Question

If $\cos {25}^0+\sin {25}^0=k,$ then $\cos {50}^0$ is equal to

Answer 1

We have,

$\cos {25}^0+\sin {25}^0=k$

On squaring both sides, we have

$(\cos {25}^0+\sin {25}^0{)}^2=k^2$

${\cos }^2{25}^0+{\sin }^2{25}^0+2\sin {25}^0\cos {25}^0=k^2$

$1+2\sin {25}^0\cos {25}^0=k^2$ $\because {\cos }^{2}x+{\sin }^{2}x=1$

$\sin {50}^0=k^2-1$ $\because 2\cos x\sin x=\sin 2x$

Since,

$\sin \theta =\frac{Height}{Hypotenuse}$

So,

$Height=k^2-1$

$Hypotenuse=1$

$Base=k\sqrt{2-k^2}$

Therefore,

$\cos {50}^0=\frac{Base}{Hypotenuse}$

$\cos {50}^0=\frac{k\sqrt{2-k^2}}{1}$

$\cos {50}^0=k\sqrt{2-k^2}$