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If cos and theta 0.65 what is the value of sin and theta?
You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.
2 cos x plus 1 cos x plus -1 equals 0?
To solve the equation 2cos(x) + cos(x) - 1 = 0, we first combine like terms to get 3cos(x) - 1 = 0. Then, we isolate the cosine term by adding 1 to both sides to get 3cos(x) = 1. Finally, we divide by 3 to solve for cos(x), which gives cos(x) = 1/3. Therefore, x = arccos(1/3) or approximately 70.53 degrees.
How do you solve double angle equations for trigonometry?
There are two ways to solve for the double angle formulas in trigonometry. The first is to use the angle addition formulas for sine and cosine. * sin(a + b) = sin(a)cos(b) + cos(a)sin(b) * cos(a + b) = cos(a)cos(b) - sin(a)sin(b) if a = b, then * sin(2a) = sin(a)cos(a) + cos(a)sin(a) = 2sin(a)cos(a) * cos(2a) = cos2(a) - sin2(b) The cooler way to solve for the double angle formulas is to use Euler's identity. eix = cos(x) + i*sin(x). Yes, that is "i" as in imaginary number. we we put 2x in for x, we get * e2ix = cos(2x) + i*sin(2x) This is the same as * (eix)2 = cos(2x) + i*sin(2x) We can substitute our original equation back in for eix. * (cos(x) + i*sin(x))2 = cos(2x) + i*sin(2x) We can distribute the squared term. * cos2(x) + i*sin(x)cos(x) + i*sin(x)cos(x) + (i*sin(x))2 = cos(2x) + i*sin(2x) And simplify. Because i is SQRT(-1), the i squared term becomes negative. * cos2(x) + 2i*sin(x)cos(x) - sin2(x) = cos(2x) + i*sin(2x) * cos2(x) - sin2(x) + 2i*sin(x)cos(x) = cos(2x) + i*sin(2x) Now you can plainly see both formulas in the equation arranged quite nicely. I don't yet know how to get rid of the i, but I'm working on it.
What is this expression as the cosine of an angle cos30cos55 plus sin30sin55?
cos(30)cos(55)+sin(30)sin(55)=cos(30-55) = cos(-25)=cos(25) Note: cos(a)=cos(-a) for any angle 'a'. cos(a)cos(b)+sin(a)sin(b)=cos(a-b) for any 'a' and 'b'.
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
