Question

Show that ${(\sin \theta +\cos \theta )}^2=1+2\sin \theta \cos \theta$

Answer 1

Let usfirstfind the value of left hand side (LHS) that is $(\sin \theta +\cos \theta {)}^2$ as shown below:

$(\sin \theta +\cos \theta {)}^2={\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta (\because (a+b{)}^2=a^2+b^2+2ab)=1+2\sin \theta \cos \theta =RHS(\because {\sin }^{2}x+{\cos }^{2}x=1)$

Since LHS=RHS,

Hence, $(\sin \theta +\cos \theta {)}^2=1+2\sin \theta \cos \theta$.