Solve the following algebraic equations
Questions:
a) Show that $$(sqrt{2} + 1)^2 = 3+2sqrt2$$
b) Express $$cos3theta$$in the form of$$costheta(a-bsin^2theta)$$
c) Solve $$sqrt2(sin2theta+costheta) + cos3theta = sin2theta + cos2theta$$ $$Theta in [0,360]$$
Answers:
a) $$(sqrt{2} + 1)^2 = 2 + 1 + 2sqrt2 = 3+2sqrt2$$
b)
begin{align*}
cos3theta &= cos(2theta + theta)\
&=cos2thetacostheta - sin2thetasintheta\
&=(cos^2theta-sin^2theta)(costheta) -2sinthetacosthetasintheta\
&=costheta(cos^2theta-sin^2theta-2sin^2theta)\
&=costheta(1-4sin^2theta)
end{align*}
c) I'm still working on this one, this was my approach
begin{align*}
sqrt2(sin2theta+costheta)+cos3theta &= sin2theta + cos2theta\
impliessqrt2(2sinthetacostheta+costheta)+costheta(1-4sin^2theta)&= 2sinthetacostheta + 1 -2sin^2theta\
sqrt2costheta(2sintheta + 1)+costheta(1-4sin^2theta) &= 2sinthetacostheta + 2cos^2theta - 1
end{align*}
Before I proceed further, I would like to know whether this would be the right approach. Also, any tips on my formatting would be a great help. I need help with spacing the equations.
algebra-precalculus
edited Nov 27 '18 at 16:08
Tianlalu
3,09621038
asked Nov 27 '18 at 15:38
01110000_01110000
75
|
show 1 more comment
Questions:
a) Show that $$(sqrt{2} + 1)^2 = 3+2sqrt2$$
b) Express $$cos3theta$$in the form of$$costheta(a-bsin^2theta)$$
c) Solve $$sqrt2(sin2theta+costheta) + cos3theta = sin2theta + cos2theta$$ $$Theta in [0,360]$$
Answers:
a) $$(sqrt{2} + 1)^2 = 2 + 1 + 2sqrt2 = 3+2sqrt2$$
b)
begin{align*}
cos3theta &= cos(2theta + theta)\
&=cos2thetacostheta - sin2thetasintheta\
&=(cos^2theta-sin^2theta)(costheta) -2sinthetacosthetasintheta\
&=costheta(cos^2theta-sin^2theta-2sin^2theta)\
&=costheta(1-4sin^2theta)
end{align*}
c) I'm still working on this one, this was my approach
begin{align*}
sqrt2(sin2theta+costheta)+cos3theta &= sin2theta + cos2theta\
impliessqrt2(2sinthetacostheta+costheta)+costheta(1-4sin^2theta)&= 2sinthetacostheta + 1 -2sin^2theta\
sqrt2costheta(2sintheta + 1)+costheta(1-4sin^2theta) &= 2sinthetacostheta + 2cos^2theta - 1
end{align*}
Before I proceed further, I would like to know whether this would be the right approach. Also, any tips on my formatting would be a great help. I need help with spacing the equations.
algebra-precalculus
edited Nov 27 '18 at 16:08
Tianlalu
3,09621038
asked Nov 27 '18 at 15:38
01110000_01110000
75
To me it seems okay.
– Akash Roy
Nov 27 '18 at 15:43
1
Is "$cos3theta$ in the form of $costheta(a-bsintheta)$"correct? You got "$=costheta(1-4sin^2theta)$" where we have a square.
– Robert Z
Nov 27 '18 at 15:47
@RobertZ Sorry, I have made a mistake in the question, I shall edit it.
– 01110000_01110000
Nov 27 '18 at 15:51
1
For the formatting, usebegin{align*} [LaTeX here, with lines separated by \ and & placed where you want to line things up]end{align*}For example,begin{align*}a &= b \&= cend{align*}produces: begin{align*}a &= b \&= cend{align*}
– user3482749
Nov 27 '18 at 15:52
@user3482749 Thanks!
