Question

If $\sin \theta -\cos \theta =0,$ then find the value of ${\sin }^4\theta +{\cos }^4\theta$

Answer 1

$sin\theta -cos=0$ _________ (1)

Now

$si{n}^4\theta +co{s}^4\theta =(si{n}^2\theta +co{s}^2\theta {)}^2-2si{n}^2\theta co{s}^2\theta$ _________ (2)

using equation (1)

$(sin\theta -cos\theta {)}^2=si{n}^2\theta +co{s}^2\theta -2sin\theta cos\theta$

$0+2sin\theta cos\theta =1$

$[sin\theta cos\theta =\frac{1}{2}]$

From equation (2) we get

$si{n}^4\theta +co{s}^4\theta =1-2(sin\theta cos\theta {)}^2$

$si{n}^4\theta +co{s}^4\theta =1-2\times \frac{1}{4}=1-\frac{1}{2}$

$si{n}^4\theta +co{s}^4\theta =\frac{1}{2}$