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Answer:

#x~~0.3398#, #x~~0.8481#, #x=2.294#, and #x=2.802# radians.

Explanation:

Strategy: Rewrite this equation as a quadratic equation using #u=sin(x)#. Solve the quadratic equation for #u# by factoring. Then replace #u# with #sin(x)# again and solve using #arcsin#.

Step 1. Rewrite this equation as a quadratic using #color(red)u=color(red)sin(x)#

You are given
#12(color(red)(sin(x)))^2-13(color(red)(sin(x)))+3=0#

Replace #color(red)(sin(x))# with #color(red)(u)#
#12color(red)(u)^2-13color(red)(u)+3=0#

Step 2. Factor the quadratic equation.
#(3u-1)(4u-3)=0#

Solving gives us
#3u-1=0# and #4u-3=0#
#u=1/3# and #u=3/4#

Step 3. Replace #u# with #sin(x)# again and solve with #arcsin#
#sin(x)=1/3# and #sin(x)=3/4#

#sin^(-1)(sin(x))=sin^(-1)(1/3)# and #sin^(-1)(sin(x))=sin^(-1)(3/4)#

#x~~0.3398# radians and #x~~0.8481# radians

These answer work because we were asked to find the solutions in #[0,2pi]~~[0,6.2832]#

However, the graph of #y=12(sin(x))^2-13(sin(x))+3# is

Desmos.com and MS Paint

Which shows four solutions in the interval #[0,2pi]#, not two.

We must find the other solutions by recognizing that sine functions are periodic with respect to #pi#

So, #x=pi-0.8481=2.294# and #x=pi-0.3398=2.802#

The other two solutions are
#x=2.294# and #x=2.802#

Answer:

#theta=tan^-1(3)+npi " or "theta=tan^-1(-2)+npi;#

#n in ZZ#

Explanation:

#sec^2theta=7+tantheta#

#color(red)(uarr sec^2theta=tan^2theta+1#

#color(red)(tan^2theta+1)=7+tanthetacolor(orange)(larr "subtract"(7+tantheta) " from both sides" #

#tan^2theta+1-7-tantheta=0#

#tan^2theta-tantheta-6=0#

Let #tantheta=u#

#u^2-u-6=0#

  • now we've got a quadratic equation that can be solved by using the quadratic formula #u=(-b+-sqrt(b^2-4ac))/(2a)#.
    we can get the values the values of #a, b, and c# by comparing our equation to the standard form #au^2+bu+c#, and the values are #a=1, b=-1, c=-6#

#u=(-b+-sqrt(b^2-4ac))/(2a)=>tantheta=(-b+-sqrt(b^2-4ac))/(2a)#

#=(-(-1)+-sqrt((-1)^2-4(1)(-6)))/(2(1))#

#tantheta=(1+-5)/2=>#

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

#tantheta=3" or "tantheta=-2#

  • a period of the function #tantheta# is #pi# therefore #tan(theta+pi)=tantheta#

Let #tantheta_1=3" and " tantheta_2=-2#

#theta_1=tan^-1(3)+npi; "where "n inZZ#

#theta_2=tan^-1(-2)+npi" where "n in ZZ#

Answer:

One does not use integration or rationalizing to prove this. One proves this by changing only 1 side of the equation, until it is identical to the other side.

Explanation:

Prove:

#(1-sin(2x)) /(1+sin(2x))=tan^2(45^@-x)#

The identity for #tan(A-B)=(tan(A)-tan(B))/(1+tan(A)tan(B))#, therefore, we may substitute #((tan(45^@)-tan(x))/(1 + tan(45^@)tan(x)))^2# for #tan^2(45^@-x)#:

#(1-sin(2x)) /(1+sin(2x))=((tan(45^@)-tan(x))/(1 + tan(45^@)tan(x)))^2#

Use the fact that #tan(45^@) = 1#:

#(1-sin(2x)) /(1+sin(2x))=((1-tan(x))/(1 + tan(x)))^2#

Write #tan(x)# as #sin(x)/cos(x)#:

#(1-sin(2x)) /(1+sin(2x))=((1-sin(x)/cos(x))/(1 + sin(x)/cos(x)))^2#

Multiply the fraction inside the square by 1 in the form of #(cos(x)/cos(x))#:

#(1-sin(2x)) /(1+sin(2x))=((cos(x)/cos(x))(1-sin(x)/cos(x))/(1 + sin(x)/cos(x)))^2#

Perform the multiplication:

#(1-sin(2x)) /(1+sin(2x))=((cos(x)-cos(x)sin(x)/cos(x))/(cos(x) + cos(x)sin(x)/cos(x)))^2#

Please observe the cancellation of the embedded denominators:

#(1-sin(2x)) /(1+sin(2x))=((cos(x)-cancel(cos(x))sin(x)/cancel(cos(x)))/(cos(x) + cancel(cos(x))sin(x)/cancel(cos(x))))^2#

Write the equation without the cancelled factors:

#(1-sin(2x)) /(1+sin(2x))=((cos(x)-sin(x))/(cos(x) + sin(x)))^2#

Perform the multiplication implied by the square:

#(1-sin(2x)) /(1+sin(2x))=(cos^2(x)-2sin(x)cos(x)+sin^2(x))/(cos^2(x) +2sin(x)cos(x)+ sin^2(x))#

Use the identity #sin^2(x) + cos^2(x) = 1#:

#(1-sin(2x)) /(1+sin(2x))=(1-2sin(x)cos(x))/(1 +2sin(x)cos(x))#

Use the identity #sin(2x) = 2sin(x)cos(x)#:

#(1-sin(2x)) /(1+sin(2x))=(1-sin(2x))/(1 +sin(2x))# Q.E.D.

