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sinx-cosx=0 --> move cosx to opposite side sinx=cosx --> square both sides sin2x=cos2x --> use pythagorean identities for (cos2x=1-sin2x) sin2x=1-sin2x --> add sin2x to both sides of equation 2sin2x=1 --> divide both sides by 2 sin2x=1/2 --> take the square root of both sides sinx= +/- (square root of 2)/2 or .7071 If giving answers in radians --> answer appears in all four quadrants, so answer would be (pi/4 + piN/2). Other answers would be (3pi/4 + piN/2), (5pi/4 + piN/2), and (7pi/4 + piN/2). Check for extraneous solutions: The answers in the first and third quadrant are extraneous. Therefore, your answer is (3pi/4 + piN), because every pi, an answer occurs. In one trip around the quadrants, both 3pi/4 and 7pi/4 are answers.
What else can I help you with?
Why sin integral is cos?
sin integral is -cos This is so because the derivative of cos x = -sin x
Prove that sin A cos A2?
Not correct. sin2alpha + cos2alpha = 1
What is cos theta times cos theta?
Cos theta squared
What is cos(33)?
If the angle is measured in degrees, then cos(33) = 0.8387, approx.
What is the integral of cosine cubed?
The integral of cosine cubed is sinx- 1/3 sin cubed x + c
What is the lim of h if it equals 0 Sinxcosh plus cosxsinh minus sinx divided by h?
lim(h→0) (sin x cos h + cos x sin h - sin x)/h As h tends to 0, both the numerator and the denominator have limit zero. Thus, the quotient is indeterminate at 0 and of the form 0/0. Therefore, we apply l'Hopital's Rule and the limit equals: lim(h→0) (sin x cos h + cos x sin h - sin x)/h = lim(h→0) (sin x cos h + cos x sin h - sin x)'/h' = lim(h→0) [[(cos x)(cos h) + (sin x)(-sin h)] + [(-sin x)(sin h) + (cos x)(cos h)] - cos x]]/0 = cosx/0 = ∞
What is the derivative of sin x?
f'[x] = lim(h->0) (f[x+h]-f[x])/h lim(h->0) (sin[x+h]-sin[x])/h By angle-addition formula, we have: lim(h->0) (sin[x]cos[h]+sin[h]cos[x]-sin[x])/h lim(h->0) (sin[x]cos[h]-sin[x])/h + lim(h->0) (sin[h]cos[x])/h sin[x]*lim(h->0) (cos[h]-1)/h + cos[x]*lim(h->0) sin[h]/h In a calculus class, it is shown that: lim(h->0) (cos[h]-1)/h = 0 and that lim(h->0) sin[h]/h is 1. So, sin[x]*lim(h->0) (cos[h]-1)/h + cos[x]*lim(h->0) sin[h]/h becomes sin[x]*0 + cos[x]*1 cos[x] So, if f[x] = sin[x], f'[x] = cos[x]
Verify the identity sinx cotx - cosx divided by tanx equals 0?
(sin(x)cot(x) - cos(x))/tan(x)(Multiply by tan(x)/tan(x))sin(x) - cos(x)tan(x)(tan(x) = sin(x)/cos(x))sinx - cos(x)(sin(x)/cos(x))(cos(x) cancels out)sin(x) - sin(x)0
Definition of 6 trigonometric functions?
sin(x) = x - x3/3! + x5/5! - x7/7! + ... and cos(x) = 1 - x2/2! + x4/4! - x6/6! + ... where x is the angle measured in radians. Then tan(x) = sin(x)/cos(x) where cos(x) is not 0 cosec(x) = 1/sin(x) where sin(x) is not 0 sec(x) = 1/cos(x) where cos(x) is not 0 and cot(x) = cos(x)/sin(x) where sin(x) is not 0
Limit when x goes to 0 tanx divided by x?
tan x = sin x / cos x, so:lim (tan x / x) = lim (sin x / x cos x). Since it is known that the limit of sin x / x = 1, you have lim 1 / cos x = 1 (since cos 0 = 1).tan x = sin x / cos x, so:lim (tan x / x) = lim (sin x / x cos x). Since it is known that the limit of sin x / x = 1, you have lim 1 / cos x = 1 (since cos 0 = 1).tan x = sin x / cos x, so:lim (tan x / x) = lim (sin x / x cos x). Since it is known that the limit of sin x / x = 1, you have lim 1 / cos x = 1 (since cos 0 = 1).tan x = sin x / cos x, so:lim (tan x / x) = lim (sin x / x cos x). Since it is known that the limit of sin x / x = 1, you have lim 1 / cos x = 1 (since cos 0 = 1).
How do you solve sin 2x equals sin x over 2?
they do have calculators for these questions you knowsin 2x = (sin x)/22 sin x cos x - (1/2)sin x = 02 sin x(cos x - 1/4) = 02 sin x = 0 or cos x - 1/4 = 0sin x = 0 or cos x = 1/4in the interval [0, 360)sin x = 0, when x = 0, 180cos x = 1/4, when x = 75.52, 284.48Check:
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
What is cosx-2sinxcosx equals 0?
cos x - 2 sin x cos x = 0 -> cos x (1 - 2 sin x) = 0 => cos x = 0 or 1 - 2 sin x = 0 cos x = 0: x = π/2 + kπ 1 - 2 sin x = 0: sin x = 1/2 -> x = π/6 + 2kπ or 5/6π + 2kπ Thus x = π/2 + kπ; x = π/6 + 2kπ; x = 5/6π + 2kπ solve the original equation.
How do you factor sin squared times x plus cos2x -cosx equals 0?
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Prove each Indentity tanx mins sinx divided by tanxsinx equals tanxsinx divided by tanx plus sinx?
(tan x - sin x)/(tan x sin x) = (tan x sin x)/(tan x + sin x)[sin x/cos x) - sin x]/[(sin x/cos x)sin x] =? [(sin x/cos x)sin x]/[sin x/cos x) + sin x][(sin x - sin x cos x)/cos x]/(sin2 x/cos x) =? (sin2 x/cos x)/[(sin x + sin x cos x)/cos x)(sin x - sin x cos x)/sin2 x =? sin2 x/(sin x + sin x cos x)[sin x(1 - cos x)]/sin2 x =? sin2 x/[sin x(1 + cos x)(1 - cos x)/sin x =? sin x/(1 + cos x)(1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[(1 + cos x)(1 - cos x)](1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[1 - cos2 x)(1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[1 - (1 - sin2 x)](1 - cos x)/sin x =? [(sin x)(1 - cos x)]/sin2 x(1 - cos x)/sin x = (1 - cos x)/sin x True
Verify cot x-180 cot x?
cot x = (cos x) / (sin x) cos (x - 180) = cos x cos 180 + sin x sin 180 = - cos x sin (x - 180) = sin x cos 180 - cos x sin 180 = - sin x cot (x - 180) = (cos (x - 180)) / (sin (x - 180)) = (- cos x) / (- sin x) = (cos x) / (sin x) = cot x
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.
