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  • Fill in the blanks If sin x + cos x = a, then (i) sin6 x + cos6x = _______ (ii) |sin x – cos x| = ______.

Fill in the blanks

If \sin x + \cos x = a, then

(i) \sin^6 x + \cos^6x = \_\_\_\_\_\_\_  

(ii)    \vert \sin x - \cos x \vert = \_\_\_\_\_\_.

Answers (1)

\\Given\: \: that \: \: \sin x+\cos x=a \\\\ \text{On squaring both sides}, \left( \sin x+\cos x \right) ^{2}=a^{2} \\\\ 1+2\sin x\cos x=a^{2} \\\\ \sin x\cos x=\frac{a^{2} - 1}{2} \\\\ \sin ^{6}x+\cos ^{6}x= \left( \sin ^{2}x \right) ^{3}+ \left( \cos ^{2}x \right) ^{3} \\\\ = \left( \sin ^{2}x+\cos ^{2}x \right) ^{3} - 3\sin ^{2}x\cos ^{2}x \left( \sin ^{2}x+\cos ^{2}x \right) \\\\ =1 - 3 \left( \frac{a^{2} - 1}{2} \right) ^{2}=1 - \frac{3 \left( a^{2} - 1 \right) ^{2}}{4}=\frac{1}{4} \left[ 4 - 3 \left( a^{2} - 1 \right) ^{2} \right] \\\\ \vert \sin x - \cos x \vert ^{2}=\sin ^{2}x+\cos ^{2}x - 2\sin x\cos x \\\\ =1 - 2 \left( \frac{a^{2} - 1}{2} \right) =1 - a^{2}+1=2 - a^{2} \\\\ \vert \sin x - \cos x \vert =\sqrt {2 - a^{2}} \\\\

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