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Answer:

# -pi/3#.

Explanation:

Recall the definition of the #sin^-1# function :

#sin^-1 x=theta, (-1 le x le 1) iff x=sintheta, -pi/2 le theta le pi/2#.

Now, #cos(5/6pi)=cos(pi-pi/6)=-cos(pi/6)=-sqrt3/2#.

#:. sin^-1(cos(5/6pi))=sin^-1(-sqrt3/2)#.

Since, #sin(-pi/3)=-sin(pi/3)=-sqrt3/2, &, -pi/3 in [-pi/2,pi/2]#,

# sin^-1(-sqrt3/2)=-pi/3#.

#:. sin^-1(cos(5/6pi))=sin^-1(-sqrt3/2)=-pi/3#.

Answer:

#2,8pi,0#

Explanation:

#"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"#

Answer:

If #a, b, c# are integers, then for some integer #k#, we have:

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

and:

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

Explanation:

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 #[-1, 1]#, so is not satisfied by any Real value of #A#.

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 #a, b, c# coordinate space.

It is insufficient to determine a unique value for #(b + c)/(a + b) #

Perhaps the question is missing some specification, e.g. that #a, b, c# are integers.

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 #a# be an integer, we need #b# to be even, i.e. #b=2k# for some integer #k#

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 #k=3# we find:

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

The other answer is fine. I'll just point out that once the student has internalized these Two Tired Triangles of Trig (30/60/90 and 45/45/90) they're done. If you review the trig questions here you'll find the vast majority use just these triangles.

Answer:

As detailed.

Explanation:

Standard form of sinusoidal function is #y = A sin (Bx -C) + D#

Given #y = 2 sin (2x - pi) + 1#

#A = 2, B = 2, C = pi, D = 1#

#Amplitude = |A| = 2#

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

#"Phase Shift " = -C / B = pi/2#, #color(red)(pi/2 " to the right"#

#"Vertical Shift " = D = 1#

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

Answer:

#x=(7pi)/6, (11pi)/6#

Explanation:

.

#-5sinx=-2cos^2x+4#

#-5sinx=-2(1-sin^2x)+4#

#-5sinx=-2+2sin^2x+4#

#2sin^2x+5sinx+2=0#

#sinx=(-5+-sqrt(25-4(2)(2)))/4=(-5+-3)/4=-2,-1/2#

#sinx=-2# is invalid because #sin# values are always between #1# and #-1#.

#sinx=-1/2, :. x=(7pi)/6, (11pi)/6#

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