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What else can I help you with?
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)
Is cos 2x 2 cos x?
No, the equation ( \cos 2x = 2 \cos x ) is not generally true. The correct identity for ( \cos 2x ) is ( \cos 2x = 2 \cos^2 x - 1 ), which shows that it is related to the square of the cosine of ( x ) rather than just ( 2 \cos x ). Thus, without additional context or constraints, the statement is incorrect.
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 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 cos x2?
cos(x^2)=cos(x times x)
What is cos2 A?
Cos(2A) = Cos(A + A) Double Angle Indentity Cos(A+A) = Cos(A)Cos(A) - Sin(A)Sin(A) => Cos^(2)[A] - SIn^(2)[A] => Cos^(2)[A] - (1 - Cos^(2)[A] => 2Cos^(2)[A] - 1
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)
Verify that sin minus cos plus 1 divided by sin plus cos subtract 1 equals sin plus 1 divided by cos?
[sin - cos + 1]/[sin + cos - 1] = [sin + 1]/cosiff [sin - cos + 1]*cos = [sin + 1]*[sin + cos - 1]iff sin*cos - cos^2 + cos = sin^2 + sin*cos - sin + sin + cos - 1iff -cos^2 = sin^2 - 11 = sin^2 + cos^2, which is true,
What is Cos 2 x divided by cos x?
cos 2x = cos2 x - sin2 x = 2 cos2 x - 1; whence, cos 2x / cos x = 2 cos x - (1 / cos x) = 2 cos x - sec x.
How do you verify 1 divided by cos to the second theta minus tan to the second theta equals cos to the second theta plus 1 divided by csc to the second theta?
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What are the double-angle and half-angle identities?
sin 2θ = 2(sin θ)(cos θ) cos 2θ = (cos θ)2 - (sin θ)2 cos 2θ = 2(cos θ)2 - 1 cos 2θ = 1 - 2(sin θ)2 tan 2θ = 2(tan θ)/[1 - (tan θ)2] sin θ/2 = ±√[(1 - (cos θ))/2] cos θ/2 = ±√[(1 + (cos θ))/2] tan θ/2 = ±√[(1 - (cos θ))/(1 + (cos θ))] ; cos θ ≠ -1 tan θ/2 = [1 - (cos θ)]/(sin θ) tan θ/2 = (sin θ)/[1 + (cos θ)]
What is cos2x times cosx?
The expression ( \cos(2x) \cdot \cos(x) ) can be simplified using the double angle identity for cosine. Specifically, ( \cos(2x) = 2\cos^2(x) - 1 ). Thus, multiplying gives ( \cos(2x) \cdot \cos(x) = (2\cos^2(x) - 1) \cdot \cos(x) = 2\cos^3(x) - \cos(x) ).
Express all the trignometric ratios in terms of COS A?
Provided that any denominator is non-zero, sin = sqrt(1 - cos^2)tan = sqrt(1 - cos^2)/cos sec = 1/cos cosec = 1/sqrt(1 - cos^2) cot = cos/sqrt(1 - cos^2)
Why did Chinese first used chop-sticks?
cos they want 2 cos they want 2 cos they want 2
Is cos 2x 2 cos x?
No, the equation ( \cos 2x = 2 \cos x ) is not generally true. The correct identity for ( \cos 2x ) is ( \cos 2x = 2 \cos^2 x - 1 ), which shows that it is related to the square of the cosine of ( x ) rather than just ( 2 \cos x ). Thus, without additional context or constraints, the statement is incorrect.
How do you factor sin squared times x plus cos2x -cosx equals 0?
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How do you show that 2 sin squared x minus 1 divided by sin x minus cos x equals sin x plus cos x?
(2 sin^2 x - 1)/(sin x - cos x) = sin x + cos x (sin^2 x + sin^2 x - 1)/(sin x - cos x) =? sin x + cos x [sin^2 x - (1 - sin^2 x)]/(sin x - cos x) =? sin x + cos x (sin^2 x - cos^2 x)/(sin x - cos x) =? sin x + cos x [(sin x - cos x)(sin x + cos x)]/(sin x - cos x) =? sin x + cos x sin x + cos x = sin x + cos x
