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How do you prove this trigonometric relationship sin3A equals 3sinA cos 2 A - sin 3 A?
sin(3A) = sin(2A + A) = sin(2A)*cos(A) + cos(2A)*sin(A)= sin(A+A)*cos(A) + cos(A+A)*sin(A) = 2*sin(A)*cos(A)*cos(A) + {cos^2(A) - sin^2(A)}*sin(A) = 2*sin(A)*cos^2(A) + sin(a)*cos^2(A) - sin^3(A) = 3*sin(A)*cos^2(A) - sin^3(A)
What is cos x2?
cos(x^2)=cos(x times x)
What is the solution to cos equals sec-sintan?
I'm not really sure what you mean by "the solution", but that equation cos = sec - sintan does simplify down to sin^2 + cos^2 = 1 which so happens to be an identity. I'm not sure if that's what you're looking for, but if it is, here are the steps in simplifying it. 1. Convert sec to 1/cos 2. Convert tan into sin/cos and multiply it by sin sintan = sin(sin/cos) = (sin^2)/cos You then have cos = 1/cos - (sin^2/cos) 3. Multiply everything by cos cos^2 = 1 - sin^2 4. And finally, send the sin^2 over to the left side by adding it (since it is being subracted on the right) You should see this sin^2 + cos^2 = 1 which is an identity.
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
Express all the trigonometric ratios of cosA?
sin = sqrt(1 - cos^2)tan = sqrt(1 - cos^2)/cossec = 1/coscosec = 1/sqrt(1 - cos^2)cot = cos/sqrt(1 - cos^2)
