60 is pretty

Geometry Level 3

P = s i n ( 20 ) × s i n ( 40 ) × s i n ( 60 ) × s i n ( 80 ) \large P= sin(20) \times sin(40) \times sin(60) \times sin(80)

If P P can be represented in the form a b \large \frac{a}{b} where a a and b b are coprime positive integers

Find a − b a-b

Note:All the Angles are in Degree measure.


The answer is -13.

Achal Jain by Achal Jain , 19, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Achal Jain
Jun 3, 2017

We can rewrite the expression as

s i n ( 60 ) [ s i n ( 20 ) × s i n ( 40 ) × s i n ( 80 ) = s i n ( 60 ) [ s i n ( 20 ) × s i n ( 60 − 20 ) × s i n ( 60 + 20 ) ] sin(60)[sin(20) \times sin(40) \times sin(80) = sin(60)[ sin(20) \times sin(60-20) \times sin(60+20)] .

Now s i n ( θ ) × s i n ( 60 − θ ) × s i n ( 60 + θ ) sin(\theta) \times sin(60-\theta) \times sin(60+\theta) = s i n ( 3 θ ) 4 \dfrac{sin(3\theta)}{4}

Then the above statement simplifies to

s i n ( 60 ) × s i n ( 60 ) 4 sin(60) \times \dfrac{sin(60)}{4} which then simplifies to 3 16 \boxed{\dfrac{3}{16}}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...