)}, . . .$

Let $t_{1} = \frac{π}{2 (p \pm q)}$ , $t_{2} = \frac{5 π}{2 (p \pm q)}$ , $t_{3} = \frac{9 π}{2 (p \pm q)}$ ,...

Now, the common difference $d = t_{2} - t_{1} = \frac{5 π}{2 (p \pm q)} - \frac{π}{2 (p \pm q)} = \frac{2 π}{(p \pm q)}$ and $t_{3} - t_{2} = \frac{9 π}{2 (p \pm q)} - \frac{5 π}{2 (p \pm q)} = \frac{2 π}{(p \pm q)}$

So, clearly the solutions are in A.P. with common difference $d = \frac{2 π}{p \pm q}$ .

Case $(2)$ : $θ = \frac{1}{(p \pm q)} (4 n - 1) \frac{π}{2}, n \in Z$

Now, if $n = 0 \Rightarrow θ = - \frac{π}{2 (p \pm q)}$

If $n = 1 \Rightarrow θ = \frac{3 π}{2 (p \pm q)}$

If $n = 2 \Rightarrow θ = \frac{7 π}{2 (p \pm q)}$ and so on.

The sequence formed as $- \frac{π}{2 (p \pm q)}, \frac{3 π}{2 (p \pm q)}, \frac{7 π}{2 (p \pm q)}, . . .$

Let $t_{1} = - \frac{π}{2 (p \pm q)}$ , $t_{2} = \frac{3 π}{2 (p \pm q)}$ , $t_{3} = \frac{7 π}{2 (p \pm q)}$ ,...

Now, the common difference $d = t_{2} - t_{1} = \frac{3 π}{2 (p \pm q)} + \frac{π}{2 (p \pm q)} = \frac{2 π}{(p \pm q)}$ and $t_{3} - t_{2} = \frac{7 π}{2 (p \pm q)} - \frac{3 π}{2 (p \pm q)} = \frac{2 π}{(p \pm q)}$

So, clearly the solutions are in A.P. with common difference $d = \frac{2 π}{p \pm q}$ .

Show that the solutions of $\cos p θ = \sin q θ$ form two series each of which is an A.P. Find also the common difference of each A.P. $(p \neq \pm q)$

Important Questions on Trigonometric Equations

Learn and Prepare for any exam you want !

Show that the solutions of cospθ=sinqθ form two series each of which is an A.P. Find also the common difference of each A.P. (p≠±q)

Important Questions on Trigonometric Equations

Learn and Prepare for any exam you want !

Show that the solutions of $\cos p θ = \sin q θ$ form two series each of which is an A.P. Find also the common difference of each A.P. $(p \neq \pm q)$