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# If $\sin a\theta + \cos b\theta = 0$ then the possible values of $\theta$ form:A. an APB. Two APsC. One GPD. Two GPs

Last updated date: 13th Jun 2024
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Hint: We know that the cosine function $\cos A$ can also be written in terms of sine function as $\sin \left( {{{90}^ \circ } - A} \right)$ . As the given trigonometric function has a sine function and cosine function, convert the cosine function into sine using the given relation. And then find the values of $\theta$ .

We are given a trigonometric equation $\sin a\theta + \cos b\theta = 0$ .
We have to find the values of $\theta$ are in AP or in GP.
$\sin a\theta + \cos b\theta = 0 \to eq\left( 1 \right)$
Here write $\cos b\theta$ in terms of sine using the relation $\cos A = \sin \left( {{{90}^ \circ } - A} \right)$ , here the value of A is $b\theta$
So therefore,
$\cos b\theta = \sin \left( {{{90}^ \circ } - b\theta } \right) \\ {90^ \circ } = \dfrac{\pi }{2}radians \\ \Rightarrow \cos b\theta = \sin \left( {\dfrac{\pi }{2} - b\theta } \right) \\$
On substituting the above value in equation 1, we get
$\sin a\theta + \sin \left( {\dfrac{\pi }{2} - b\theta } \right) = 0 \\ \Rightarrow \sin a\theta = - \sin \left( {\dfrac{\pi }{2} - b\theta } \right) \\$
Sending the minus inside in the above right hand side term, which is $- \sin \left( {\dfrac{\pi }{2} - b\theta } \right)$
$\to \sin a\theta = \sin \left( {b\theta - \dfrac{\pi }{2}} \right)$
Both the sides, the functions are sine, so equate their angle measures.
$a\theta = b\theta - \dfrac{\pi }{2} \\ \Rightarrow \dfrac{\pi }{2} = b\theta - a\theta \\ \Rightarrow \theta \left( {a - b} \right) = \dfrac{\pi }{2} \\ \Rightarrow \theta = \dfrac{\pi }{{2\left( {a - b} \right)}} \\ \Rightarrow \theta = \dfrac{\pi }{{2\left( {a - b} \right)}} + 2n\pi \\$
When n is equal to 0, $\theta = \dfrac{\pi }{{2\left( {a - b} \right)}} + 0 = \dfrac{\pi }{{2\left( {a - b} \right)}}$
When n is equal to 1, $\theta = \dfrac{\pi }{{2\left( {a - b} \right)}} + 2\pi$
When n is equal to 2, $\theta = \dfrac{\pi }{{2\left( {a - b} \right)}} + 4\pi$
When n is equal to 3, $\theta = \dfrac{\pi }{{2\left( {a - b} \right)}} + 6\pi$
As we can see, for two every two consecutive values of $\theta$ , there is a difference of $2\pi$ .
So this difference can also be called a common difference.
Therefore, we can say that the possible values of $\theta$ are in an $AP$.
So, the correct answer is “Option A”.

Note: Here we have added $2n\pi$ to the value of $\theta$ , because sine is a periodic function and its value repeats after every $2n\pi$ radians. And an AP is a sequence in which every term starting from the second term is obtained by adding a fixed value to its previous term; this fixed value is called common difference whereas a GP is a sequence in which every term starting from the second term is obtained by multiplying a fixed value to its previous term; this fixed value is called common ratio. So do not confuse an AP with a GP.