Rearrange the equation so m is the independent variable. 7m+2=7n−5
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We'll use the following properties of sine and cosine to prove this:
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
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
a) A line can intersect a parabola at 2 points, 1 point, or no points respectively. The only possible way at which it intersects at 1 point, would be if it were tangent to the parabola. [Check first attachment]
b) Two different parabolas can intersect at 2 points, 1 point or no points respectively. [Check second attachment]
It is possible for 2 parabolas to intersect at 4 and even 3 points. But one of the parabolas would be on it's side. You can consider those cases.
