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:

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

Answer 3

Let's simplify the expression:

1. First, we can use the trigonometric identities:

   • sin^2(π/4) = 1/2
   • cos^2(π/4) = 1/2
   • tan(π/4) = 1

2. Now, we can substitute these values into the expression:

   sin^4(π/4) - cos^4(π/4) + tan^2(π/4)

   (1/2)^2 - (1/2)^2 + 1^2

   1/4 - 1/4 + 1

   0 + 1

So, the value of the expression sin^4(π/4) - cos^4(π/4) + tan^2(π/4) is equal to 1.