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Answer:

See below:

Explanation:

With a triangle with #sectheta=2#, let's first remember that:

#sectheta="hypotenuse"/"adjacent"=2/1#

We could run through the pythagorean theorem for the third side, but we can also remember that these two measurements are part of a 30-60-90 triangle, and so the third side is #sqrt3#. To prove it, let's go ahead and do Pythagorean Theorem:

#a^2+b^2=c^2#

#1^2+(sqrt3)^2=2^2#

#1+3=4#

This gives us the 6 trig ratios:

#sintheta=sqrt3/2#

#costheta=1/2#

#tantheta=(sqrt3/2)/(1/2)=(sqrt3/2)(2/1)=sqrt3#

#csctheta=2/sqrt3=(2sqrt3)/3#

#sectheta=2/1=2#

#cottheta=1/sqrt3=sqrt3/3#

The angle of the triangle we're looking at is #60=pi/3#, with the opposite being the middle length of #sqrt3#, the adjacent length of 1, and hypotenuse of 2.

freemathhelp.com

Answer:

The angle #-3.4°# lies in quadrant 4 (#Q_"IV"#).

Explanation:

When we talk about the quadrant an angle is in, we're really talking about the quadrant the terminal arm of the angle is in, when the angle is drawn with its initial arm along the #+x# axis (a.k.a. standard position).

To draw an angle, imagine a clock centered at the origin, with the minute hand pointing to the right (at the "3"). Then, rotate the hand counterclockwise (towards the "12") by the amount of the angle. Since the angle #-3.4°# is negative, we end up rotating clockwise (towards the "6") by 3.4 degrees.

Now, #-3.4°# is quite small compared to a full #360°# circle, so we will only rotate a little bit (not even to the "4" on our clock). So we have our terminal arm in the bottom-right quadrant.

Which quadrant is this? The quadrants are named in increasing order, starting with quadrant 1 (#Q_"I"#) in the top right, and moving counterclockwise to #Q_"II"# in the top left, etc. In this way, higher numbered quadrants contain bigger angles (until #360°#, of course).

enter image source here

So the angle #-3.4°# lies in #Q_"IV"#.

Answer:

#x_1=npi, n=2k+-1, k in ZZ#

#x_2=n/3pi,mod(n,2)=0#

Explanation:

#cos2x+3cosx=-2#

Use the double angle formula for cosine to expand #cos2x# and rewrite the equation in standard form

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

#2cos^2x+3cosx+1=0#

Let #cosx=y#

#y=(-3+-sqrt(9-4(2)(1)))/(2(2))=(-3+-1)/4#

#y=-1# or #y=-1/2#

#cosx=-1#

#x=npi, n=2k+-1, k in ZZ#

#cosx=-1/2#

#x_2=n/3pi,mod(n,2)=0#

drawn

The given function representing the height #h# in cm of the person above the ground with time #t# is given by

#h=15cos((2pi)/3t)+65#

(a) h is a cosine function of t, So it will have maximum value when #cos((2pi)/3t)=1=cos0=>t=0# s
and the maximum height of the swing becomes

#h_(max)=15cos((2pi)/3xx0)+65=80cm#

(b) it takes #t=0#sec to reach maximum height after the person starts.

c) The minimum height of the swing will be achieved when #cos((2pi)/3t)=-1#

Minimum height

#h_(min)=15xx(-1)+65=50cm#

(d). Again #cos((2pi)/3t)=-1=cos(pi)=>t=3/2=1.5#s

Here #t=1.5#s represents the minimum time required to achieve the minimum height after start.

e) For #h=60# the equation becomes

#60=15cos((2pi)/3t)+65#

#=>60-65=15cos((2pi)/3t)#

#=>cos((2pi)/3t)=-1/3#

#=>t=[1/120(2nxx180pmcos^-1(-1/3))] sec" where " n in ZZ#

#=>t=(3npm0.912)#sec

when #n=0, t=0.912s# this is the minimum time to reach at height #h=60 cm # after start. just after that height #h=60 cm # will be achieved again for #n=1 #

#=>t=(3xx1-0.912)=2.088#sec

So in #(2.088-0.912)=1.176s# sec (the minimum time difference) within one cycle the swing will be less than 60 cm above the ground.

f) The height of the swing at 10 s can be had by inserting #t=10#s in the given equation

#h_(10)=15cos((2pi)/3xx10)+65=57.5#cm

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#

#tan 5^@ + 2tan 10^@ + 4tan 20^@ + 8tan40^@ + 16cot 80^@#

#=tan 5^@ + 2tan 10^@ + 4tan 20^@ + 8tan40^@ + 16/tan 80^@#

#=tan 5^@ + 2tan 10^@ + 4tan 20^@ + 8tan40^@ + (16(1-tan^2 40^@))/(2tan 40^@)#

#=tan 5^@ + 2tan 10^@ + 4tan 20^@ + 8tan40^@ + (8(1-tan^2 40^@))/(tan 40^@)#

#=tan 5^@ + 2tan 10^@ + 4tan 20^@ + (8tan^2 40^@+8-8tan^2 40^@)/(tan 40^@)#

#=tan 5^@ + 2tan 10^@ + 4tan 20^@ + 8/(tan 40^@)#

#=tan 5^@ + 2tan 10^@ + 4tan 20^@ + (8(1-tan^2 20^@))/(2tan 20^@)#

#=tan 5^@ + 2tan 10^@ + (4tan^2 20^@ + 4-4tan^2 20^@)/(tan 20^@)#

#=tan 5^@ + 2tan 10^@ + 4/(tan 20^@)#

#=tan 5^@ + 2tan 10^@ + (4(1-tan^2 10^@))/(2tan 10^@)#

#=tan 5^@ + (2tan^2 10^@ + 2-2tan^2 10^@)/(tan 10^@)#

#=tan 5^@ + 2/(tan 10^@)#

#=tan 5^@ + (2(1-tan^2 5^@))/(2tan 5^@)#

#=(tan^2 5^@ + 1-tan^2 5^@)/(tan 5^@)#

#=(1/tan5^@) =cot5^@-># option (d) is correct

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