Question

if cos tita + sin tita = root 2 cos tita, show that cos tita - sin tita = root 2 sin tita

Answer 1

$Given \: cos \theta + sin \theta = \sqrt{2} cos \theta$

/* On squaring both sides, we get */

$\implies (cos \theta + sin \theta)^{2} = ( \sqrt{2}cos \theta )^{2}$

$\implies cos^{2} \theta + sin^{2} \theta + 2cos \theta sin \theta = 2cos^{2} \theta$

$\implies cos^{2} \theta + sin^{2} \theta- cos^{2}\theta = - 2cos \theta sin \theta + cos^{2} \theta$

$\implies sin^{2} \theta = cos^{2} \theta - 2cos\theta sin \theta$

$\implies sin^{2} \theta + sin^{2} \theta = cos^{2} \theta + sin^{2} \theta - 2cos\theta sin \theta$

$\implies 2sin^{2} \theta = (cos \theta - sin \theta )^{2}$

/* Do the square root both sides , we get */

$\implies \sqrt{2sin^{2}\theta } = \sqrt{(cos \theta - sin \theta )^{2}}$

$\implies \sqrt{2} sin \theta = cos \theta - sin \theta$

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