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Answer:

See below

Explanation:

There are several definitions of the function #sinx#. I will give three of the most common ones below.

(i) Probably the simplest is in relation to any right #triangle ABC# where #angle C# is a right angle (#90^0 -= pi/2 rad#). The sides of #triangle ABC# are #a, b, c# opposite their respective angle. Here side #c# is called the hypotenuse as it is opposite the right angle.

#sin theta# is defined as the ratio: #"opposite / hypotenuse"#.

So, in the case above #sin A = a/c; sin B =b/c#

(ii) Consider the unit circle centred at the origin #(O)#. A point #P (x,y)# moves around the circumference in a counter clockwise direction from the point #(1,0)#. Let the angle formed between #OP# and the positive #x-#axis be #theta#

#sin theta# is then defined as the #y-#component of #P#

NB: This definition is equivalent to (i) above for #theta in (0, pi/2)# and generalises the function for #theta in [0,2pi]# and, by extension, to all real numbers.

(iii) #sinx# is continuous and differentiable for all real numbers - #(forall x in RR)#

Hence, it can be defined by a Taylor series at #x=0#. This series expansion is:

#sinx = x-x^3/(3!)+x^5/(5!)-x^7/(7!)+....#

#= sum_"n=0"^oo (-1)^n/((2n+1)!) x^(2n+1)#

Finally, to "represent" #y=sinx# as requested in the question, the graph of #y=sinx# is shown below.

graph{sinx [-6.244, 6.244, -3.12, 3.124]}

Answer:

Yes, but only for complex values of #x#

#x = -i ln (1/2(3+-sqrt(5))i) + 2npi" "n in ZZ#

Explanation:

As a real valued function of real numbers, #sin x# maps #(-oo, oo)# onto #[-1, 1]#, so does not take the value #3/2#.

So there is no real number #x# such that #sin x = 3/2#

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 #sin x# for complex values of #x#.

Then we want to solve:

#3/2 = sin x = (e^(ix)-e^(-ix))/(2i)#

Multiply both ends by #2# to get:

#3 = e^(ix)/i - 1/(ie^(ix)) = e^(ix)/i + i/e^(ix) = t+1/t#

where #t= e^(ix)/i#

Multiply both ends by #4t# to get:

#12t = 4t^2+4#

Subtract #12t# from both sides to get:

#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#

#LHS=cos^2 (pi /11) + cos^2((2 pi )/11 )+ cos^2 ((3 pi)/ 11) + cos^2(( 4 pi)/ 11) + cos^2(( 5 pi )/11) )#

#=1/2(2cos^2 (pi /11) + 2cos^2((2 pi )/11 )+ 2cos^2 ((3 pi)/ 11) + 2cos^2(( 4 pi)/ 11) + 2cos^2(( 5 pi )/11) )#

#=1/2(5+cos ((2pi) /11) + cos((4 pi )/11 )+ cos((6pi)/ 11) + cos(( 8pi)/ 11) + cos(( 10pi )/11) )# #color(red)(["using "2cos^2theta=(1+cos2theta)])#

#=1/2(5+cos ((6pi) /11) + cos((4 pi )/11 )+cos((2pi)/ 11) + cos(( 8pi)/ 11) +cos((pi-pi /11) ))#

#=1/2(5 + 2cos((5 pi )/11 )cos((pi)/ 11) + 2cos(( 5pi )/11) cos((3pi)/11)-cos(pi/11))#

#=1/2(5 + 2cos((5 pi )/11 )(cos((pi)/ 11) + cos((3pi)/11))-cos(pi/11))#

#=1/2(5 + 2cos((5 pi )/11 )*2cos((2pi)/ 11) cos((pi)/11)-cos(pi/11))#

#=1/2(5 + cos((5 pi )/11 )/sin(pi/11)*2cos((2pi)/ 11) *2sin(pi/11)cos((pi)/11)-cos(pi/11))#

#=1/2(5 + cos((5 pi )/11 )/sin(pi/11)*2cos((2pi)/ 11) sin((2pi)/11)-cos(pi/11))#

#=1/2(5 + (2cos((5 pi )/11 )sin((4pi)/11))/(2sin(pi/11))-cos(pi/11))#

#=1/2(5 + (sin((9 pi )/11 )-sin((pi)/11))/(2sin(pi/11))-cos(pi/11))#

#=1/2(5 + sin(pi-(2 pi )/11 )/(2sin(pi/11))-sin((pi)/11)/(2sin(pi/11))-cos(pi/11))#

#=1/2(5 + sin((2 pi )/11 )/(2sin(pi/11))-1/2-cos(pi/11))#

#=1/2(5 + (2sin(pi /11 )cos(pi/11))/(2sin(pi/11))-1/2-cos(pi/11))#

#=1/2(5 + cos(pi/11)-1/2-cos(pi/11))#

#=1/2(5-1/2)#

#=1/2*9/2=9/4=RHS#

Answer:

#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^@#

Explanation:

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 #10# from both ends to get:

#2cot^2 x + cot x - 8 = 0#

This is in the form:

#at^2+bt+c = 0#

with #t = cot x#, #a=2#, #b=1# and #c=-8#.

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):

  • #cot x# is continuous and monotonically decreasing on #(0^@, 180^@)# with range #(-oo, oo)#.

  • The range of #cot^(-1)(y)# is #(0^@, 180^@)#.

  • #cot x# has period #180^@#.

Hence all the solutions in #(0^@, 360^@)# are:

#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 #cot^(-1) ((1-sqrt(65))/4)# (namely #~~ -29.527^@#). If it does then just add #180^@#.

If your calculator does not have #cot^(-1)# then take the reciprocal and use #tan^(-1)#, but note that this will give values in the range #(-90^@, 90^@)#, so you will need to add an initial #180^@# to the #~~ -29.527^@# value to get a solution in the requested range.

Answer:

Plot #cos(x)#
then reflect in the X-axis to get #-cos(x)#
and finally shift all point up #3# units to get #3-cos(x)#

Explanation:

I have assumed that you are familiar with the graph for #color(red)(y=cos(x))#:
enter image source here

Reflecting this in the X-axis causes every #y# coordinate to be come the negative of what it was for #color(grey)(y=cos(x))#;
that is we get #color(red)(y=-cos(x))#
enter image source here

Then adding #3# to every #y# coordinate increases each #y# value of #color(grey)(y=-cos(x))# by #3# to give #color(red)(y=3-cos(x))#
enter image source here

Answer:

#C=27.1^@# or #C=153^@#

Explanation:

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:

I'm not artist, but it's the thought that counts, right?

The way we calculate the angle is the same regardless. We use the sine rule:

#sinA/a=sinC/c#.

#sinC=(csinA)/a#

#sinC=(40sin20)/30#

#sinC=0.45602...#

Leave this value in your calculator.

One of the answers for the angle will be the inverse sine of this.

#C=27.131...#
#C=27.1^@#

C can also be #180-C#, (I will show why later).

so #C=180-27.131...#
#C=152.868...#
#C=153^@#

So why does this work? Consider the graph of #y=sinx# for #0^@<=x<=180^@#

y=sinx and y=4/3(sin20)

The red curve is the graph of #y=sinx#, and the green line is the line #y=4/3sin20#, which is what we obtained from the sine rule earlier.

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 #27.131^@,# and one at #153.869^@#. This means that our angle C can take either of these values, whether it is acute or obtuse.

For any angle, #theta, sintheta=sin(180-theta)#. It is necessary to consider both angles.

This does not arise for the cosine rule. The graph of #y=cosx# does not give the same value for the range #0<=x<=180#, so such a problem does not arise.

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