Questions
Question type
Use these controls to find questions to answer
See below
There are several definitions of the function
(i) Probably the simplest is in relation to any right
So, in the case above
(ii) Consider the unit circle centred at the origin
NB: This definition is equivalent to (i) above for
(iii)
Hence, it can be defined by a Taylor series at
Finally, to "represent"
graph{sinx [-6.244, 6.244, -3.12, 3.124]}
Yes, but only for complex values of
As a real valued function of real numbers,
So there is no real number
However, consider the following:
#e^(itheta) = cos theta + i sin theta#
So:
#e^(itheta) - e^(-itheta) = (cos theta + i sin theta) - (cos(-theta) + i sin(-theta))#
#color(white)(e^(itheta) - e^(-itheta)) = (cos theta + i sin theta) - (cos(theta) - i sin(theta))#
#color(white)(e^(itheta) - e^(-itheta)) = 2i sin theta#
So we find:
#sin x = (e^(ix)-e^(-ix))/(2i)#
We can use this definition of
Then we want to solve:
#3/2 = sin x = (e^(ix)-e^(-ix))/(2i)#
Multiply both ends by
#3 = e^(ix)/i - 1/(ie^(ix)) = e^(ix)/i + i/e^(ix) = t+1/t#
where
Multiply both ends by
#12t = 4t^2+4#
Subtract
#0 = 4t^2-12t+4#
#color(white)(0) = (2t)^2-2(2t)(3)+3^2-5#
#color(white)(0) = (2t-3)^2-(sqrt(5))^2#
#color(white)(0) = ((2t-3)-sqrt(5))((2t-3)+sqrt(5))#
#color(white)(0) = (2t-3-sqrt(5))(2t-3+sqrt(5))#
Hence:
#e^(ix)/i = t = 1/2(3+-sqrt(5))#
So:
#e^(ix) = 1/2(3+-sqrt(5))i#
So:
#ix = ln (1/2(3+-sqrt(5))i) + 2npii" "n in ZZ#
So:
#x = -i ln (1/2(3+-sqrt(5))i) + 2npi" "n in ZZ#
#cot^(-1) ((1+sqrt(65))/4) ~~ 23.816^@#
#cot^(-1) ((1-sqrt(65))/4) ~~ 150.473^@#
#cot^(-1) ((1+sqrt(65))/4) + 180^@ ~~ 203.816^@#
#cot^(-1) ((1-sqrt(65))/4) + 180^@ ~~ 330.473^@#
Note that:
#csc^2 x = 1/sin^2 x = (cos^2 x+sin^2 x)/sin^2 x = 1+cot^2 x#
So:
#10 = 2csc^2 x + cot x#
#color(white)(10) = 2(cot^2 x + 1) + cot x#
#color(white)(10) = 2cot^2 x + cot x + 2#
Subtract
#2cot^2 x + cot x - 8 = 0#
This is in the form:
#at^2+bt+c = 0#
with
So using the quadratic formula, we have:
#cot x = (-b+-sqrt(b^2-4ac))/(2a)#
#color(white)(cot x) = (-color(blue)(1)+-sqrt(color(blue)(1)^2-4(color(blue)(2))(color(blue)(-8))))/(2(color(blue)(2)))#
#color(white)(cot x) = (-1+-sqrt(65))/4#
Note that (expressed in degrees):
The range of
Hence all the solutions in
#cot^(-1) ((1+sqrt(65))/4) ~~ 23.816^@#
#cot^(-1) ((1-sqrt(65))/4) ~~ 150.473^@#
#cot^(-1) ((1+sqrt(65))/4) + 180^@ ~~ 203.816^@#
#cot^(-1) ((1-sqrt(65))/4) + 180^@ ~~ 330.473^@#
Note that your calculator may give a negative value for
If your calculator does not have
Plot
then reflect in the X-axis to get
and finally shift all point up
I have assumed that you are familiar with the graph for
Reflecting this in the X-axis causes every
that is we get
Then adding
This question concerns itself with the ambiguous case of the sine rule.
Sometimes with the sine rule, you can get two values for the angle. One is acute, the other is obtuse.
This is partly because there are two triangles we can draw for the information given:
The way we calculate the angle is the same regardless. We use the sine rule:
Leave this value in your calculator.
One of the answers for the angle will be the inverse sine of this.
C can also be
so
So why does this work? Consider the graph of
The red curve is the graph of
Since we know that our angle is in a triangle, it cannot be greater than 180, and cannot be less than 0.
However, in this range, note how there are two solutions to these equations? One at
For any angle,
This does not arise for the cosine rule. The graph of
Use these controls to find questions to answer