Question

If $f(x) = \dfrac{{1 - \sin 2\theta + \cos 2\theta }}{{2\cos 2\theta }},$ find the value of $8f({11^ \circ }) \times f({34^ \circ })$ is

Answer 1

Hint: In this question first we simplify the given function into the simplest form by using trigonometric formulas of ${\text{sin2}}\theta$,$\cos {\text{2}}\theta$ . Then find the values of $f({34^ \circ }){\text{ and f(1}}{{\text{1}}^ \circ })$in terms of simplest form and finally put in $8f({11^ \circ }) \times f({34^ \circ })$ to get answer.

Complete step-by-step answer:
We know
$\Rightarrow {\text{sin2}}\theta = 2\sin \theta \cos \theta {\text{eq}}{\text{.1}} \\\ \Rightarrow \cos {\text{2}}\theta = {\cos ^2}\theta - {\sin ^2}\theta {\text{eq}}{\text{.2}} \\\$
$\Rightarrow f(x) = \dfrac{{1 - \sin 2\theta + \cos 2\theta }}{{2\cos 2\theta }} \\\ \Rightarrow f(x) = \dfrac{{({{\cos }^2}\theta {\text{ + }}{{\sin }^2}\theta ) - 2\sin \theta \cos \theta + {{\cos }^2}\theta - {{\sin }^2}\theta }}{{2({{\cos }^2}\theta - {{\sin }^2}\theta )}}{\text{ \\{ }}{\cos ^2}\theta - {\sin ^2}\theta = 1\\} \\\ \Rightarrow f(x) = \dfrac{{({{\cos }^2}\theta {\text{ + }}{{\sin }^2}\theta - 2\sin \theta \cos \theta {\text{)}} + {{\cos }^2}\theta - {{\sin }^2}\theta }}{{2({{\cos }^2}\theta - {{\sin }^2}\theta )}} \\\ \Rightarrow f(x) = \dfrac{{{{(\cos \theta - \sin \theta {\text{)}}}^2} + {{\cos }^2}\theta - {{\sin }^2}\theta }}{{2({{\cos }^2}\theta - {{\sin }^2}\theta )}} \\\$
$\because {\cos ^2}\theta - {\sin ^2}\theta = (\cos \theta {\text{ + }}\sin \theta {\text{)(}}\cos \theta - \sin \theta {\text{)}} \\\$
$\Rightarrow f(x) = \dfrac{{{{(\cos \theta - \sin \theta {\text{)}}}^2} + (\cos \theta {\text{ + }}\sin \theta {\text{)(}}\cos \theta - \sin \theta {\text{)}}}}{{2(\cos \theta {\text{ + }}\sin \theta {\text{)(}}\cos \theta - \sin \theta {\text{)}}}} \\\$
On cancelling the common ${\text{(}}\cos \theta - \sin \theta {\text{)}}$ from above expression, we get
$\Rightarrow f(x) = \dfrac{{\cos \theta - \sin \theta + \cos \theta {\text{ + }}\sin \theta }}{{2(\cos \theta {\text{ + }}\sin \theta {\text{)}}}} \\\ \Rightarrow f(x) = \dfrac{{2\cos \theta }}{{2(\cos \theta {\text{ + }}\sin \theta {\text{)}}}} \\\ \Rightarrow f(x) = \dfrac{{\cos \theta }}{{\cos \theta {\text{ + }}\sin \theta }} \\\$
Now, on dividing numerator and denominator by $\cos \theta$, we get,
$\Rightarrow f(x) = \dfrac{1}{{{\text{1 + tan}}\theta }} \to (3) {\because \text{\\{ tan}}\theta {\text{ = }}\dfrac{{{\text{sin}}\theta }}{{{\text{cos}}\theta }}\\}$
Now, we find $f({11^ \circ }){\text{ and }}f({34^ \circ }){\text{ }}$
$\Rightarrow f({11^ \circ }) = \dfrac{1}{{1 + \tan ({{11}^ \circ })}}{\text{ eq}}{\text{.4}}$
And
$\Rightarrow f({34^ \circ }) = \dfrac{1}{{1 + \tan ({{34}^ \circ })}}$ eq.5
We can rewrite $\tan ({34^ \circ }) = \tan {(45 - 11)^ \circ }$ eq.6
Now using formula $\tan (a - b) = \dfrac{{\tan a - \tan b}}{{1 + \tan a.\tan b}}$,we can write above equation as
$\Rightarrow \tan {(45 - 11)^ \circ } = \dfrac{{\tan {{45}^ \circ } - \tan {{11}^ \circ }}}{{1 + \tan {{45}^ \circ }.\tan {{11}^ \circ }}} \\\ \Rightarrow \tan {(45 - 11)^ \circ } = \dfrac{{1 - \tan {{11}^ \circ }}}{{1 + \tan {{11}^ \circ }}}{\text{ \\{ }}\tan {45^ \circ } = 1\\} {\text{ eq}}{\text{.7}} \\\$
Now using eq.6 and eq.7 we can rewrite eq.5 as
$\Rightarrow f({34^ \circ }) = \dfrac{1}{{1 + \tan ({{34}^ \circ })}} \\\ \Rightarrow f({34^ \circ }) = \dfrac{1}{{1 + (\dfrac{{1 - \tan {{11}^ \circ }}}{{1 + \tan {{11}^ \circ }}})}} \\\ \Rightarrow f({34^ \circ }) = \dfrac{{1 + \tan {{11}^ \circ }}}{2}{\text{ eq}}{\text{.8}} \\\$
Now ,put values $f({34^ \circ }){\text{ and f(1}}{{\text{1}}^ \circ })$ from eq.8 and eq.4 in
$\Rightarrow 8f({11^ \circ }) \times f({34^ \circ }){\text{ = 8}} \times \dfrac{1}{{1 + \tan ({{11}^ \circ })}} \times \dfrac{{1 + \tan {{11}^ \circ }}}{2} \\\ \Rightarrow 8f({11^ \circ }) \times f({34^ \circ }){\text{ = }}4 \\\$
Hence, the value of $8f({11^ \circ }) \times f({34^ \circ })$ is $4$.

Note: Whenever you get this type of problem the key concept of solving this is to check what trick or pattern is involved in it like in this question relation between ${34^ \circ }{{\text{(45 - 11)}}^ \circ }{\text{ and 1}}{{\text{1}}^ \circ }$ . You have to learn as much as possible trigonometric formulas like in this we use trigonometric formulas of $\sin 2\theta ,\cos 2\theta ,$ $\tan (a - b)$.