Question

If $\alpha + \beta = {90^0}$ , then the maximum value of $\sin \alpha \sin \beta $ is ?
$\left( A \right)1$
$\left( B \right)\dfrac{1}{2}$
$\left( C \right)2$
$\left( D \right)$ None of these

Answer 1

Hint: The given question deals with basic simplification of trigonometric functions. We should always remember the standard trigonometric identities to solve this type of question. In the given question they have asked the value of $\sin \alpha \sin \beta $. The standard formula for $\sin \alpha \sin \beta $ is;
$ 2\sin A\sin B = \cos \left( {A - B} \right) - \cos \left( {A + B} \right)$ . Basic algebraic rules and operations have to be kept in mind while doing simplification for the given problem.

Complete step by step answer:
We can also write the given question as;
$ \Rightarrow \sin \alpha \sin \beta = \dfrac{1}{2}\left( {2\sin \alpha \sin \beta } \right)$
(by multiplying and dividing the R.H.S by $2$ , so that we can apply the standard formula)
As stated by the above formula, $2\sin \alpha \sin \beta $can be replaced by;
$ \Rightarrow \sin \alpha \sin \beta = \dfrac{1}{2}\left[ {\cos \left( {\alpha - \beta } \right) - \cos \left( {\alpha + \beta } \right)} \right]$
$\left( {\because \alpha + \beta = {{90}^0}} \right)$ ( as given in the question)
Put the value $\alpha + \beta = {90^0}$ , in the above formula, we get;
$ \Rightarrow \sin \alpha \sin \beta = \dfrac{1}{2}\left[ {\cos \left( {\alpha - \beta } \right) - \cos \left( {{{90}^0}} \right)} \right]$
We know that the value of the cosine function at ${90^0}$ is zero , i.e.
$\left( {\because \cos {{90}^0} = 0} \right)$
Therefore, we get;
$ \Rightarrow \sin \alpha \sin \beta = \dfrac{1}{2}\cos \left( {\alpha - \beta } \right)$
According to the question we need the maximum value of $\sin \alpha \sin \beta $ , that is possible only ;
If $\cos \left( {\alpha - \beta } \right) = 1$ ;
If this is true, than we get;
$ \Rightarrow \sin \alpha \sin \beta = \dfrac{1}{2} \times \left( 1 \right)$
Then,
$ \Rightarrow \sin \alpha \sin \beta = \dfrac{1}{2}$
Therefore, the maximum value of $\sin \alpha \sin \beta = \dfrac{1}{2}$ when $\alpha + \beta = {90^0}$.
Hence option $\left( B \right)$ is the correct answer.

Note: In the question they have asked about the maximum value of $\sin \alpha \sin \beta $ for $\alpha + \beta = {90^0}$. The maximum value of a given function indicates its highest point (in terms of amplitude) on a graph. The amplitude of a function is basically the measure of its height, and this height is changing continuously, therefore amplitude is different from point to point. For example:-The maximum value of the sine function ranges from $ - 1$ to $1$.