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sin x can have any value between -1 and 1; therefore, 3 sin x has three times this range (from -3 to 3).

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โˆ™ 2012-08-04 11:57:00
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Algebra

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A polynomial of degree zero is a constant term

The grouping method of factoring can still be used when only some of the terms share a common factor A True B False

The sum or difference of p and q is the of the x-term in the trinomial

A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials

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Q: What is the range of y equals 3 sin x?
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Related questions

What is the range of y equals -3 sin x?

[-3,3]


What is the range of y equals -sin x?

The range of -sin x depends on the domain of x. If the domain of x is unrestricted then the range of y is [-1, 1].


If sin x - cos x equals 1 over 3 what is sin x?

Sin[x] = Cos[x] + (1/3)


What is x if the Sin of x equals square root 3 divided by 2?

1.5


2 sin x - 3 equals 0?

2 sin(x) - 3 = 0 2 sin(x) = 3 sin(x) = 1.5 No solution. The maximum value of the sine function is 1.0 .


What is the range y equals - sin x?

Assuming a large enough domain, the range is -1 to 1.


Need help with working this Trig problem out. A sin alpha plus cos alpha equals 1 B sin alpha - cos alpha equals 1 Solution is AB equals 1?

A*sin(x) + cos(x) = 1B*sin(x) - cos(x) = 1Add the two equations: A*sin(x) + B*sin(x) = 2(A+B)*sin(x) = 2sin(x) = 2/(A+B)x = arcsin{2/(A+B)}That is the main solution. There may be others: depending on the range for x.


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).


What is the period of y equals 3 sin pi x?

2


What is the range of the graph of y equals sin x?

the range is greater then -1 but less than 1 -1<r<1


Find the derivative of y equals 3cosx?

y=3 cos(x) y' = -3 sin(x)


What is the range of y equals sin2x?

range of y=sin(2x) is [-1;1] and in generally when is y=sin(k*x) (k=....-1,0,1....) range is always [-1;1] and the period is w=(2pi)/k


Find the domain and range of sin x - cos x?

sin(x)-cos(x) = (1)sin(x)+(-1)cos(x) so the range is sqrt((1)^2+(-1)^2)=1 and the domain is R <><><><><> The domain of sin x - cos x is [-infinity, +infinity]. The range of sin x - cos x is [-1.414, +1.414].


1 over sin x equals what?

1/sin x = csc x


If Sin equals x and Cos equals y then x squared equals what function of y?

If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²


Sin x Tan x equals Sin x?

No. Tan(x)=Sin(x)/Cos(x) Sin(x)Tan(x)=Sin2(x)/Cos(x) Cos(x)Tan(x)=Sin(x)


How do you prove that 2 sin 3x divided by sin x plus 2 cos 3x divided by cos x equals 8 cos 2x?

You need to know the trigonometric formulae for sin and cos of compound angles. sin(x+y) = sin(x)*cos(y)+cos(x)*sin(y) and cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y) Using these, y = x implies that sin(2x) = sin(x+x) = 2*sin(x)cos(x) and cos(2x) = cos(x+x) = cos^2(x) - sin^2(x) Next, the triple angle formulae are: sin(3x) = sin(2x + x) = 3*sin(x) - 4*sin^3(x) and cos(3x) = 4*cos^3(x) - 3*cos(x) Then the left hand side = 2*[3*sin(x) - 4*sin^3(x)]/sin(x) + 2*[4*cos^3(x) - 3*cos(x)]/cos(x) = 6 - 8*sin^2(x) + 8cos^2(x) - 6 = 8*[cos^2(x) - sin^2(x)] = 8*cos(2x) = right hand side.


How do you prove sin x tan x equals cos x?

You can't. tan x = sin x/cos x So sin x tan x = sin x (sin x/cos x) = sin^2 x/cos x.


Is sin 2x equals 2 sin x cos x an identity?

Yes. sin(A+B) = sin A cos B + cos A sin B If A = B = x, this becomes: sin(x+x) = sin x cos x + cos x sin x → sin 2x = 2 sin x cos x


Y equals 4 sin x for x equals pi?

sin(pi) = 0 so 4*sin(pi) = 0 so Y = 0


What is the solution to the equation 4 sin x - 3 equals 0 to the nearest degree over the domain 0degrees360degrees?

4 sin(x) - 3 = 0 Therefore sin(x) = 3/4 And so the primary solution is x = sin-1(3/4) = 49 deg The second solution in the domain is 180 - 49 = 131 deg.


What is the range of y equals 3 sin x plus 3?

The multiplication by 3 increases the magnitude, and the + 3 shifts the graph upward to be "centred" at y = 3. The graph now oscillates between 6 and 0 (3 + 3, 3 - 3).


The function y equals sin ฮธ is not a function beacause sin 30 equals sin 150?

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How do you use the Euler's formula to obtain the sin3X in terms of cosX and sinX?

Using Euler's Formula, you use (cos(x) + i sin(x))^n = cos (nx) + i sin(nx) Now you let n=3 (cos(x) + i sin (x))3 = cos(3x) + i sin (3x) (cos(x))3 + 3(cos(x))2 * i sin(x) + 3cos(x) * i2 (sin(x))3 = cos(3x)+ i sin(3x) (cos(x))3 + i(3sin(x)(cos (x))2) - 3cos(x)(sin(x)2) - i(sin(x))3 = cos (3x) + i sin(3x) Now only use the terms with i in them to figure out what sin(3x) is... 3sin(x)(cos(x))2 - (sin(x))3 = sin(3x) Hope this helps! :D


How do you show that sinxcosxtanx equals 1-cos2x?

-1