Question

Find the maximum and minimum value of trigonometric ratio’s \[cosA \times cosB\] , if $A + B = {90^ \circ }$ ?

Answer 1

Hint: We need to learn different trigonometric ratios and identities. We will substitute the value of A in terms of B. we will try to convert them in a trigonometric ratio whose value can be easily identified. To find it’s maxima and minima we have to be very careful about its range and domain.

Complete step-by-step answer:
We have given $A + B = {90^ \circ }$ ,
So, $A = {90^ \circ } - B$
Now, we have to find the maximum and minimum value of \[cosA \times cosB\]
We assumed
\[f(x) = cosA \times cosB\]
We substitute the value of A,
\[f(x) = cos\left( {{{90}^ \circ } - B} \right) \times cosB\]
We have substituted \[\sin \theta = \cos \left( {{{90}^ \circ } - \theta } \right)\]
\[f(x) = \sin B \times cosB\]\[\]
We know that \[\sin 2\theta = 2 \times \sin \theta \times \cos \theta \]
\[f(x) =\dfrac{{\sin 2B}}{2}\]
Now, we have to find the maximum value of the above function and to maximize it we have to put the maximum value of the sin function. Similarly, to minimize it, we have to put the minimum value of sin function.
We know that the maximum and minimum value of sin function is 1 and -1 respectively.
So, maximum value of \[f(x) =\dfrac{1}{2}\]
Minimum value of \[f(x) =\dfrac{{ - 1}}{2}\]
The value of \[cosA \times cosB\] belongs to \[\left( {\dfrac{1}{2},\dfrac{{ - 1}}{2}} \right)\]

Note: The key for solving this type of problem is to be familiar with the formulae of trigonometric terms of the form \[\sin \theta = \cos \left( {{{90}^ \circ } - \theta } \right)\] and \[\sin 2\theta = 2 \times \sin \theta \times \cos \theta \] , as well as other relevant formulae of this type. The method used to solve these problems, particularly trigonometric problems, can make a significant difference. By solving more problems of this type, the approach taken can be improved.