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Featured 3 weeks ago

Recall the **definition** of the

Now,

Since,

Featured 2 weeks ago

#"the standard form of the sine function is"#

#color(red)(bar(ul(|color(white)(2/2)color(black)(y=asin(bx+c)+d)color(white)(2/2)|)))#

#"amplitude "=|a|," period "=(2pi)/b#

#"phase shift "=-c/b" and vertical shift "=d#

#"here "a=2,b=1/4,c=d=0#

#"amplitude "=|2|=2," period "=(2pi)/(1/4)=8pi#

#"there is no phase shift"#

Featured 2 weeks ago

If

#(a, b, c) = (1 - k, 2k, 4-k)#

and:

#(b + c)/(a + b) = (k+4)/(k+1)#

Given:

#sinA + sin^2A=1#

We have:

#0 = 4(sin^2A + sin A-1)#

#color(white)(0) = 4sin^2A+4sinA+1-5#

#color(white)(0) = (2sinA+1)^2-(sqrt(5))^2#

#color(white)(0) = (2sinA+1-sqrt(5))(2sinA+1+sqrt(5))#

So:

#sinA = (-1+sqrt(5))/2" "# or#" "sinA = (-1-sqrt(5))/2#

We can discount the latter since it gives a value out of the range

So:

#sinA = -1/2+sqrt(5)/2#

Then:

#sin^2A = (-1/2+sqrt(5)/2)^2 = 1/4-sqrt(5)/2+5/4 = 3/2-sqrt(5)/2#

So:

#cos^2A = 1-sin^2A = 1-(3/2-sqrt(5)/2) = -1/2+sqrt(5)/2#

Also:

#cos^4A = (cos^2A)^2 = (-1/2+sqrt(5)/2)^2 = 3/2-sqrt(5)/2#

#cos^6A = (cos^2A)(cos^4A)#

#color(white)(cos^6A) = (-1/2+sqrt(5)/2)(3/2-sqrt(5)/2)#

#color(white)(cos^6A) = -3/4+sqrt(5)/4+3sqrt(5)/4-5/4#

#color(white)(cos^6A) = -2+sqrt(5)#

#cos^8A = (cos^4 A)^2 = (3/2-sqrt(5)/2)^2 = 9/4-(3sqrt(5))/2+5/4 = 7/2-(3sqrt(5))/2#

#cos^12A = (cos^6A)^2 = (-2+sqrt(5))^2 = 4-4sqrt(5)+5 = 9-4sqrt(5)#

Then:

#0 = 2(a cos^12A+ b cos^8A + c cos^6A-1)#

#color(white)(0) = 2(a(9-4sqrt(5))+b(7/2-3/2sqrt(5))+c(-2+sqrt(5))-1)#

#color(white)(0) = (18a+7b-4c-2)+(-8a-3b+2c)sqrt(5)#

This defines a plane of irrational slope in

It is insufficient to determine a unique value for

Perhaps the question is missing some specification, e.g. that

What integer lattice points lie on the plane?

They must satisfy:

#{ (18a+7b-4c-2=0), (-8a-3b+2c=0) :}#

Adding twice the second equation to the first, we find:

#2a+b-2=0#

So in order that

Then:

#a=1/2(2-b) = 1/2(2-2k) = 1-k#

and we find:

#c = 1/2(8a+3b) = 1/2(8-8k+6k) = 4-k#

Then:

#(b + c)/(a + b) = (2k+(4-k))/((1-k)+2k) = (k+4)/(k+1)#

For example, with

#a=-2# ,#b=6# ,#c=1# and#(b+c)/(a+b) = (k+4)/(k+1) = 7/4#

Featured 2 weeks ago

Featured 5 days ago

As detailed.

Standard form of sinusoidal function is

Given

graph{2 sin(2x -pi) + 1 [-10, 10, -5, 5]}

Featured 1 week ago

.

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