The more reasonable measurement of the distance between the tracks on a CD is 1.6 ⋅ 10^-3 mm. This is also equal to 0.0016mm. <span>1.6 ⋅ 10^3 mm is too large for a CD track</span>
Answer:
how many does he have???
Step-by-step explanation:
The first composite shape we would need to find is the rectangle. We would use the formula length multiplied by width. The second composite shape is a hemisphere. The area of a hemisphere is Pi multiplied by the radius squared and then divide it by 2. To find the area of both of these figures, add up the area for each one and then you will get the
area of both.
Let's work on the left side first. And remember that
the<u> tangent</u> is the same as <u>sin/cos</u>.
sin(a) cos(a) tan(a)
Substitute for the tangent:
[ sin(a) cos(a) ] [ sin(a)/cos(a) ]
Cancel the cos(a) from the top and bottom, and you're left with
[ sin(a) ] . . . . . [ sin(a) ] which is [ <u>sin²(a)</u> ] That's the <u>left side</u>.
Now, work on the right side:
[ 1 - cos(a) ] [ 1 + cos(a) ]
Multiply that all out, using FOIL:
[ 1 + cos(a) - cos(a) - cos²(a) ]
= [ <u>1 - cos²(a)</u> ] That's the <u>right side</u>.
Do you remember that for any angle, sin²(b) + cos²(b) = 1 ?
Subtract cos²(b) from each side, and you have sin²(b) = 1 - cos²(b) for any angle.
So, on the <u>right side</u>, you could write [ <u>sin²(a)</u> ] .
Now look back about 9 lines, and compare that to the result we got for the <u>left side</u> .
They look quite similar. In fact, they're identical. And so the identity is proven.
Whew !