STEP I 2007 question 3 solution

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STEP I 2007 question 3 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 2007 question 3 solution

$\cos ^{4} { \theta } - \sin ^{4} { \theta } \equiv \left( \cos ^{2} { \theta } - \sin ^{2} { \theta } \right) \left( \cos ^{2} { \theta } + \sin ^{2} { \theta } \right) \equiv \cos ^{2} { \theta } - \sin ^{2} { \theta } = \cos{ \left( 2 \theta \right) }$ .

$\displaystyle \cos^{4} { \theta } + \sin ^{4} { \theta }$

$\equiv (1 - \sin ^{2} { \theta })( \cos ^{2} { \theta } ) + \sin ^{4} { \theta }$

$\equiv \cos ^{2} { \theta } - \cos ^{2} { \theta } \sin ^{2} { \theta } + \sin ^{4} { \theta }$

$\equiv \cos ^{2} { \theta } - \frac{1}{4} \sin ^{2} { \left( 2 \theta \right) } + (1 - \cos ^{2} { \theta } )( \sin ^{2} { \theta } )$

$\equiv ( \sin ^{2} { \theta } + \cos ^{2} { \theta } ) - \frac{1}{2} \sin ^{2} { \left( 2 \theta \right) }$

$\equiv 1 - \frac{1}{2} \sin ^{2} { \left( 2 \theta \right) }$ .

Let $\displaystyle \int _{0} ^{ \frac{ \pi }{2} } \cos ^{4} { \theta } \,d \theta = I$

Let $\displaystyle \int _{0} ^{ \frac{ \pi }{2} } \sin ^{4} { \theta } \,d \theta = J$

Hence:

$\displaystyle I + J = \int _{0} ^{ \frac{ \pi }{2} } \left( \sin ^{4} { \theta } + \cos ^{4} { \theta } \right) \,d \theta = \int _{0} ^{ \frac{ \pi }{2} } \left( 1 - \frac{1}{2} \sin ^{2} { \left( 2 \theta \right) } \right) \,d \theta$ .

Hence:

$\displaystyle I + J = \int _{0} ^{ \frac{ \pi }{2} } \left( 1 - \frac{1}{2} \sin ^{2} { \left( 2 \theta \right) } \right) \,d \theta = \int _{0} ^{ \frac{ \pi }{2} } \left( 1 - \frac{1}{4} \left( 1 - \cos{ \left( 4 \theta \right) } \right) \right) \,d \theta$

Hence:

$\displaystyle I + J = \left[ \theta - \frac{1}{4} \theta + \frac{1}{16} \sin{ \left( 4 \theta \right) } \right] _{0} ^{ \frac{ \pi }{2} } = \frac{ \pi }{2} - \frac{ \pi }{8} = \frac{ 3 \pi }{8}$ .

See that sine and cosine enclose equivalent areas between the limits of 0 and $\frac{ \pi }{2}$ .

Therefore $I = J = \frac{1}{2} (I + J)$

Hence:

$J = I = \frac{1}{2}(I+J) = \frac{ 3 \pi }{16}$ .

Now consider:

$\cos ^{6} { \theta } - \sin ^{6} { \theta }$

$\left( \cos^{4} { \theta } + \sin ^{4} { \theta } \right) \left( \cos ^{2} { \theta } + \sin ^{2} { \theta } \right) \equiv \cos ^{6} { \theta } - \sin ^{6} { \theta } - \sin ^{2} { \theta } \cos ^{2} { \theta } \left( \sin^{2} { \theta } - \cos^{2} { \theta } \right)$

Hence:

$\cos ^{6} { \theta } - \sin ^{6} { \theta } = ( \cos{ \left( 2 \theta \right) } )( 1 ) + \sin ^{2} { \theta } \cos ^{2} { \theta } \left( - \cos{ \left( 2 \theta \right) } \right)$

Hence:

$\cos ^{6} { \theta } - \sin ^{6} { \theta } = \cos{ \left( 2 \theta \right) } + \frac{1}{4} \sin ^{2} { \left( 2 \theta \right) } ( - \cos{ \left( 2 \theta \right) } )$

Hence:

$\cos ^{6} { \theta } - \sin ^{6} { \theta } = \cos{ \left( 2 \theta \right) } \left( 1 - \frac{1}{4} \sin ^{2} { \left( 2 \theta \right) } \right)$ .

Let $\displaystyle I = \int _{0} ^{ \frac{ \pi }{2} } \cos ^{6} { \theta } \,d \theta$

Let $\displaystyle J = \int _{0} ^{ \frac{ \pi }{2} } \sin ^{6} { \theta } \,d \theta$

See that sine and cosine enclose equivalent areas between the limits of 0 and $\frac{ \pi }{2}$ .

Therefore $I = J = \frac{1}{2} (I + J)$

Now consider:

$\cos ^{6} { \theta } + \sin ^{6} { \theta } = ( \cos ^{4} { \theta } - \sin ^{4} { \theta } )( \cos ^{2} { \theta } - \sin ^{2} { \theta } ) + \cos ^{2} { \theta } \sin ^{2} { \theta } \left( \sin ^{2} { \theta } + \cos ^{2} { \theta } \right)$

Hence:

$\cos ^{6} { \theta } + \sin ^{6} { \theta } = ( \cos{ \left( 2 \theta \right) } )( \cos{ \left( 2 \theta \right) } ) + \frac{1}{4} \sin ^{2} { \left( 2 \theta \right) }$

Hence:

$\cos ^{6} { \theta } + \sin ^{6} { \theta } = \cos ^{2} { \left( 2 \theta \right) } + \frac{1}{4} \sin ^{2} { \left( 2 \theta \right) }$

Hence:

$\displaystyle I + J = \int _{0} ^{ \frac{ \pi }{2}} \left( \cos ^{2} { \left( 2 \theta \right) } + \frac{1}{4} \sin ^{2} { \left( 2 \theta \right) } \right) \,d \theta$

$\displaystyle \implies I + J = \int _{0} ^{ \frac{ \pi }{2}} \left( \cos ^{2} { \left( 2 \theta \right) } \right) \, d \theta + \frac{1}{4} \int _{0} ^{ \frac{ \pi }{2}} \left( \sin ^{2} { \left( 2 \ theta \right) } \right) \,d \theta$

$\displaystyle \implies I + J = \frac{1}{2} \int _{0} ^{ \frac{ \pi }{2}} \left( 1 + \cos{ \left( 4 \theta \right) } \right) \,d \theta + \frac{1}{8} \int _{0} ^{ \frac{ \pi }{2}} \left( 1 - \cos{ \left( 4 \theta \right) } \right) \,d \theta$

$\displaystyle \implies I + J = \frac{1}{2} \left[ \theta + \frac{1}{4} \sin{ \left( 4 \theta \right) } \right] _{0} ^{ \frac{ \pi }{2}} + \frac{1}{8} \left[ \theta - \frac{1}{4} \sin{ \left( 4 \theta \right) } \right] _{0} ^{ \frac{ \pi }{2}}$

$\displaystyle \implies I + J = \frac{ \pi }{4} + \frac{ \pi }{16} = \frac{5 \pi }{16}$ .

Hence, by the symmetrical relation identified above:

$I = J = \frac{1}{2}(I + J) = \frac{5 \pi }{32}$ .

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