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(a)

#sqrt15sin(2x)+sqrt5cos(2x)#

#=sqrt20(sqrt15/sqrt20sin(2x)+sqrt5/sqrt20cos(2x))#

[Taking #sqrt15/sqrt20=cosalpha and sqrt5/sqrt20=sinalpha#]

#=sqrt20(cosalphasin(2x)+sinalphacos(2x))#

#=2sqrt5sin(2x+alpha)#

Where #alpha=tan^-1(sqrt5/sqrt15)=tan^-1(1/sqrt3)=pi/6#

(b)

(i)
#f(x)=2/(5+sqrt15sin(2x)+sqrtcos(2x))#

#=>f(x)=2/(5+2sqrt5sin(2x+alpha))#

The value of #f(x)# will be maximum when #sin(2x+alpha)# is minimum i.e. #sin(2x+alpha)=-1#

So

#f(x)_"max"=2/(5-2sqrt5)#

#=>f(x)_"max"=(2(5+2sqrt5))/((5-2sqrt5)(5+2sqrt5))#

#=>f(x)_"max"=(10+4sqrt5)/(25-20)=2+4/5sqrt5#

(ii)Here #f(x)# is maximum when

#sin(2x+alpha)=-1#

#=>sin(2x+pi/6)=sin((3pi)/2)#

#=>2x+pi/6=(3pi)/2#

#=>2x=(3pi)/2-pi/6=(4pi)/3#

#=>x=(2pi)/3#

Answer:

#f_0 = 1/(2pi)" Hz"#

Explanation:

Given: #f(t)= sin(4t) - cos(7t)# where t is seconds.

Use this reference for Fundamental Frequency

Let #f_0# be the fundamental frequency of the combined sinusoids, in Hz (or #"s"^-1#).

#omega_1 = 4" rad/s"#
#omega_2 = 7" rad/s"#

Using the fact that #omega = 2pif#

#f_1 = 4/(2pi) = 2/pi " Hz"# and #f_2 = 7/(2pi)" Hz"#

The fundamental frequency is the greatest common divisor of the two frequencies:

#f_0 = gcd(2/pi " Hz", 7/(2pi)" Hz")#

#f_0 = 1/(2pi)" Hz"#

Here is a graph:

graph{y = sin(4x) - cos(7x) [-10, 10, -5, 5]}

Please observe that it repeats every #2pi#

Answer:

#"The Solution Set is "phi.#

Explanation:

#costheta/(1+sintheta)+sintheta/(1+costheta)=2sectheta.#

#:. {costheta(1+costheta)+sintheta(1+sintheta)}/{(1+sintheta)(1+costheta)}=2/costheta.#

#:.(costheta+cos^2theta+sintheta+sin^2theta)/{(1+sintheta)(1+costheta)}=2/costheta.#

#:. (1+costheta+sintheta)costheta=2(1+sintheta)(1+costheta).#

#:. costheta+cos^2theta+sinthetacostheta=2+2sintheta+2costheta+2sinthetacostheta.#

#:. cos^2theta=2+2sintheta+costheta+sinthetacostheta.#

#:.1-sin^2theta=2(1+sintheta)+costheta(1+sintheta).#

#:.(1+sintheta)(1-sintheta)=(1+sintheta)(2+costheta).#

#:.(1+sintheta){(1-sintheta)-(2+costheta)}=0.#

#:.(1+sintheta)(-1-sintheta-costheta)=0.#

#:. (1+sintheta)=0, or, costheta+sintheta=-1.#

#"But, "because (1+sintheta)=0,# makes the given eqn. meaningless,

#:. costheta+sintheta=-1,.#

Remembering that, #0^@ < theta < 90^@,# we have,

# 0 < sintheta < 1, and, 0 < costheta <1,#

and, adding these give,

# 0 < sintheta+costheta < 2.#

Obviously, no such #theta in (0^@,90^@)# exists.

Hence, #"The Solution Set is "phi.#

Enjoy Maths.!

Answer:

#sin 45^@/(cos 30^@+sin 60^@) = sqrt(6)/6#

Explanation:

This is essentially a guess, but note that all of these angles have exact algebraic expressions for the their trigonometric values. So the question probably wants you to express this quotient as an exact algebraic expression.

Since all of the angles are positive but less than #90^@#, they can be treated directly as angles in right angled triangles.

To find the value of #sin 45^@#, consider a right angled triangle with sides #1, 1, sqrt(2)#, being half of a unit square ...

enter image source here

Then:

#sin 45^@ = "opposite"/"hypotenuse" = 1/sqrt(2) = sqrt(2)/2#

For the other two trigonometric values, consider a right angled triangle with sides #1, sqrt(3), 2#, which is one half of an equilateral triangle ...

enter image source here

We have:

#cos 30^@ = "adjacent"/"hypotenuse" = sqrt(3)/2#

#sin 60^@ = "opposite"/"hypotenuse" = sqrt(3)/2#

#color(white)()#
Having obtained our trigonometric values, here's the algebra:

#sin 45^@/(cos 30^@+sin 60^@) = (sqrt(2)/2)/(sqrt(3)/2+sqrt(3)/2)#

#color(white)(sin 45^@/(cos 30^@+sin 60^@)) = sqrt(2)/(2sqrt(3))#

#color(white)(sin 45^@/(cos 30^@+sin 60^@)) = (sqrt(2)sqrt(3))/6#

#color(white)(sin 45^@/(cos 30^@+sin 60^@)) = sqrt(6)/6#

#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#

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)#

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