Question

sin theta -cos theta=0,find the value of sin^4theta + cos^4 theta

Answer 1

sin theta-cos theta=0

${sin}^{4}theta+ {cos}^{4}theta = {sin}^{4}theta + {cos}^{4}theta - 2 {sin}^{2}theta \times {cos}^{2}theta + 2 {sin}^{2} theta\times {cos}^{2}theta$
=》
$( {sin}^{2}theta - {cos}^{2}theta ) + 2 {sin}^{2}theta {cos}^{2}theta$
$= 0 + 2 {sin}^{2}theta {cos}^{2}theta$
=》
$2 {sin}^{2}theta {cos}^{2}theta$
answer

Answer 2

take this ( || )as theta,
sin || - cos || =0
sin ||=cos ||
sin ||/cos ||=1
tan ||=1
TAN ||=P/B
P=1,B=1
pythagoras theorem,
H^2=P^2+B^2
H^2=1^2 +1^2
H^2=2
H=√2

Therefore,
sin^4 || + cos^4 ||
=(p/h)^4 +(b/h)^4
=(1/√2)^4 +(1/√2)^4
=1/4 + 1/4
=2/4
=1/2 (answer)