Question

If $\cos {20}^{\circ }-\sin {20}^{\circ }=p$ then $\cos {40}^{\circ }$ is equal to:

• A. $-p\sqrt{2-p^2}$
• B. $p\sqrt{2-p^2}$
• C. $p+\sqrt{2-p^2}$
• D. none of these

Answer 1

The correct option is B $p\sqrt{2-p^2}$

We have, $\cos 20-\sin 20=p$ ...(1)
Squaring equation (1) we get,
${\cos }^{2}20+{\sin }^{2}20-2\sin 20\cos 20=p^2$

$1+\sin (2\times 20)=p^2$

$\Rightarrow \sin 40=1-p^2\{\because 2\sin \theta \cos \theta =\sin 2\theta \}$
$\cos 40=\sqrt{1-{\sin }^{2}40}$

$\Rightarrow \cos 40=\sqrt{1-{(1-p^2)}^2}=p\sqrt{2-p^2}$
Ans: B