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Question: If \( \sin a\theta + \cos b\theta = 0 \) then the possible values of \( \theta \) form: A. an AP ...

If sin⁡aθ+cos⁡bθ=0\sin a\theta + \cos b\theta = 0 then the possible values of θ\theta form:
A. an AP
B. Two APs
C. One GP
D. Two GPs

Explanation

Solution

Hint : We know that the cosine function cos⁡A\cos A can also be written in terms of sine function as sin⁡(90∘−A)\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 θ\theta .

Complete step-by-step answer :
We are given a trigonometric equation sin⁡aθ+cos⁡bθ=0\sin a\theta + \cos b\theta = 0 .
We have to find the values of θ\theta are in AP or in GP.
sin⁡aθ+cos⁡bθ=0→eq(1)\sin a\theta + \cos b\theta = 0 \to eq\left( 1 \right)
Here write cos⁡bθ\cos b\theta in terms of sine using the relation cos⁡A=sin⁡(90∘−A)\cos A = \sin \left( {{{90}^ \circ } - A} \right) , here the value of A is bθb\theta
So therefore,
cos⁡bθ=sin⁡(90∘−bθ) 90∘=π2radians ⇒cos⁡bθ=sin⁡(π2−bθ)  \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⁡aθ+sin⁡(π2−bθ)=0 ⇒sin⁡aθ=−sin⁡(π2−bθ)  \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⁡(π2−bθ)- \sin \left( {\dfrac{\pi }{2} - b\theta } \right)
→sin⁡aθ=sin⁡(bθ−π2)\to \sin a\theta = \sin \left( {b\theta - \dfrac{\pi }{2}} \right)
Both the sides, the functions are sine, so equate their angle measures.
aθ=bθ−π2 ⇒π2=bθ−aθ ⇒θ(a−b)=π2 ⇒θ=π2(a−b) ⇒θ=π2(a−b)+2nπ  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, θ=π2(a−b)+0=π2(a−b)\theta = \dfrac{\pi }{{2\left( {a - b} \right)}} + 0 = \dfrac{\pi }{{2\left( {a - b} \right)}}
When n is equal to 1, θ=π2(a−b)+2π\theta = \dfrac{\pi }{{2\left( {a - b} \right)}} + 2\pi
When n is equal to 2, θ=π2(a−b)+4π\theta = \dfrac{\pi }{{2\left( {a - b} \right)}} + 4\pi
When n is equal to 3, θ=π2(a−b)+6π\theta = \dfrac{\pi }{{2\left( {a - b} \right)}} + 6\pi
As we can see, for two every two consecutive values of θ\theta , there is a difference of 2π2\pi .
So this difference can also be called a common difference.
Therefore, we can say that the possible values of θ\theta are in an APAP.
So, the correct answer is “Option A”.

Note : Here we have added 2nπ2n\pi to the value of θ\theta , because sine is a periodic function and its value repeats after every 2nπ2n\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.