Trigonometric Ratio and Identites

Given $0 < \alpha < {\pi \over 4}$

and $0 < \beta < {\pi \over 4}$

$ \therefore $ $0 > - \beta > - {\pi \over 4}$

$ \therefore $ $0 < \alpha + \beta < {\pi \over 2}$

and $ - {\pi \over 4} < \alpha - \beta < {\pi \over 4}$

As cos($\alpha $ + $\beta $) = 3/5

so ${\tan \left( {\alpha + \beta } \right) = {4 \over 3}}$

As sin( $\alpha $ - $\beta $) = 5/13

so ${\tan \left( {\alpha - \beta } \right) = {5 \over {12}}}$

Now tan(2$\alpha $) = tan($\alpha $ + $\beta $ + $\alpha $ - $\beta $)

= ${{\tan \left( {\alpha + \beta } \right) + \tan \left( {\alpha - \beta } \right)} \over {1 - \tan \left( {\alpha + \beta } \right)\tan \left( {\alpha - \beta } \right)}}$

= ${{{4 \over 3} + {5 \over {12}}} \over {1 - {4 \over 3} \times {5 \over {12}}}}$ = ${{63} \over {16}}$

y = 3cos$\theta $ + 5 $\left( {\sin \theta {{\sqrt 3 } \over 2} - \cos \theta {1 \over 2}} \right)$

${{5\sqrt 3 } \over 2}$ sin$\theta $ + ${1 \over 2}$cos$\theta $

y_max = $\sqrt {{{75} \over 4} + {1 \over 4}} $ = $\sqrt {19} $

Given $\cos {\pi \over {{2^2}}}.\cos {\pi \over {{2^3}}}\,.....\cos {\pi \over {{2^{10}}}}.\sin {\pi \over {{2^{10}}}}$

Let ${\pi \over {{2^{10}}}}\, = \,\theta $

$ \therefore $ ${\pi \over {{2^9}}}\, = \,2\theta $

${\pi \over {{2^8}}}\, = \,{2^2}\theta $

${\pi \over {{2^7}}}\, = \,{2^3}\theta $
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${\pi \over {{2^2}}}\, = \,{2^8}\theta $

So given term becomes,

$\cos {2^8}\theta .\cos {2^7}\theta .....\cos \theta $$.\sin {\pi \over {{2^{10}}}}$

= $(\cos \theta .\cos 2\theta ......\cos {2^8}\theta )\sin {\pi \over {{2^{10}}}}$

= ${{\sin {2^9}\theta } \over {{2^9}\sin \theta }}.\sin {\pi \over {{2^{10}}}}$

= ${{\sin {2^9}\left( {{\pi \over {{2^{10}}}}} \right)} \over {{2^9}\sin {\pi \over {{2^{10}}}}}}.\sin {\pi \over {{2^{10}}}}$

= ${{\sin \left( {{\pi \over 2}} \right)} \over {{2^9}}}$

= ${1 \over {{2^9}}}$ = ${1 \over {512}}$

Note :

$(\cos \theta .\cos 2\theta ......\cos {2^{n - 1}}\theta )$ = ${{\sin {2^n}\theta } \over {{2^n}\sin \theta }}$

Given,

3(sin$\theta $ $-$ cos$\theta $)⁴ + 6(sin$\theta $ + cos$\theta $)² + 4sin⁶$\theta $

= 3[(sin$\theta $ $-$ cos$\theta $)²]² + 6 (sin²$\theta $ + cos²$\theta $ + 2sin$\theta $cos$\theta $) + 4sin⁶$\theta $

= 3[sin²$\theta $ + cos²$\theta $ $-$2sin$\theta $cos$\theta $]² + 6(1 + sin2$\theta $) + 4sin⁶$\theta $

= 3(1 $-$ sin2$\theta $)² + 6(1 + sin2$\theta $) + 4sin⁶$\theta $

= 3 (1 $-$ 2 sin2$\theta $ + sin²2$\theta $) + 6 + 6sin2$\theta $ + 4sin⁶$\theta $

= 3 $-$ 6sin2$\theta $ + 3sin²2$\theta $ + 6 + 6sin2$\theta $ + 4sin⁶$\theta $

= 9 + 3sin²2$\theta $ + 4 sin⁶$\theta $

= 9 + 3(2sin$\theta $cos$\theta $)² + 4(1 $-$ cos²$\theta $)³

= 9 + 12sin²$\theta $ cos²$\theta $ + 4 (1 $-$ cos⁶$\theta $ $-$ 3cos²$\theta $ + 3cos⁴$\theta $)

= 13 + 12 (1 $-$ cos²$\theta $ $-$ 4cos⁶$\theta $ $-$ 12cos$\theta $ + 12 cos⁴$\theta $