– 01110000_01110000
Nov 27 '18 at 16:00
|
show 1 more comment
Questions:
a) Show that $$(sqrt{2} + 1)^2 = 3+2sqrt2$$
b) Express $$cos3theta$$in the form of$$costheta(a-bsin^2theta)$$
c) Solve $$sqrt2(sin2theta+costheta) + cos3theta = sin2theta + cos2theta$$ $$Theta in [0,360]$$
Answers:
a) $$(sqrt{2} + 1)^2 = 2 + 1 + 2sqrt2 = 3+2sqrt2$$
b)
begin{align*}
cos3theta &= cos(2theta + theta)\
&=cos2thetacostheta - sin2thetasintheta\
&=(cos^2theta-sin^2theta)(costheta) -2sinthetacosthetasintheta\
&=costheta(cos^2theta-sin^2theta-2sin^2theta)\
&=costheta(1-4sin^2theta)
end{align*}
c) I'm still working on this one, this was my approach
begin{align*}
sqrt2(sin2theta+costheta)+cos3theta &= sin2theta + cos2theta\
impliessqrt2(2sinthetacostheta+costheta)+costheta(1-4sin^2theta)&= 2sinthetacostheta + 1 -2sin^2theta\
sqrt2costheta(2sintheta + 1)+costheta(1-4sin^2theta) &= 2sinthetacostheta + 2cos^2theta - 1
end{align*}
Before I proceed further, I would like to know whether this would be the right approach. Also, any tips on my formatting would be a great help. I need help with spacing the equations.
algebra-precalculus
edited Nov 27 '18 at 16:08
Tianlalu
3,09621038
asked Nov 27 '18 at 15:38
01110000_01110000
75
Questions:
a) Show that $$(sqrt{2} + 1)^2 = 3+2sqrt2$$
b) Express $$cos3theta$$in the form of$$costheta(a-bsin^2theta)$$
c) Solve $$sqrt2(sin2theta+costheta) + cos3theta = sin2theta + cos2theta$$ $$Theta in [0,360]$$
Answers:
a) $$(sqrt{2} + 1)^2 = 2 + 1 + 2sqrt2 = 3+2sqrt2$$
b)
begin{align*}
cos3theta &= cos(2theta + theta)\
&=cos2thetacostheta - sin2thetasintheta\
&=(cos^2theta-sin^2theta)(costheta) -2sinthetacosthetasintheta\
&=costheta(cos^2theta-sin^2theta-2sin^2theta)\
&=costheta(1-4sin^2theta)
end{align*}
c) I'm still working on this one, this was my approach
begin{align*}
sqrt2(sin2theta+costheta)+cos3theta &= sin2theta + cos2theta\
impliessqrt2(2sinthetacostheta+costheta)+costheta(1-4sin^2theta)&= 2sinthetacostheta + 1 -2sin^2theta\
sqrt2costheta(2sintheta + 1)+costheta(1-4sin^2theta) &= 2sinthetacostheta + 2cos^2theta - 1
end{align*}
Before I proceed further, I would like to know whether this would be the right approach. Also, any tips on my formatting would be a great help. I need help with spacing the equations.
algebra-precalculus
algebra-precalculus
edited Nov 27 '18 at 16:08
Tianlalu
3,09621038
asked Nov 27 '18 at 15:38
01110000_01110000
75
edited Nov 27 '18 at 16:08
Tianlalu
3,09621038
asked Nov 27 '18 at 15:38
01110000_01110000
75
edited Nov 27 '18 at 16:08
Tianlalu
3,09621038
edited Nov 27 '18 at 16:08
Tianlalu
3,09621038
edited Nov 27 '18 at 16:08
Tianlalu
3,09621038
3,09621038
asked Nov 27 '18 at 15:38
01110000_01110000
75
asked Nov 27 '18 at 15:38
01110000_01110000
75
asked Nov 27 '18 at 15:38
01110000_01110000
75
75
To me it seems okay.