Answer:

In order to answer this I have assumed a vertical shift of #+7#

#color(red)(3cos(2theta)+7)#

Explanation:

The standard cos function #color(green)(cos(gamma))# has a period of #2pi#

If we want a period of #pi# we need to replace #gamma# with something that will cover the domain "twice as fast" e.g. #2theta#.
That is #color(magenta)(cos(2theta))# will have a period of #pi#.

To get an amplitude of #3# we need to multiply all values in the Range generated by #color(magenta)(cos(2theta))# by #color(brown)3# giving
#color(white)("XXX")color(brown)(3cos(2theta))#

There is to be no horizontal shift, so the argument for #cos# will not be modified by any further addition/subtraction.

In order to achieve the vertical shift (that I assumed would be #color(red)(+7)# [substitute your own value]) we will need to add #color(red)7# to all values in our modified range:
#color(white)("XXX")color(red)(3 cos(2theta) +7)#

enter image source here

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:

The way that one proves an identity is to make substitutions to only one side until it is identical to the other side:

Explanation:

Given: #sin^-1(tanh(x)) = tan^-1(sinh(x))#

Use the property #u = sin^-1(sin(u))# on the right side and mark as equation [1]:

#sin^-1(tanh(x)) = sin^-1(sin(tan^-1(sinh(x)))" [1]"#

Digress and prove that #sin(tan^-1(sinh(x)) = tanh(x)#

An alternate form for #tan^-1(u) = i/2ln(1-iu)-i/2ln(1+iu)#

Substitute #sinh(x)# for u:

#tan^-1(sinh(x)) = i/2ln(1-isinh(x))-i/2ln(1+isinh(x))#

An alternate form for #sin(v) = i/2e^(-iv)-i/2e^(iv)#

#sin(tan^-1(sinh(x))) = i/2e^(-i(i/2ln(1-isinh(x))-i/2ln(1+isinh(x)))-i/2e^(i(i/2ln(1-isinh(x))-i/2ln(1+isinh(x))))#

Distribute through #-i#:

#sin(tan^-1(sinh(x))) = i/2e^((-i^2/2ln(1-isinh(x))+i^2/2ln(1+isinh(x))))-i/2e^(i(i/2ln(1-isinh(x))-i/2ln(1+isinh(x))))#

Distribute through #i#:

#sin(tan^-1(sinh(x))) = i/2e^((-i^2/2ln(1-isinh(x))+i^2/2ln(1+i(sinh(x))))-i/2e^((i^2/2ln(1-isinh(x))-i^2/2ln(1+isinh(x))))#

use the property #i^2 = -1#:

#sin(tan^-1(sinh(x))) = i/2e^((1/2ln(1-isinh(x))-1/2ln(1+isinh(x))))-i/2e^((-1/2ln(1-isinh(x))+1/2ln(1+isinh(x))))#

Write the #1/2# in the exponent as a square root:

#sin(tan^-1(sinh(x))) = i/2e^((ln(sqrt(1-isinh(x)))-ln(sqrt(1+isinh(x)))))-i/2e^((-ln(sqrt(1-isinh(x)))+ln(sqrt(1+isinh(x)))))#

Factor out #i/2#:

#sin(tan^-1(sinh(x))) = i/2{e^((ln(sqrt(1-isinh(x)))-ln(sqrt(1+isinh(x))))-e^((-ln(sqrt(1-isinh(x)))+ln(sqrt(1+isinh(x)))))}#

Use the property of logarithms #ln(a) - ln(b) = ln(a/b)#:

#sin(tan^-1(sinh(x))) = i/2{e^((ln((sqrt(1-isinh(x)))/(sqrt(1+isinh(x))))))-e^((ln((sqrt(1+isinh(x)))/(sqrt(1-isinh(x))))))}#

Use the property #e^ln(u) = u#:

#sin(tan^-1(sinh(x))) = i/2{(sqrt(1-isinh(x)))/(sqrt(1+isinh(x)))-(sqrt(1+isinh(x)))/(sqrt(1-isinh(x)))}#

When we make a common denominator we obtain:

#sin(tan^-1(sinh(x))) = i/2{(1-isinh(x))/(sqrt(1+sinh^2(x)))-(1+isinh(x))/(sqrt(1+sinh^2(x)))}#

Combine over the common denominator:

#sin(tan^-1(sinh(x))) = i/2{(-2isinh(x))/(sqrt(1+sinh^2(x)))}#

The leading coefficient multiplied into the numerator becomes 1:

#sin(tan^-1(sinh(x))) = sinh(x)/(sqrt(1+sinh^2(x)))#

Use the identity #1 + sinh^2(x) = cosh^2(x)#:

#sin(tan^-1(sinh(x))) = sinh(x)/(sqrt(cosh^2(x)))#

#sin(tan^-1(sinh(x))) = sinh(x)/cosh(x)#

#sin(tan^-1(sinh(x))) = tanh(x)#

Substitute this into equation [1]:

#sin^-1(tanh(x)) = sin^-1(tanh(x))# Q.E.D.

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