Question

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

Answer 1

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 $ .

Complete step-by-step answer:
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.