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 !
the ball bpunces about 5 ft when dropped from the 100 inch height
8050 is the answer to "what does 805 tens="
The two consecutive odd numbers that have a product of 15 are 3 and 5
There are 100 random 2 digits numbers [00 , 99]
There are 34 divisible by 3 {00, 03, 06, 09, … 93, 96, 99}
There are 20 divisible by 5 {00, 05, 10, … 90, 95}
However we must avoid counting numbers twice so we need to subtract those divisible by 15.
There are 7 divisible by 15 {00, 15, 30, 45, 60, 75, 90}
This means there are 34 + 20 - 7 = 47 2-digit numbers that are divisible by 3 or 5.
If you pick a random 2-digit number then P(divisible by 3 or 5) = 47/100 = 0.47
