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3

#A+B+C=180^@#

#=>cot(A+B)=cot(180^@-C)#

#=>(cotAcotB-1)/(cotB+cotA)=-cotC#

#=>(cotAcotB-1)=-(cotB+cotA)*cotC#

#=>(cotAcotB+cotBcotC+cotCcotA=1#

Now

#cot^2A+cot^2B+cot^2C-1#

#=cot^2A+cot^2B+cot^2C-(cotAcotB+cotBcotC+cotCcotA)#

#=1/2(2cot^2A+2cot^2B+2cot^2C-2cotAcotB-2cotBcotC-2cotCcotA)#

#=1/2[(cotA-cotB)^2+(cotB-cotC)^2+(cotC-cotA)^2]#
This being sum of three squared quantities each of which is difference of two quantities. So this sum is #>=0#

Hence

#cot^2A+cot^2B+cot^2C-1>=0#

#=>cot^2A+cot^2B+cot^2C>=1#

3

We wish to show that:

# arcsin ( (x-1)/(x+1) ) = 2arctan sqrt(x) - pi/2 #

Define a function #f(x)# by:

# f(x) = arcsin ( (x-1)/(x+1) ) - 2arctan sqrt(x) + pi/2 # ..... [A]

Put #x=0# and we get:

# f(0) = arcsin ( -1 ) - 2arctan 0 + pi/2 #
# \ \ \ \ \ \ \ = -pi/2 - 0 + pi/2 #
# \ \ \ \ \ \ \ = 0 #

We can use the standard derivatives:

# {: (ul("Function"), ul("Derivative"), ul("Notes")), (f(x), f'(x),), (sin^(-1)x, 1/sqrt(1-x^2), ), (tan^(-1)x, 1/(1+x^2), ) :} #

So if we differentiate [A] and apply the chain rule, and quotient rule, we have:

# f'(x) = 1/sqrt(1-((x-1)/(x+1))^2)d/dx ((x-1)/(x+1)) - 2/(1+(sqrt(x))^2)d/dx(sqrt(x)) #

# \ \ \ \ \ \ \ = 1/sqrt(1-(x-1)^2/(x+1)^2) ( (x+1)(1) - (x-1)(1) ) / (x+1)^2 - 2/(1+x) 1/(2sqrt(x)) #

# \ \ \ \ \ \ \ = 1/sqrt( ( (x+1)^2 -(x-1)^2) /(x+1)^2) ( x+1 - x + 1 ) / (x+1)^2 - 1/((1+x)sqrt(x)) #

# \ \ \ \ \ \ \ = sqrt( (x+1)^2 / ( (x+1)^2 -(x-1)^2) ) * ( 2 ) / (x+1)^2 - 1/((1+x)sqrt(x)) #

# \ \ \ \ \ \ \ = sqrt( (x+1)^2 / ( (x+1+x-1)(x+1-x+1) ) ) * ( 2 ) / (x+1)^2 - 1/((1+x)sqrt(x)) #

# \ \ \ \ \ \ \ = sqrt( (x+1)^2 / ( (2x)(2) ) ) * ( 2 ) / (x+1)^2 - 1/((1+x)sqrt(x)) #

# \ \ \ \ \ \ \ = (x+1)/(2sqrt(x)) * ( 2 ) / (x+1)^2 - 1/((1+x)sqrt(x)) #

# \ \ \ \ \ \ \ = 1/(sqrt(x)) * ( 1 ) / (x+1) - 1/((1+x)sqrt(x)) #

# \ \ \ \ \ \ \ = 0 #

So the function #f(x)# has a zero derivative everywhere, and therefore integrating we have:

# f(x) = c #

We established earlier that #f(0)=0 => f(0) = c = 0#, Hence:

# c = 0 #

Hence we have:

# \ \ \ \ arcsin ( (x-1)/(x+1) ) - 2arctan sqrt(x) + pi/2 = 0 #

# :. arcsin ( (x-1)/(x+1) ) = 2arctan sqrt(x) - pi/2 # QED

3

#6(tanx)^2-4(sinx)^2=1#

#=>6(tan^2x)xxcos^2x-4sin^2x xxcos^2x=cos^2x

#=>6sin^2x-4sin^2x xxcos^2x=cos^2x#

#=>6(1-cos^2x)-4(1-cos^2x) cos^2x=cos^2x#

#=>6-6cos^2x-4cos^2x+4cos^4x=cos^2x#

#=>4cos^4x-11cos^2x+6=0#

#=>4cos^4x-8cos^2x-3cos^2x+6=0#

#=>4cos^2x(cos^2x-2)-3(cos^2x-2)=0#

#=(cos^2x-2)(4cos^2x-3)=0#

when #cos^2x-2=0=>cosxpmsqrt2#

but #-1<=cosx <=+1#
so this solution is not acceptable.

when

#(4cos^2x-3)=0#

#cosx=pmsqrt3/2#

For

#cosx=+sqrt3/2=cos(pi/6)#

#=>x=2npipmpi/6" where " n in ZZ#

For

#cosx=-sqrt3/2=-cos(pi/6)=cos((5pi)/6)#

#=>x=2npipm(5pi)/6" where " n in ZZ#

1

Answer:

Graph #y=cos(x)# and shift everything to the left by #pi/6#

Desmos

Explanation:

We know that #sin# and #cos# has a period of #2pi#. That is to say that it repeats itself every #2pi# units.
I would assume you know how to graph a #f(x)=cos(x)# functions, if not, it should look like this:

Desmos

Now, you need to graph #f(x)=cos(x+pi/6)#.

