Question

Using a trig identity, write y(t)= -2cos(4t)-sin(4t) using only one cosine function

Answer 1

Use auxilliary angles to write it in the form of
\[\large R\cos(\epsilon+4t)\]
where
\[\large \epsilon is the angle you are finding.
\[\large R=\sqrt{(-2)^2+(-1)^2}\]
\[\large R=\sqrt{5}\]
\[\large \sqrt{5}[-\frac{2}{\sqrt{5}}\cos(4t)-\frac{1}{\sqrt{5}}\sin(4t)]\]
Therefore
\[\large \cos (\epsilon) =-\frac{2}{\sqrt{5}}\]
\[\large \epsilon=153`26'\]
NOTE:` denotes a degree sign. I don't have one on my keyboard.
THEREFORE:
\[\huge -2\cos(4t)-\sin(4t)= \sqrt{5}\cos(153`26'+4t)\]

Answer 2

\[\huge y(t)=\sqrt{5}\cos(153`26'+4t)\]

Answer 3

\[put r \cos \theta=-2,r \sin \theta=1\]
square and add
\[r ^{2}\left( \cos ^{2}\theta+\sin ^{2} \theta\right)=4+1=5\]
\[r ^{2}=5,r=\sqrt{5}\]
divide,
\[\tan \theta=-\frac{ 1 }{ 2 },\theta=\tan^{-1} \left( \frac{ -1 }{ 2 } \right)\]
\[y \left( t \right)=r \cos \theta \cos \left( 4t \right)-r \sin \theta \sin \left( 4t \right)=r \cos \left( 4t+theta \right)\]

Answer 4

put the values of r & theta ,we get the result

Answer 5

y"+y'=2

\[\frac{ d }{ dx }\left( \frac{ dy }{dx } \right)+\frac{ dy }{dx }=2\]
integrating w.r.t. x,
\[\frac{ dy }{dx }+y=2x+c\]
it is an exact diff. eq.
\[I.F=e ^{\int\limits1 dx}=e ^{x}\]
\[c.s. is y*e ^{x}=\int\limits \left( 2x+c \right)e ^{x}dx+c1\]
\[y e ^{x}=c \int\limits e ^{x} dx+2\int\limits x e ^{x} dx+c1\]
\[y e ^{x}=c e ^{x}+2\int\limits x e ^{x} dx+c1\]
\[\int\limits xe ^{x}dx=first function *\left( \int\limits of second function \right) \[-\int\limits \frac{ d }{ dx } first function*\int\limits second function dx dx\]
\]
\[\int\limits x e ^{x} dx=x \int\limits e ^{x}dx-1\int\limits e ^{x}dx=x e ^{x}-e ^{x}=\left( x-1 \right)e ^{x}\]
substitute and get the complete solution.