Question

The value of ${\cos }^4\theta +{\sin }^4\theta +2{\cos }^2\theta {\sin }^2\theta$ when $\theta ={45}^0$ is

• A. $1$
• B. $2$
• C. $\frac{1}{\sqrt{2}}$
• D. $2\sqrt{2}$

Answer 1

Answer: (A)

$1$

$Given\theta ={45}^{\circ }\therefore \cos \theta =\sin \theta =\frac{1}{\sqrt{2}}\therefore {\cos }^4\theta +{\sin }^4\theta +2{\cos }^2\theta {\sin }^2\theta ={\left(\frac{1}{\sqrt{2}}\right)}^4+{\left(\frac{1}{\sqrt{2}}\right)}^4+2{\left(\frac{1}{\sqrt{2}}\right)}^2{\left(\frac{1}{\sqrt{2}}\right)}^2=\frac{1}{4}+\frac{1}{4}+\frac{1}{2}=1$