Imagine you have a function #f(x)# and another function #g(x)=f(x+1)#.

What this means is that for any point #(x, y)# on the graph #g(x)#, it will take #x+1# units for #f(x)# to reach that same #y# value.
That is what this #g(x)=f(x+1)# is saying.

This means that all points on #g(x)# is occurring 1 unit earlier than #f(x)# so we shift #f(x)# to the left by 1 unit to obtain #g(x)#.

To generalize:
If #g(x)=f(x+n)# we shift #f(x) #n# units to the **left** to get #g(x)#. If #g(x)=f(x-n)# we shift #f(x) #n# units to the right to get #g(x)#.

Now, we can apply it to this question:

We have #f(x)=cos(x+pi/6)# which is basically saying we should shift #cos(x)# to the left by #pi/6# units.

Desmos

The blue curve is your #y=cos(x+pi/6)#
The red curve is your #y=cos(x)#

3

#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#

3

#LHS=(1+cos(pi/10))(1+cos((3pi)/10))(1+cos((7pi)/10))(1+cos((9pi)/10))#

Puttingv#pi/10=2theta# we have

#LHS=(1+cos(2theta))(1+cos(6theta))(1+cos(14theta))(1+cos(18theta))#

#=(2cos^2(theta))(2cos^2(3theta))(2cos^2(7theta))(2cos^2(9theta))#

#=[(2cos(theta)cos(9theta))(2cos(3theta)cos(7theta))]^2#

#=[(cos(10theta)+cos(8theta))(cos(10theta)+cos(4theta))]^2#

#=[(cos(pi/2)+cos(8theta))(cos(pi/2)+cos(4theta))]^2#

#=[cos(8theta)cos(4theta)]^2#

#=[(4cos(8theta)cos(4theta)sin(4theta))/(4sin(4theta))]^2#

#=1/16[sin(16theta)/sin(4theta)]^2#

#=1/16[sin((8 pi)/10)/sin((2pi)/10)]^2#

#=1/16[sin(pi-(2pi)/10)/sin((2pi)/10)]^2#

#=1/16[sin((2pi)/10)/sin((2pi)/10)]^2#

#=(1/16)=RHS#

2

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.

1

Answer:

#p ~~ 51.4 "ft"#

Explanation:

See the diagram below.

enter image source here

In the diagram, #alpha# represents the angle the pole makes with the ground, while point #P# marks the edge of the shadow on the ground.

From inspection, we can see that the angles #alpha#, #beta#, and #55^o# must all sum to #180^o#. Furthermore, since the pole makes an angle of #7^o# to the vertical, we know that #alpha + 7^o = 90^o#, and thus #alpha = 83^o#.

This means that we can calculate #beta#:

#alpha + beta + 55^o = 180^o #

#83^o + beta + 55^o = 180^o #

#beta = 42^o#

Since we now have #/_P#, #beta#, and #b#, and we are looking for #p#, the Law of Sines can be helpful here:

#sin P/p = sin beta/b#

#sin 55^o/p = sin 42^o/42#

#(42*sin 55^o)/(sin 42^o) = p#

#p ~~ 51.4 "ft"#

2

Please help?

Jason K.
Jason K.
Featured 3 weeks ago

Answer:

#~~ 24.0"km"#

Explanation:

In the picture, the unknown you are looking for (the distance between the two cities) is unlabeled, but is opposite of the angle of #2.1^o# that is shown at the satellite. For convenience sake I will label the angle of #2.1^o# as #/_C#, and the unknown distance between the cities as #c#. Furthermore, I will arbitrarily label the "top" distance of 370km as #a# and the "bottom" distance of 350km as #b#.

In this triangle problem, we have 2 sides (#a# and #b#) and the angle between those sides (#/_C#), and we are solving for the opposite side (#c#), which means the Law of Cosines is best suited for this problem:

#c^2 = a^2 + b^2 -2abcos C#

Filling in the knowns as listed above, we have to solve for #c#, which should be a simple matter so long as care is taken to calculate all values properly. (This is particularly true when using a calculator when some questions may be in radians and others in degrees!)

#c^2 = a^2 + b^2 -2abcos C#

#c^2 = (370)^2 + (350)^2 - 2(370)(350)(cos 2.1) #

# c^2 ~~ 136900 + 122500 - 259000(0.999) #

# c^2 ~~ 259400 - 258826.054 #

# c^2 ~~ 573.946 #

# c ~~ sqrt(573.946) ~~ 24.0"km" #

1

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