2002 Paper 3 Q7

Year: 2002
Paper: 3
Question Number: 7

Course: LFM Pure
Section: Introduction to trig

Difficulty: 1700.0 Banger: 1484.0

trigonometry complementary angles Pythagorean identity coordinate geometry cosine rule inequality geometric proof if and only if

Problem

Given that $\alpha$ and $\beta$ are acute angles, show that $\alpha + \beta = \tfrac{1}{2}\pi$ if and only if $\cos^2 \alpha + \cos^2 \beta = 1$. In the $x$--$y$ plane, the point $A$ has coordinates $(0,s)$ and the point $C$ has coordinates $(s,0)$, where $s>0$. The point $B$ lies in the first quadrant ($x>0$, $y>0$). The lengths of $AB$, $OB$ and $CB$ are respectively $a$, $b$ and $c$. Show that \[ (s^2 +b^2 - a^2)^2 + (s^2 +b^2 -c^2)^2 = 4s^2b^2 \] and hence that \[ (2s^2 -a^2-c^2)^2 + (2b^2 -a^2-c^2)^2 =4a^2c^2\;. \] Deduce that $$ \l a - c \r^2 \le 2b^2 \le \l a + c \r^2\;. $$ %Show, %by considering the case $a=1+\surd2\,$, $b=c=1\,$, % that the condition $\l \ast \r\,$ %is not sufficient to ensure that $B$ lies in the first quadrant.

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Difficulty Rating: 1700.0

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Banger Rating: 1484.0

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Problem source

Given that  $\alpha$ and $\beta$ are acute angles,  show that
$\alpha + \beta = \tfrac{1}{2}\pi$ if and only if $\cos^2 \alpha + \cos^2 \beta = 1$.
In the $x$--$y$ plane, the point $A$ has coordinates $(0,s)$ and the point
$C$ has coordinates $(s,0)$, where $s>0$. The point $B$ lies in the 
first quadrant ($x>0$, $y>0$). The lengths of $AB$, $OB$ and $CB$
are respectively $a$, $b$ and $c$.
Show that
\[
(s^2 +b^2 - a^2)^2 + (s^2 +b^2 -c^2)^2 = 4s^2b^2
\]
and hence that
\[
(2s^2 -a^2-c^2)^2 + (2b^2 -a^2-c^2)^2 =4a^2c^2\;.
\]

Deduce that
$$
\l a - c \r^2 \le 2b^2 \le \l a + c \r^2\;. 
$$

%Show,
%by considering the case $a=1+\surd2\,$, $b=c=1\,$,
% that the condition $\l \ast \r\,$ 
%is not sufficient to ensure that $B$ lies in the first quadrant.

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