Question

If $\sin \theta +\cos \theta =a$, then ${\sin }^4\theta +{\cos }^4\theta =$

• A. $1-\frac{1}{2}(a^2+1{)}^2$
• B. $1-\frac{1}{2}(a^2-1{)}^2$
• C. $1+\frac{1}{2}(a^2+1{)}^2$
• D. $1+\frac{1}{2}(a^2-1{)}^2$

Answer 1

The correct option is B $1-\frac{1}{2}(a^2-1{)}^2$

$\sin \theta +\cos \theta =a$
${\sin }^4\theta +{\cos }^4\theta ={({\sin }^2\theta +{\cos }^2\theta )}^2-2{\sin }^2\theta {\cos }^2\theta$
= $1$ - $2{\sin }^2\theta {\cos }^2\theta$
Now, ${(\sin \theta +\cos \theta )}^2={\sin }^2\theta +{\cos }^2\theta +2sin\theta \cos \theta$
$a^2=1+2\sin \theta \cos \theta$
$\sin \theta \cos \theta =\frac{1}{2}(a^2-1)$
Hence, substituting the value of $\sin \theta \cos \theta$, we get,
${\sin }^4\theta +{\cos }^4\theta =1-\frac{1}{2}{(a^2-1)}^2$