= 13 + 12 cos²$\theta $ $-$ 12 cos⁴$\theta $ $-$ 4cos⁶$\theta $ $-$ 12 cos²$\theta $ + 12 cos⁴$\theta $

= 13 $-$ 4 cos⁶$\theta $

Given that,

$5\left( {{{\tan }^2}x - {{\cos }^2}x} \right) = 2\cos 2x + 9$

$ \Rightarrow 5\left( {{{{{\sin }^2}x} \over {{{\cos }^2}x}} - {{\cos }^2}x} \right) = 2\left( {2{{\cos }^2}x - 1} \right) + 9$

Let ${\cos ^2}x = t,$ then we have

$5\left( {{{1 - t} \over t} - t} \right) = 2\left( {2t - 1} \right) + 9$

$ \Rightarrow 5\left( {{{1 - t - {t^2}} \over t}} \right) = 4t - 2 + 9$

$ \Rightarrow 5 - 5t - 5{t^2} = 4{t^2} + 7t$

$ \Rightarrow 9{t^2} + 12t - 5 = 0$

$ \Rightarrow 9{t^2} + 15t - 3t - 5 = 0$

$ \Rightarrow 3t\left( {3t + 5} \right) - 1\left( {3t + 5} \right) = 0$

$ \Rightarrow \left( {3t + 5} \right)\left( {3t - 1} \right) = 0$

$\therefore$ $t = {1 \over 3}$ and $t = - {5 \over 3}$

If $t = - {5 \over 3}$ then ${\cos ^2}x$ is negative

So , $t$ can not be $ - {5 \over 3}$.

So, correct value of $t = {1 \over 3}$ then $\cos {}^2x = t = {1 \over 3}$

$\therefore\,\,\,$ $\cos 4x$

$ = 2{\cos ^2}2x - 1$

$ = 2{\left[ {2{{\cos }^2}x - 1} \right]^2} - 1$

$ = 2.{\left[ {2.{1 \over 3} - 1} \right]^2} - 1$

$ = 2.{\left( { - {1 \over 3}} \right)^2} - 1$

$ = {2 \over 9} - 1$

$ = - {7 \over 9}$

Given,

4 + ${1 \over 2}$ sin² 2x $-$ 2cos⁴ x

= 4 + ${1 \over 2}$ (2sinx cosx)² $-$ 2cos⁴x

= 4 + ${1 \over 2}$ $ \times $ 4 sin²x cos²x $-$ 2cos⁴ x

= 4 + 2 (1 $-$ cos²x) cos²x $-$ 2cos⁴ x

= 4 + 2 cos²x $-$ 4cos⁴x

= $-$ 4 $\left\{ {\cos } \right.$⁴x $-$ $\left. {{{{{\cos }^2}x} \over 2} - 1} \right\}$

= $-$ 4 $\left\{ {\cos } \right.$⁴x $-$ 2 . ${1 \over 4}$ . cos²x + $\left. {{1 \over {16}} - {1 \over {16}} - 1} \right\}$

= $-$ 4 $\left\{ {{{\left( {{{\cos }^2}x - {1 \over 4}} \right)}^2} - {{17} \over {16}}} \right\}$

We know,

O $ \le $ cos²x $ \le $ 1

$ \Rightarrow $   $ - {1 \over 4}$ $ \le $cos²x $ - {1 \over 4}$ $ \le $ ${3 \over 4}$

$ \Rightarrow $   O $ \le $ ${\left( {{{\cos }^2}x - {1 \over 4}} \right)^2}$ $ \le $ ${9 \over {16}}$

$ \Rightarrow $   $-$ ${17 \over {16}}$ $ \le $ ${\left( {{{\cos }^2}x - {1 \over 4}} \right)^2}$ $-$ ${{17} \over {16}}$ $ \le $ ${9 \over {16}}$ $-$ ${{17} \over {16}}$

$ \Rightarrow $    $-$ ${{17} \over {16}}$ $ \le $ ${\left( {{{\cos }^2}x - {1 \over 4}} \right)^2}$ $-$ ${{17} \over {16}}$ $ \le $ $-$ ${{1} \over {2}}$

$ \Rightarrow $   ${{17} \over {4}}$ $ \ge $ $-$ 4 $\left\{ {{{\left( {{{\cos }^2}x - {1 \over 4}} \right)}^2} - {{17} \over {16}}} \right\} \ge 2$

$ \therefore $   Maximum value, M = ${{17} \over 4}$

      Minimum value, m = $2$

$ \therefore $   M $-$ m = ${{17} \over 4}$ $-$ $2$ = ${{9} \over 4}$

Let ${f_k}\left( x \right) = {1 \over k}\left( {{{\sin }^k}x + {{\cos }^k}.x} \right)$

