Featured Answers

23
Active contributors today

Answer:

#=csc2theta#

Explanation:

We know that,

#(1)tan2x=(2tanx)/(1-tan^2x)#

#(2)sin2x=(2tanx)/(1+tan^2x)#

Using #(1)and(2)#

#cot2theta+tantheta=1/tan(2theta)+tantheta#

#=1/((2tantheta)/(1-tan^2theta))+tantheta#

#=(1-tan^2theta)/(2tantheta)+tantheta#

#=(1-tan^2theta+2tan^2theta)/(2tantheta)#

#=(1+tan^2theta)/(2tantheta)#

#=1/((2tantheta)/(1+tan^2theta))#

#=1/(sin2theta)#

#=csc2theta#

Answer:

As below.

Explanation:

#y = A sin (Bx - C) + D#

#color(brown)(Amplitude = |A|, " Period " = (2pi) / |B|, " Phase Shift " = -C / B, " Vertical Shift " = D#

#"Given Equation is " y = sin ((1/3)x + (pi/2)) -2#

#Amplitude = |A| = 3#

#"Period " = (2pi) / |B| = (2pi) / (1/3) = 6pi#

#"Phase Shift " = -C / B = -(pi/2) / (1/3) = - (3pi)/2, color(crimson)((3pi)/2 " to the LEFT"#

#"Vertical Shift " = D = -2, color(crimson)(" 2 down"##

graph{3 sin ((1/3)x + (pi/2)) - 2 [-10, 10, -5, 5]}

Answer:

#rarrx=2npi+-pi/3# OR #x=2npi+-pi/2#

Explanation:

#rarr3-2sin^2x=3cosx#

#rarr3-2(1-cos^2x)=3cosx#

#rarr3-2+2cos^2x=3cosx#

#rarr2cos^2x-3cosx+1=0#

#rarr2cos^2x-2cosx-cosx+1=0#

#rarr2cosx(cosx-1)-1(cosx-1)=0#

#rarr(2cosx-1)(cosx-1)=0#

Either, #2cosx-1=0#

#rarrcosx=1/2=cos(pi/3)#

#rarrx=2npi+-pi/3# where #nrarrZ#

OR, #cosx-1=0#

#rarrcosx=1=cos(pi/2)#

#rarrx=2npi+-pi/2# where #nrarrZ#

Answer:

As below.

Explanation:

#"Standard form of equation for a sine function is "#

#y = A sin (Bx - C) + D#

#"Given equation is " y = - 2 sin x + 1#

#A = -2, B = 1, C = 0, D = 1#

#Amplitude = |A| = 2#

#"Period " = P = (2pi) / |B| = 2pi#

#"Phase Shift " = -C / B = 0#

#"Vertical Shift " = D = +1#

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

Answer:

#(-sqrt(2) - sqrt(6))/4#

Explanation:

#sin((19pi)/12) = sin((15pi)/12 + (4pi)/12)#
# = sin((5pi)/4 + pi/3)#
# = sin((5pi)/4)cos(pi/3) + cos((5pi)/4)sin(pi/3)#
# = (-sqrt(2)/2)(1/2) + (-sqrt(2)/2)(sqrt(3)/2)#
# = (-sqrt(2)/4) + (-sqrt(6)/4)#
# = (-sqrt(2) - sqrt(6))/4#

Answer:

#arcsin 2 = (pi/2 + 2pi k ) pm i ln(2 - sqrt{3}) quad# integer #k#

Explanation:

Obviously no regular angle, no real number, has a sine of #2#. We're getting in the realm of complex numbers.

Euler's Formula is valid for complex numbers too:

# e^{iz} = cos z + i sin z #

We want to know about #e^{i(-z)} = e^{-iz}.# If #z# is real that would be the conjugate, but it's not for complex #z#. But even for complex #z# we want cosine to be even, #cos z = cos(-z),# and sine to be odd, #sin z = -sin(-z).# So

#e^{-iz} = e^{i(-z)} = cos(-z) + i sin(-z) = cos z - i sin z #

Subtracting,

# e^{iz} - e^{-iz} = 2 i sin z #

#sin z = 1/{2i}(e^{iz} - e^{-iz}) #

From here out we'll use that essentially as the definition of #sin z.#

It's actually easier to get a relatively nice closed form for the inverse sine of a real number greater than one than of a regular sine.

We want to solve

#sin z = a # for real #a>1.#

# 1/{2i}(e^{iz} - e^{-iz}) = a #

Let #y=e^{iz}.# Then #e^{-iz}=1/y.#

#y - 1/y = 2i a #

#y^2 - 2i a y - 1 = 0#

I call it the Shakespeare Quadratic Formula (#2b# or #-2b#): #x^2-2bx+c# has zeros #x=b pm sqrt{b^2-c}.#

#y = ia pm sqrt{(-ia)^2+1} = ia pm sqrt{1-a^2} #

Since #a>1# the radicand is negative. Let's make that explicit.

#y = ia pm sqrt{ (a^2-1) (-1) } = i (a pm sqrt{a^2-1} ) #

#e^{iz} = i (a pm sqrt{a^2-1} ) #

Since #i=e^{i pi/2}# and #e^{2pi k i} = 1 quad # integer #k,#

#e^{iz} = e^{i pi /2 } e^{2pi k i} e^{ln (a pm sqrt{a^2-1} ) } #

#iz = i pi/2 + 2pi k i + ln ( a pm sqrt{a^2-1} ) #

#z = (pi/2 + 2 pi k) - i ln ( a pm sqrt{a^2-1} ) #

That's the general solution to #sin z = a # for real #a>1.# There are a few things to note about it. Let's look at the two imaginary parts first.

# (a + sqrt{a^2 - 1} ) (a - sqrt{a^2 - 1 }) = a^2 - (a^2-1) =1 #

# a - sqrt{a^2 - 1 } = 1/{ a + sqrt{a^2 - 1} } #

#ln (a - sqrt{a^2 - 1 } ) = -ln ( a + sqrt{a^2 - 1} ) #

#z = (pi/2 + 2 pi k) pm i ln ( a - sqrt{a^2-1} ) #

The two imaginary parts are negations of each other! That means the #z#s come in complex conjugate pairs.

Considering #k=0#, when we add the two roots together we'll get #pi.# In other words those two roots, even though complex, are supplementary angles who share the same sine, just like real angles. Just like real angles, we can add or subtract #2pi# as many times as we like and get another number with the same sine.

OK, let's plug in #a=2# for the big finish.

#arcsin 2 = (pi/2 + 2pi k ) pm i ln(2 - sqrt{3}) quad# integer #k#

View more
Questions
Ask a question Filters
Loading...
  • Double-check the answer
    Nutter Butter answered · 3 hours ago
This filter has no results, see all questions.
×
Question type

Use these controls to find questions to answer

Unanswered
Need double-checking
Practice problems
Conceptual questions