Question

Sin^4(п/4) -cos^4(п/4) +tg^2(п/4)

Answer 1

$\sin^4\dfrac{\pi }{4} -\cos^4\dfrac{\pi }{4}+\mathrm{tg}^2\dfrac{\pi }{4}=\left(\dfrac{\sqrt{2} }{2} \right)^4-\left(\dfrac{\sqrt{2} }{2} \right)^4+1^2=1^2=1$

Answer 2

Ответ:

$sin^4\dfrac{\pi}{4}-cos^4\dfrac{\pi}{4}+tg^2\dfrac{\pi}{4}=\Big(sin^2\dfrac{\pi}{4}-cos^2\dfrac{\pi}{4}\Big)\Big(\underbrace{sin^2\dfrac{\pi}{4}+cos^2\dfrac{\pi}{4}}_{1}\Big)+1^2=\\\\\\=\Big(\dfrac{\sqrt2}{2}\Big)^2-\Big(\dfrac{\sqrt2}{2}\Big)^2+1=\dfrac{2}{4}-\dfrac{2}{4}+1=1$