– Akash Roy
Nov 27 '18 at 15:43
1
Is "$cos3theta$ in the form of $costheta(a-bsintheta)$"correct? You got "$=costheta(1-4sin^2theta)$" where we have a square.
– Robert Z
Nov 27 '18 at 15:47
@RobertZ Sorry, I have made a mistake in the question, I shall edit it.
– 01110000_01110000
Nov 27 '18 at 15:51
1
For the formatting, usebegin{align*} [LaTeX here, with lines separated by \ and & placed where you want to line things up]end{align*}For example,begin{align*}a &= b \&= cend{align*}produces: begin{align*}a &= b \&= cend{align*}
– user3482749
Nov 27 '18 at 15:52
@user3482749 Thanks!
– 01110000_01110000
Nov 27 '18 at 16:00
|
show 1 more comment
To me it seems okay.
– Akash Roy
Nov 27 '18 at 15:43
1
Is "$cos3theta$ in the form of $costheta(a-bsintheta)$"correct? You got "$=costheta(1-4sin^2theta)$" where we have a square.
– Robert Z
Nov 27 '18 at 15:47
@RobertZ Sorry, I have made a mistake in the question, I shall edit it.
– 01110000_01110000
Nov 27 '18 at 15:51
1
For the formatting, usebegin{align*} [LaTeX here, with lines separated by \ and & placed where you want to line things up]end{align*}For example,begin{align*}a &= b \&= cend{align*}produces: begin{align*}a &= b \&= cend{align*}
– user3482749
Nov 27 '18 at 15:52
@user3482749 Thanks!
– 01110000_01110000
Nov 27 '18 at 16:00
To me it seems okay.
– Akash Roy
Nov 27 '18 at 15:43
To me it seems okay.
– Akash Roy
Nov 27 '18 at 15:43
1
1
Is "$cos3theta$ in the form of $costheta(a-bsintheta)$"correct? You got "$=costheta(1-4sin^2theta)$" where we have a square.
– Robert Z
Nov 27 '18 at 15:47
Is "$cos3theta$ in the form of $costheta(a-bsintheta)$"correct? You got "$=costheta(1-4sin^2theta)$" where we have a square.
– Robert Z
Nov 27 '18 at 15:47
@RobertZ Sorry, I have made a mistake in the question, I shall edit it.
– 01110000_01110000
Nov 27 '18 at 15:51
@RobertZ Sorry, I have made a mistake in the question, I shall edit it.
– 01110000_01110000
Nov 27 '18 at 15:51
1
1
For the formatting, use
begin{align*} [LaTeX here, with lines separated by \ and & placed where you want to line things up]end{align*} For example, begin{align*}a &= b \&= cend{align*} produces: begin{align*}a &= b \&= cend{align*}– user3482749
Nov 27 '18 at 15:52
For the formatting, use
begin{align*} [LaTeX here, with lines separated by \ and & placed where you want to line things up]end{align*} For example, begin{align*}a &= b \&= cend{align*} produces: begin{align*}a &= b \&= cend{align*}– user3482749
Nov 27 '18 at 15:52
@user3482749 Thanks!
– 01110000_01110000
Nov 27 '18 at 16:00
@user3482749 Thanks!
– 01110000_01110000
Nov 27 '18 at 16:00
|
show 1 more comment
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To me it seems okay.
– Akash Roy
Nov 27 '18 at 15:43
1
Is "$cos3theta$ in the form of $costheta(a-bsintheta)$"correct? You got "$=costheta(1-4sin^2theta)$" where we have a square.
– Robert Z
Nov 27 '18 at 15:47
@RobertZ Sorry, I have made a mistake in the question, I shall edit it.
– 01110000_01110000
Nov 27 '18 at 15:51
1
For the formatting, use
begin{align*} [LaTeX here, with lines separated by \ and & placed where you want to line things up]end{align*}For example,begin{align*}a &= b \&= cend{align*}produces: begin{align*}a &= b \&= cend{align*}– user3482749
Nov 27 '18 at 15:52
@user3482749 Thanks!
– 01110000_01110000
Nov 27 '18 at 16:00