Question

If $\cos {25}^{\circ }+\sin {25}^{\circ }=p$, then $\cos {50}^{\circ }$ is

• A. $\sqrt{2-p^2}$
• B. $-\sqrt{2-p^2}$
• C. $p\sqrt{2-p^2}$
• D. $-p\sqrt{2-p^2}$

Answer 1

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

Given, $cos{25}^{\circ }+sin{25}^{\circ }=p$

Squaring both sides

$(cos{25}^{\circ }+sin{25}^{\circ }{)}^2=p^2$

${cos}^2{25}^{\circ }+{sin}^2{25}^{\circ }+2cos{25}^{\circ }sin{25}^{\circ }=p^2$ ........... Using $(a+b{)}^2=a^2+2ab+b^2$

$\because [sin²\theta +cos²\theta =1]$ and $[\sin 2x=2\sin x\cos x]$

$1+sin{50}^{\circ }=p^2$

$sin{50}^{\circ }=p^2-1$

$\because cos²\theta =1-sin²\theta ⟹cos\theta =\sqrt{1-sin²\theta }$

$cos{50}^{\circ }=\sqrt{1-{(p^2-1)}^2}=\sqrt{1-p^4-1+2p^2}=\sqrt{2p^2-p^4}$

$cos{50}^{\circ }=p\sqrt{2-p^2}$