Prove that sin20sin40sin60sin80=3/16

1 answer:

5 0

We'll use the following properties of sine and cosine to prove this:

$sin(a)\cdot sin(b) = \frac12(cos(a-b) - cos(a+b))$

$cos(a)\cdot cos(b) = \frac12(cos(a+b) + cos(a-b))$

$cos(180+a) = cos(180-a)$

Then it's just a matter of filling it in...

sin20sin40 * sin60sin80 = 1/2(cos20 - cos60) * 1/2 (cos20 - cos140) =

1/8( cos40 + 1 - cos160 - cos120 - cos40 - cos80 + cos80 + cos200) =

1/8(1 - cos160 - cos120 + cos200) =

1/8(1 - cos160 - cos120 + cos160) =

1/8(1 - cos120 ) = 1/8( 1 + 1/2 ) = 3/16

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The first thing we must do for this case is find the scale factor.
We have then that for the larger side of both triangle, the scale factor is:
$k = \frac{RP}{HF}$
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For QR
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Answer:
You have that the lengths for the other two sides of triangle PQR are:
$PQ = 3

QR = 6$