The answer in itself is 1/128 and here is the procedure to prove it:
cos(A)*cos(60+A)*cos(60-A) = cos(A)*(cos²60 - sin²A)
<span>= cos(A)*{(1/4) - 1 + cos²A} = cos(A)*(cos²A - 3/4) </span>
<span>= (1/4){4cos^3(A) - 3cos(A)} = (1/4)*cos(3A) </span>
Now we group applying what we see above
<span>cos(12)*cos(48)*cos(72) = </span>
<span>=cos(12)*cos(60-12)*cos(60+12) = (1/4)cos(36) </span>
<span>Similarly, cos(24)*cos(36)*cos(84) = (1/4)cos(72) </span>
<span>Now the given expression is: </span>
<span>= (1/4)cos(36)*(1/4)*cos(72)*cos(60) = </span>
<span>= (1/16)*(1/2)*{(√5 + 1)/4}*{(√5 - 1)/4} [cos(60) = 1/2; </span>
<span>cos(36) = (√5 + 1)/4 and cos(72) = cos(90-18) = </span>
<span>= sin(18) = (√5 - 1)/4] </span>
<span>And we seimplify it and it goes: (1/512)*(5-1) = 1/128</span>
Answer:
Step-by-step explanation:
d/dn(2 n^2 + 12 n + 18) = 4 (n + 3)
Indefinite integral:
integral(18 + 12 n + 2 n^2) dn = (2 n^3)/3 + 6 n^2 + 18 n + constant
Global minimum:
min{2 n^2 + 12 n + 18} = 0 at n = -3
Answer:6/35
Step-by-step explanation:
<u>Answer:</u>
-
The length of the hypotenuse is 40 mm.
<u>Step-by-step explanation:</u>
<em>We can solve for 'c' using Pythagoras theorem. Let's solve it.</em>
-
c² = 32² + 24²
- => c² = 1024 + 576
- => c² = 1600
- => c = 40
Hence, the length of the hypotenuse is 40 mm.
Hoped this helped.
