Question

Dado que

(1+\sin t)(1+\cos t)=\frac54

y

\begin{aligned}(1-\sin t)&(1-\cos t)\\&=\frac mn-\sqrt k,\end{aligned}

donde

k,

m,

y

n

son enteros positivos y

m

y

n

son coprimos, halla

k+m+n.

Given that

(1+\sin t)(1+\cos t)=\frac54

and

\begin{aligned}(1-\sin t)&(1-\cos t)\\&=\frac mn-\sqrt k,\end{aligned}

where

k,

m,

and

n

are positive integers with

m

and

n

relatively prime, find

k+m+n.

Respuesta

Define

u=\sin t+\cos t

y expresa

\sin t\cos t

en términos de

u

Set

u=\sin t+\cos t

and express

\sin t\cos t

in terms of

u

Los dos productos dados se convierten en la mitad de un cuadrado perfecto en

u

Both given products become half of a perfect square in

u

Sea

u=\sin t+\cos t.

Como

\sin t\cos t=\frac{u^2-1}{2},

\begin{aligned}(1+\sin t)&(1+\cos t)\\&=\frac{(u+1)^2}{2}=\frac54.\end{aligned}

La elección de signo compatible da

u+1=\sqrt{\frac{5}{2}}.

Por tanto,

\begin{aligned}(1-\sin t)&(1-\cos t)\\&=\frac{(1-u)^2}{2}\\&=\frac{13}{4}-\sqrt{10}.\end{aligned}

Así,

k+m+n=10+13+4=27.

Let

u=\sin t+\cos t.

Since

\sin t\cos t=\frac{u^2-1}{2},

\begin{aligned}(1+\sin t)&(1+\cos t)\\&=\frac{(u+1)^2}{2}=\frac54.\end{aligned}

The feasible sign gives

u+1=\sqrt{\frac{5}{2}}.

Therefore

\begin{aligned}(1-\sin t)&(1-\cos t)\\&=\frac{(1-u)^2}{2}\\&=\frac{13}{4}-\sqrt{10}.\end{aligned}

Thus

k+m+n=10+13+4=27.