Question

(cos theta + sin theta)^2 + ( cos theta - sin theta)^2 is equal to​

Answer 1

Answer:

2

Step-by-step explanation:

(cosθ+sinθ)^2 +(cosθ−sinθ)^2

=(cos^2  θ+sin^2  θ + 2sinθcosθ)+(cos^2  θ+sin ^ 2 θ−2sinθcosθ)

=2(cos^ 2 θ+sin^ 2 θ)

=2(1)=2

Answer 2

${\cos0 \: + \sin0 = \sqrt2 \cos0}$

${squaring \: on \: both \: the \: sides \: we \: get}$

${ cos^{2} + sin^{2}0 + 2 sin0 cos0 = 2\cos^{2}0 }$

$\cos^{2}0 - \sin0 = 2 \cos0 \sin0$

$( \cos0 + \sin0)( \cos0 - \sin0) = 2 \cos0 \sin0$

$\sqrt{2} \cos0( \cos0 - \sin0) = 2 \cos0 \sin0$

$(given \: cos0 + sin0 = \sqrt{2}cos0)$

$\infty \: cos0 - sin0 = \sqrt{2}sin0$

$(hence \: proved)$