Q25

Q25

"Given that $(1 + \sin t)(1 + \cos t) = \frac{5}{4}$ and $(1 - \sin t)(1 - \cos t) = \frac{m}{n} - \sqrt{k}$, where $k, m$ and $n$ are positive integers with $m$ and $n$ relatively prime, find $k + m + n + 3$"

Answer: 30

Solution sketch

Let $S=\sin t$ and $C=\cos t$.

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1. Use the first condition

$$(1+S)(1+C)=\tfrac54$$ $$\Longrightarrow\;1+S+C+SC=\tfrac54\;\;\; \Rightarrow\;\; S+C+SC=\tfrac14. \tag{A}$$

Define

$$u=S+C,\quad v=SC .$$

Then (A) says $u+v=\tfrac14$.

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2. Relate $u$ and $v$ with $S^2+C^2=1$

$$S^2+C^2=(S+C)^2-2SC=u^2-2v=1. \tag{B}$$

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3. Solve the system

From $v=\tfrac14-u$ (by A), substitute in (B):

$$u^2-2\!\left(\tfrac14-u\right)=1 \;\;\Longrightarrow\;\; u^2+2u-\tfrac32=0$$ $$2u^2+4u-3=0 \quad\Rightarrow\quad u=\frac{-2+\sqrt{10}}{2}\;\;(\text{the other root is extraneous}).$$

Hence

$$u=\frac{\sqrt{10}-2}{2},\qquad v=\frac14-u=\frac{5-2\sqrt{10}}{4}.$$

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4. Compute $(1-\sin t)(1-\cos t)$

$$(1-S)(1-C)=1-(S+C)+SC =1-u+v =\frac{4-\sqrt{10}}{2}+\frac{5-2\sqrt{10}}{4} =\frac{13}{4}-\sqrt{10}.$$

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5. Match the required form

$$(1-\sin t)(1-\cos t)=\frac{13}{4}-\sqrt{10} \;=\;\frac{m}{n}-\sqrt{k},$$

so $m=13,\;n=4\;( \gcd(13,4)=1),\;k=10$.

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6. Final value

$$k+m+n+3=10+13+4+3=\boxed{{\displaystyle 30}}\mathrm{.}$$