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Answer:

The amplitude is #=2#. The period is #=8pi# and the phase shift is #=0#

Explanation:

We need

#sin(a+b)=sinacosb+sinbcosa#

The period of a periodic function is #T# iif

#f(t)=f(t+T)#

Here,

#f(x)=2sin(1/4x)#

Therefore,

#f(x+T)=2sin(1/4(x+T))#

where the period is #=T#

So,

#sin(1/4x)=sin(1/4(x+T))#

#sin(1/4x)=sin(1/4x+1/4T)#

#sin(1/4x)=sin(1/4x)cos(1/4T)+cos(1/4x)sin(1/4T)#

Then,

#{(cos(1/4T)=1),(sin(1/4T)=0):}#

#<=>#, #1/4T=2pi#

#<=>#, #T=8pi#

As

#-1<=sint<=1#

Therefore,

#-1<=sin(1/4x)<=1#

#-2<=2sin(1/4x)<=2#

The amplitude is #=2#

The phase shift is #=0# as when #x=0#

#y=0#

graph{2sin(1/4x) [-6.42, 44.9, -11.46, 14.2]}

Answer:

Please see the proof below

Explanation:

We need

#cos(2x)=2cos^2x-1=1-2sin^2x#

Therefore,

#cos(x/2)=sqrt((1+cosx)/2)#

#sin(x/2)=sqrt((1-cosx)/2)#

#tan(x/2)=sin(x/2)/cos(x/2)=sqrt((1-cosx)/(1+cosx))#

#=sqrt(((1-cosx)(1-cosx))/((1+cosx)(1-cosx)))#

#=sqrt((1-cosx)^2/(1-cos^2x))#

#=sqrt((1-cosx)^2/(sin^2x))#

And finally,

#tan(x/2)=(1-cosx)/sinx#

Here,

#x=225#

#cos(225)=-1/sqrt2#

#sin(225)=-1/sqrt2#

#tan(225/2)=(1-(-1/sqrt2))/(-1/sqrt2)=-(sqrt2+1)#

Given,

#rarr(x/a)costheta+(y/b)sintheta=1.....(1)#

#rarr(x/a)sintheta-(y/b)costheta=1......(2)#

Squaring and adding equation #(1)# and #(2)#,we get

#rarr(x^2)/(a^2)cos^2thetacancel(+2((xy)/(ab))costheta*sintheta)+(y^2)/(b^2)sin^2theta+(x^2)/(a^2)sin^2thetacancel(-2((xy)/(ab))sintheta*costheta)+(y^2)/(b^2)cos^2theta=1^2+1^2#

#rarr(x^2)/(a^2)(cos^2theta+sin^2theta)+(y^2)/(b^2)(sin^2theta+cos^2theta)=2#

#therefore(x^2)/(a^2)+(y^2)/(b^2)=2#

Answer:

As below.

Explanation:

#y = sin (x - (pi/4)) + 2#

Standard Form of a sine function is #y = A sin (Bx - C) + D#

#:. A = 1, B = 1, C = pi/4, D = 2#

#Amplitude = |A| = 1#

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

Phase Shift #= -C / B = pi/4#, #pi/4# to the right#

Vertical Shift #= D = 2#

graph{sin(x-(pi/4)) + 2 [-10, 10, -5, 5]}

Answer:

Kindly see a Proof in the Explanation.

Explanation:

#(sectheta-1)/(sectheta+1)#,

#=(1/costheta-1)-:(1/costheta+1)#,

#=(1-costheta)/costheta-:(1+costheta)/costheta#,

#=(1-costheta)/(1+costheta)#,

#=(1-costheta)/(1+costheta)xx(1+costheta)/(1+costheta)#,

#=(1-cos^2theta)/(1+costheta)^2#,

#=sin^2theta/(1+costheta)^2#, as desired!

Answer:

#"Screen height " = 15.2 " meters"#

Explanation:

Anjali G

Using the variable definitions of #a# and #b# as shown in the diagram above, we can use trigonometry to solve for the height of the screen, #a+b#.

#tan(13^@) = a/11#

#a = 11tan(13^@)#

#tan(49^@) = b/11#

#b = 11tan(49^@)#

Therefore, the height of the screen #a+b# is:

#a + b = 11tan(13^@)+11tan(49^@)#

#a+b = 2.5396 + 12.6541#

#a+b = 15.1936 " meters"#

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