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It seems there might be a typo or missing information in your equation, as "cos x sin 20 x" does not form a complete equation. If you meant to set it equal to something (e.g., cos x sin 20x = 0), you would solve for x by determining when either cos x = 0 or sin 20x = 0. For example, cos x = 0 when x = (π/2) + nπ, where n is any integer. Please clarify if there is a specific equation or condition you had in mind.
What else can I help you with?
What is the exact value of sin 255?
To find the exact value of sin 255°, we can use the sine subtraction formula. Since 255° = 270° - 15°, we can express it as: [ \sin(255°) = \sin(270° - 15°) = \sin(270°) \cos(15°) - \cos(270°) \sin(15°. ] Knowing that (\sin(270°) = -1) and (\cos(270°) = 0), we have: [ \sin(255°) = -1 \cdot \cos(15°). ] Thus, the exact value of (\sin(255°) = -\cos(15°)).
How do you finish the equation sin 2 theta?
(/) = theta sin 2(/) = 2sin(/)cos(/)
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.
Sin x divided by 1 minus cos x equals csc plus cot x?
It helps to convert this kind of equation into one that has only sines and cosines, by using the basic definitions of the other functions in terms of sines and cosines. sin x / (1 - cos x) = csc x + cot x sin x / (1 - cos x) = 1 / sin x + cos x / sin x Now it should be easy to do some simplifications: sin x / (1 - cos x) = (1 + cos x) / sin x Multiply both sides by 1 + cos x: sin x (1 + cos) / ((1 - cos x)(1 + cos x)) = (1 + cos x)2 / sin x sin x (1 + cos) / (1 - cos2x) = (1 + cos x)2 / sin x sin x (1 + cos) / sin2x = (1 + cos x)2 / sin x sin x (1 + cos x) / sin x = (1 + cos x)2 1 + cos x = (1 + cos x)2 1 = 1 + cos x cos x = 0 So, cos x can be pi/2, 3 pi / 2, etc. In some of the simplifications, I divided by a factor that might be equal to zero; this has to be considered separately. For example, what if sin x = 0? Check whether this is a solution to the original equation.
What is the value of cos2 67-sin2 23?
To find the value of ( \cos^2 67^\circ - \sin^2 23^\circ ), we can use the identity ( \cos^2 \theta = 1 - \sin^2 \theta ). Since ( \sin 23^\circ = \cos 67^\circ ) (because ( 23^\circ + 67^\circ = 90^\circ )), we have ( \sin^2 23^\circ = \cos^2 67^\circ ). Thus, ( \cos^2 67^\circ - \sin^2 23^\circ = \cos^2 67^\circ - \cos^2 67^\circ = 0 ). Therefore, the value is ( 0 ).
What is sin squared x minus cos squared x?
Sin squared, cos squared...you removed the x in the equation.
What is the value of sin A plus cos Acos A (1-cos A) when tan A 815?
When tan A = 815, sin A = 0.9999992 and cos A = 0.0012270 so that sin A + cos A*cos A*(1-cos A) = 1.00000075, approx.
In and 8710ABC if sin A and tan A then what is cos A?
In a right triangle, if we know (\sin A) and (\tan A), we can find (\cos A) using the identity (\tan A = \frac{\sin A}{\cos A}). Rearranging this gives us (\cos A = \frac{\sin A}{\tan A}). Therefore, if you have specific values for (\sin A) and (\tan A), you can substitute them into this equation to find (\cos A).
What is the exact value of sin 255?
To find the exact value of sin 255°, we can use the sine subtraction formula. Since 255° = 270° - 15°, we can express it as: [ \sin(255°) = \sin(270° - 15°) = \sin(270°) \cos(15°) - \cos(270°) \sin(15°. ] Knowing that (\sin(270°) = -1) and (\cos(270°) = 0), we have: [ \sin(255°) = -1 \cdot \cos(15°). ] Thus, the exact value of (\sin(255°) = -\cos(15°)).
How do you finish the equation sin 2 theta?
(/) = theta sin 2(/) = 2sin(/)cos(/)
How do you work this out sin 2x sin x equals cos x?
I would start by looking up the formulae for multiple angles, and convert that to simgle angles. In this case, sin 2x = 2 sin x cos x, so your equation becomes:2 sin x cos x sin x = cos x2 sin2x cos x = cos xNext divide both sides by cos x; note that you must consider the possibility that cos x = 0 (this may give additional solutions to the equation).
How to solve sin of cos inverse of x?
sin[cos-1(x)] is an expression; it is not an equation (nor inequality). An expression cannot be solved.
Find the value of a if tan 3a is equal to sin cos 45 plus sin 30?
If tan 3a is equal to sin cos 45 plus sin 30, then the value of a = 0.4.
Cos x tan x to sin x Transform the expression on the left to the one on the right?
cosx * tanx = sinx cos x . sin x/ cosx = sinx cos x . sin x/cosx = sinx (cos x is taken out of the equation, leaving sin x = sin x).
How do you solve the trigonomic equation sin2x equals negative cosx?
Sin(2x) = -cos(x)But sin(2x) = 2 sin(x) cos(x)Substitute it:2 sin(x) cos(x) = -cos(x)Divide each side by cos(x):2 sin(x) = -1sin(x) = -1/2x = 210°x = 330°
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.
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,
