Question

sin of power 4 + cos of power 4 = 1+ 4k × sin sqare theeta × cos sqare theeta . then k is equal to .

Answer 1

$Given \: \: \: Question \: \: Is \: \: \\ \\ \sin {}^{4} (x) + \cos {}^{4} (x) = 1 + 4k \sin {}^{2} (x) \cos {}^{2} (x) \\ \\ Answer \: \: \\ \\ ( \sin {}^{2} (x) ) {}^{2} + ( \cos {}^{2} (x) ) {}^{2} = 1 + 4k \sin {}^{2} (x) \cos {}^{2} (x) \\ \\ ( \sin {}^{2} (x) + \cos {}^{2} (x) ) {}^{2} - 2 \sin {}^{2} (x) \cos {}^{2} (x) \\ = 1 + 4k \sin {}^{2} (x) \cos {}^{2} (x) \\ \\ 1 - 2 \sin {}^{2} (x) \cos {}^{2} (x) = 1 + 4k \sin {}^{2} (x) \cos {}^{2} (x) \\ \\ - 2 \sin {}^{2} (x) \cos {}^{2} (x) = 4k \sin {}^{2} (x) \cos {}^{2} (x) \\ \\ 4k \: = \frac{ - 2 \sin {}^{2} (x) \cos {}^{2} (x) }{ \sin {}^{2} (x) \cos {}^{2} (x) } \\ \\ 4k = - 2 \\ \\ k = \frac{ - 2}{4} \\ \\ k = \frac{ - 1}{2} \\ \\ Note \: \\ \\ 1) \: \: \: \alpha {}^{4} + \beta {}^{4} = ( \alpha {}^{2} + \beta {}^{2} ) {}^{2} - 2( \alpha \beta ) {}^{2} \\ \\ 2) \: \: \: \sin {}^{2} (x) + \cos {}^{2} (x) = 1$

Answer 2

hope it helps u ..........