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What is the trig function of sin with a period of pi an amplitude of 1 and a vertical shift up of 1?
If this is a homework question, please consider trying to answer it on your own first, otherwise the value of reinforcement of the lesson will be lost on you. To determine the trigonometry function of sin, with a period of pi, and amplitude of 1, and a vertical shift of +1, start simple and expand. The period of sin(x) is 2 pi, so to halve that period you need sin(2x). The amplitude of sin(2x) is 2, so to halve that amplitude you need 1/2 sin(2x). To shift any function up by 1, simply add 1 to it, so the final answer is 1/2 sin(2x) + 1. Note: This is very simple when you take it step by step.
How do you convert this polar equation r2 sin 2 theta equals 8 to cartesian form?
It's possible
What is the amplitude of ycos 5x?
The amplitude of ycos5x is 2*abs(y).
What is 6 sin 30 equals?
The sin of 30 is sqr(3)/2 so 6 * sqr(30) is 6 * sqr(3)/2 this is also 6/2 * sqr(3) which simplifies to 3 * sqr(3) which is approximately 5.19615242
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.
