Question

Evaluate sin^2(π/4)+sin^2(3π/4)+sin^2(5π/4)+sin^2(7π/4)

Answer 1

Please see the attachment

Answer 2

$\sin^2(\frac{\pi}{4})+\sin^2(\frac{3\pi}{4})+\sin^2(\frac{5\pi}{4})+\sin^2(\frac{7\pi}{4})=1$

Step-by-step explanation:

Given : Expression $\sin^2(\frac{\pi}{4})+\sin^2(\frac{3\pi}{4})+\sin^2(\frac{5\pi}{4})+\sin^2(\frac{7\pi}{4})$

To find : Evaluate the expression ?

Solution :

$\sin^2(\frac{\pi}{4})+\sin^2(\frac{3\pi}{4})+\sin^2(\frac{5\pi}{4})+\sin^2(\frac{7\pi}{4})$

$=\sin^2(\frac{\pi}{4})+\sin^2(\pi-\frac{\pi}{4})+\sin^2(\pi+\frac{\pi}{4})+\sin^2(2\pi-\frac{\pi}{4})$

$=\sin^2(\frac{\pi}{4})+\sin^2(\frac{\pi}{4})+\sin^2(\frac{\pi}{4})+\sin^2(\frac{\pi}{4})$

$=4\sin^2(\frac{\pi}{4})$

$=4(\frac{1}{\sqrt2})^2$

$=4(\frac{1}{4})$

$=1$

Therefore, $\sin^2(\frac{\pi}{4})+\sin^2(\frac{3\pi}{4})+\sin^2(\frac{5\pi}{4})+\sin^2(\frac{7\pi}{4})=1$

#Learn more

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