Consider

${f_4}\left( x \right) - {f_6}\left( x \right) $

$=$ ${1 \over 4}\left( {{{\sin }^4}x + {{\cos }^4}x} \right) - {1 \over 6}\left( {{{\sin }^6}x + {{\cos }^6}x} \right)$

$ = {1 \over 4}\left[ {1 - 2{{\sin }^2}x{{\cos }^2}x} \right] - {1 \over 6}\left[ {1 - 3{{\sin }^2}x{{\cos }^2}x} \right]$

$ = {1 \over 4} - {1 \over 6} = {1 \over {12}}$

Given expression can be written as

${{\sin A} \over {\cos A}} \times {{sin\,A} \over {\sin A - \cos A}} + {{\cos A} \over {\sin A}} \times {{\cos A} \over {\cos A - sin\,A}}$

(As $\tan A = {{\sin A} \over {\cos A}}$ and $\cot A = {{\cos A} \over {\sin A}}$ )

$ = {1 \over {\sin A - \cos A}}\left\{ {{{{{\sin }^3}A - {{\cos }^3}A} \over {\cos A\sin A}}} \right\}$

$ = {{{{\sin }^2}A + \sin A\cos A + {{\cos }^2}\,A} \over {\sin A\cos A}}$

$ = 1 + \sec\, A{\mathop{\rm cosec}\nolimits} \,A$

$A = {\sin ^2}x + {\cos ^4}x$

$ = {\sin ^2}x + {\cos ^2}x\left( {1 - {{\sin }^2}x} \right)$

$ = {\sin ^2}x + {\cos ^2}x - {1 \over 4}{\left( {2\sin x.\cos x} \right)^2}$

$ = 1 - {1 \over 4}{\sin ^2}\left( {2x} \right)$

Now $0 \le {\sin ^2}\left( {2x} \right) \le 1$

$ \Rightarrow 0 \ge - {1 \over 4}{\sin ^2}\left( {2x} \right) \ge - {1 \over 4}$

$ \Rightarrow 1 \ge 1 - {1 \over 4}{\sin ^2}\left( {2x} \right) \ge 1 - {1 \over 4}$

$ \Rightarrow 1 \ge A \ge {3 \over 4}$

$\cos \left( {\alpha + \beta } \right) = {4 \over 5} \Rightarrow \tan \left( {\alpha + \beta } \right) = {3 \over 4}$

$\sin \left( {\alpha - \beta } \right) = {5 \over {13}} \Rightarrow \tan \left( {\alpha - \beta } \right) = {5 \over {12}}$

$\tan 2\alpha = \tan \left[ {\left( {\alpha + \beta } \right) + \left( {\alpha - \beta } \right)} \right]$

$ = {{{3 \over 4} + {5 \over {12}}} \over {1 - {3 \over 4}.{5 \over {12}}}} = {{56} \over {33}}$

$\cos x + \sin x = {1 \over 2}$

$ \Rightarrow {\left( {\cos x + {\mathop{\rm sinx}\nolimits} } \right)^2} = {1 \over 4}$

$ \Rightarrow {\cos ^2}x + {\sin ^2}x + 2\cos x\sin x = {1 \over 4}$
$\left[ \because {{{\cos }^2}x + {{\sin }^2}x = 1\, \,and \,\,2\cos x\sin x = \sin 2x} \right]$

$ \Rightarrow 1 + \sin 2x = {1 \over 4}$

$ \Rightarrow \sin 2x = - {3 \over 4},$ so $x$ is obtuse and

${{2\tan x} \over {1 + {{\tan }^2}x}} = - {3 \over 4}$

$ \Rightarrow 3{\tan ^2}x + 8\tan x + 3 = 0$

$\therefore$ $\tan x = {{ - 8 \pm \sqrt {64 - 36} } \over 6}$

$ = {{ - 4 \pm \sqrt 7 } \over 3}$

as $\tan x < 0\,$

$\therefore$ $\tan x = {{ - 4 - \sqrt 7 } \over 3}$

Given $u = \sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } $$+ \sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } $

$\therefore$ ${u^2} = {a^2}{\cos ^2}\theta + {b^2}{\sin ^2}\theta + {a^2}{\sin ^2}\theta + {b^2}{\cos ^2}\theta $
              $ + 2\sqrt {\left( {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } \right)} \times \sqrt {\left( {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } \right)} $

$ \Rightarrow {u^2} = $ ${a^2} + {b^2}$$ + $$2\sqrt {\left( {{a^4} + {b^4}} \right){{\cos }^2}\theta {{\sin }^2}\theta + {a^2}{b^2}\left( {{{\cos }^4}\theta + {{\sin }^4}\theta } \right)} $

