0
1 + cos(x) = sin(x)
First of all, since you didn't include the argument of the sin or cos in the question,
we sadly suspect that you have little clue as to what you're looking for, or what a
solution will look like.
==> You need to find an angle whose sine is 1 greater than its cosine.
The numerical values of both the sine and cosine functions range from -1 to +1.
No angle has a sin or cosine less than -1 or greater than +1. That'll help us put
some constraints on the equation, and see what may be going on.
The equation also says: sin(x) - cos(x) = 1
This would be a great place to flash a sketch of the graphs of the sin(x) and cos(x)
functions up on the screen, and see where they differ by roughly 1, with the sine
being the greater one. It's too bad that we can't do that. The best we can do is to
draw them on our scratch pad here, look at them, and tell you what we see:
-- The sine is greater than the cosine only between 45° and 225°,
so any solutions must be in that range of angles.
-- At 90°, the sine is 1 and the cosine is zero, so we have [ 1 + 0 = 1 ], and 90° definitely works.
-- At 180°, the sine is zero and the cosine is -1, we have [ 1 + -1 = 0 ], and 180° works.
-- If there were any range between 45° and 225° where the graphs of the sine
and cosine functions were parallel curves, then any angle in that range might
also be a solution.
But there isn't any such place. 90° and 180° are the only points where the values
are different by 1 and the sine is greater, so those are the only principle solutions
(answers between zero and 360°.)
Add your answer:
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Sin 15 plus cos 105 equals?
Sin 15 + cos 105 = -1.9045
How do you solve Sin x plus sin 3x plus sin 5x plus sin 7x equals?
sin7x-sin6x+sin5x
How do you solve sin x plus cos x equals zero?
The statement of the problem is equivalent to sin x = - cos x. This is true for x = 135 degrees and x = -45 degrees, and also for (135 + 180n) degrees, where n is any integer.
How do you solve sinx divided by 1 plus cosx plus cosx divided by sinx?
sin x/(1+cos x) + cos x / sin x Multiply by sin x (1+cos x) =[(sin^2 x + cos x(1+cos x) ] / sin x (1+cos x) = [(sin^2 x + cos x + cos^2 x) ] / sin x (1+cos x) sin^2 x + cos^2 x = 1 = (1+cos x) / sin x (1+cos x) = 1/sin x
How do you solve Sin x sec x equals tan x?
Cos x = 1 / Sec x so 1 / Cos x = Sec x Then Tan x = Sin x / Cos x = Sin x * (1 / Cos x) = Sin x * Sec x
Sin 15 plus cos 105 equals?
Sin 15 + cos 105 = -1.9045
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,
If a cos theta plus b sin theta equals 8 and a sin theta - b cos theta equals 5 show that a squared plus b squared equals 89?
Well, darling, if we square the first equation and the second equation, add them together, and do some algebraic magic, we can indeed show that a squared plus b squared equals 89. It's like a little math puzzle, but trust me, the answer is as sassy as I am.
How do you solve Sin x plus sin 3x plus sin 5x plus sin 7x equals?
sin7x-sin6x+sin5x
How do you solve sin x plus cos x equals zero?
The statement of the problem is equivalent to sin x = - cos x. This is true for x = 135 degrees and x = -45 degrees, and also for (135 + 180n) degrees, where n is any integer.
How do you solve sinx divided by 1 plus cosx plus cosx divided by sinx?
sin x/(1+cos x) + cos x / sin x Multiply by sin x (1+cos x) =[(sin^2 x + cos x(1+cos x) ] / sin x (1+cos x) = [(sin^2 x + cos x + cos^2 x) ] / sin x (1+cos x) sin^2 x + cos^2 x = 1 = (1+cos x) / sin x (1+cos x) = 1/sin x
Who formulated eix equals cos x plus you sin x?
leonhard euler
How do you solve Sin x sec x equals tan x?
Cos x = 1 / Sec x so 1 / Cos x = Sec x Then Tan x = Sin x / Cos x = Sin x * (1 / Cos x) = Sin x * Sec x
How do you solve csc x sin x equals cos x cot x plus?
Suppose csc(x)*sin(x) = cos(x)*cot(x) + y then, ince csc(x) = 1/sin(x), and cot(x) = cos(x)/sin(x), 1 = cos(x)*cos(x)/sin(x) + y so y = 1 - cos2(x)/sin(x) = 1 - [1 - sin2(x)]/sin(x) = [sin2(x) + sin(x) - 1]/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°
Cos 0 sin 0 plus 1 equals 0?
No. sin(0) = 0 So cos(0)*sin(0) = 0 so the left hand side = 1
When does cos x equal -sin x?
The derivative of cos(x) equals -sin(x); therefore, the anti-derivative of -sin(x) equals cos(x).
