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What is the exact value using a sum or difference formula of the expression cos 11pi over 12?
11pi/12 = pi - pi/12
cos(11pi/12) = cos(pi - pi/12)
cos(a-b) = cos(a)cos(b)+sin(a)sin(b)
cos(pi -pi/12) = cos(pi)cos(pi/12) + sin(pi)sin(pi/12)
sin(pi)=0
cos(pi)=-1
Therefore, cos(pi -pi/12) = -cos(pi/12)
pi/12=pi/3 -pi/4
cos(pi/12) = cos(pi/3 - pi/4)
= cos(pi/3)cos(pi/4)+sin(pi/3) sin(pi/4)
cos(pi/3)=1/2
sin(pi/3)=sqrt(3)/2
cos(pi/4)= sqrt(2)/2
sin(pi/4) = sqrt(2)/2
cos(pi/3)cos(pi/4)+sin(pi/3) sin(pi/4)
= (1/2)(sqrt(2)/2 ) + (sqrt(3)/2)( sqrt(2)/2)
= sqrt(2)/4 + sqrt(6) /4
= [sqrt(2)+sqrt(6)] /4
Therefore, cos(pi/12) = (sqrt(2)+sqrt(6))/4
-cos(pi/12) = -(sqrt(2)+sqrt(6))/4
cos(11pi/12) = -(sqrt(2)+sqrt(6))/4
What else can I help you with?
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How do you factor this expression completely 9a4 - b2?
You can start by using the formula for the difference of two squares. Actually, after that I don't think you can factor it any further.
How is using a formula similar to evaluating an expression?
Beacause with a formula you are finding out a problem. Just like evaluating means to find out or to solve.
What is a to the third power minus b to the third power?
The expression "a to the third power minus b to the third power" can be represented mathematically as ( a^3 - b^3 ). This difference of cubes can be factored using the formula ( a^3 - b^3 = (a - b)(a^2 + ab + b^2) ). This factorization shows that the expression can be expressed as the product of the difference of the two bases and a quadratic expression involving both bases.
How would you factorise x2-49?
To factorise ( x^2 - 49 ), you can recognize it as a difference of squares. This expression can be rewritten as ( (x)^2 - (7)^2 ). Using the difference of squares formula, ( a^2 - b^2 = (a - b)(a + b) ), we factor it as ( (x - 7)(x + 7) ).
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