$ \Rightarrow {u^2} = $ ${a^2} + {b^2}$$ + $$2\sqrt {\left( {{a^4} + {b^4}} \right){{\cos }^2}\theta {{\sin }^2}\theta + {a^2}{b^2}\left( {1 - 2{{\cos }^2}\theta {{\sin }^2}\theta } \right)} $

$ \Rightarrow {u^2} = $ ${a^2} + {b^2}$$ + $$2\sqrt {\left( {{a^4} + {b^4} - 2{a^2}{b^2}} \right){{\cos }^2}\theta {{\sin }^2}\theta + {a^2}{b^2}} $

$ \Rightarrow {u^2} = $ ${a^2} + {b^2}$$ + $$2\sqrt {{{\left( {{a^2} - {b^2}} \right)}^2}{{{{\left( {2\cos \theta \sin \theta } \right)}^2}} \over 4} + {a^2}{b^2}} $

$ \Rightarrow {u^2} = $ ${a^2} + {b^2}$$ + $$2\sqrt {{{\left( {{a^2} - {b^2}} \right)}^2}{{{{\sin }^2}2\theta } \over 4} + {a^2}{b^2}} $

We know $0 \le {\sin ^2}2\theta \le 1$

$\therefore$ $0 \le {\left( {{a^2} - {b^2}} \right)^2}{{{{\sin }^2}2\theta } \over 4} \le {{{{\left( {{a^2} - {b^2}} \right)}^2}} \over 4}$

$ \Rightarrow $ ${a^2}{b^2} \le $ ${\left( {{a^2} - {b^2}} \right)^2}{{{{\sin }^2}2\theta } \over 4} + {a^2}{b^2}$$ \le {{{{\left( {{a^2} - {b^2}} \right)}^2}} \over 4} + {a^2}{b^2}$

$\therefore$ Min value of ${u^2} = {a^2} + {b^2}$ $ + 2\sqrt {{a^2}{b^2}} $ = ${\left( {a + b} \right)^2}$

and Max value of ${u^2} = {a^2} + {b^2}$ $ + 2\sqrt {{{{{\left( {{a^2} - {b^2}} \right)}^2}} \over 4} + {a^2}{b^2}} $

$= {a^2} + {b^2}$ $ + 2\sqrt {{{{{\left( {{a^2} - {b^2}} \right)}^2} + 4{a^2}{b^2}} \over 4}} $

$= {a^2} + {b^2}$ $ + 2\sqrt {{{{{\left( {{a^2} + {b^2}} \right)}^2}} \over 4}} $

$= {a^2} + {b^2}$ $+\, {a^2} + {b^2}$

${ = 2\left( {{a^2} + {b^2}} \right)}$

Max of ${u^2}$ - Min of ${u^2}$ = ${2\left( {{a^2} + {b^2}} \right)}$ - ${\left( {a + b} \right)^2}$

= ${2\left( {{a^2} + {b^2}} \right)}$ - ${\left( {{a^2} + {b^2} + 2ab} \right)}$

= $\sqrt {{{ = 2\left( {{a^2} + {b^2} - 2ab} \right)} \over 4}} $

= ${\left( {a - b} \right)^2}$

Given $sin{\mkern 1mu} \alpha + \sin \beta = - {{21} \over {65}}$ .........(1)

and $\cos \alpha + \cos \beta = - {{27} \over {65}}$ ........(2)

Square and add (1) and (2) you will get

$2\left( {1 + \cos \alpha \cos \beta + \sin \alpha \sin \beta } \right)$$ = {{{{\left( {21} \right)}^2} + {{\left( {27} \right)}^2}} \over {{{\left( {65} \right)}^2}}}$

$ \Rightarrow $ $2\left( {1 + \cos \left( {\alpha - \beta } \right)} \right) = {{1170} \over {{{\left( {65} \right)}^2}}}$

$ \Rightarrow $ $4{\cos ^2}{{\alpha - \beta } \over 2}$$ = {{1170} \over {{{\left( {65} \right)}^2}}}$

$ \Rightarrow $ ${\cos ^2}{{\alpha - \beta } \over 2}$$ = {9 \over {130}}$

$\therefore$ $\cos {{\alpha - \beta } \over 2} = \pm {3 \over {\sqrt {130} }}$

[ But $\cos {{\alpha - \beta } \over 2} \ne + {3 \over {\sqrt {130} }}$

as $\pi < \alpha - \beta < 3\pi $

$ \Rightarrow $ ${\pi \over 2} < {{\alpha - \beta } \over 2} < {{3\pi } \over 2}$

$ \Rightarrow $ $\cos {{\alpha - \beta } \over 2} < 0$ ]

So $\cos {{\alpha - \beta } \over 2} = - {3 \over {\sqrt {130} }}$