Solve the following algebraic equations
up vote
1
down vote
favorite
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 at 16:08
Tianlalu
3,01021038
asked Nov 27 at 15:38
01110000_01110000
75
|
show 1 more comment
up vote
1
down vote
favorite
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 at 16:08
Tianlalu
3,01021038
asked Nov 27 at 15:38
01110000_01110000
75
To me it seems okay.
– Akash Roy
Nov 27 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 at 15:47
@RobertZ Sorry, I have made a mistake in the question, I shall edit it.
– 01110000_01110000
Nov 27 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 at 15:52
@user3482749 Thanks!
– 01110000_01110000
Nov 27 at 16:00
|
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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 at 16:08
Tianlalu
3,01021038
asked Nov 27 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 at 16:08
Tianlalu
3,01021038
asked Nov 27 at 15:38
01110000_01110000
75
edited Nov 27 at 16:08
Tianlalu
3,01021038
asked Nov 27 at 15:38
01110000_01110000
75
edited Nov 27 at 16:08
Tianlalu
3,01021038
edited Nov 27 at 16:08
Tianlalu
3,01021038
edited Nov 27 at 16:08
Tianlalu
3,01021038
3,01021038
asked Nov 27 at 15:38
01110000_01110000
75
asked Nov 27 at 15:38
01110000_01110000
75
asked Nov 27 at 15:38
01110000_01110000
75
75
To me it seems okay.
– Akash Roy
Nov 27 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 at 15:47
@RobertZ Sorry, I have made a mistake in the question, I shall edit it.
– 01110000_01110000
Nov 27 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 at 15:52
@user3482749 Thanks!
– 01110000_01110000
Nov 27 at 16:00
|
show 1 more comment
To me it seems okay.
– Akash Roy
Nov 27 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 at 15:47
@RobertZ Sorry, I have made a mistake in the question, I shall edit it.
– 01110000_01110000
Nov 27 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 at 15:52
@user3482749 Thanks!
– 01110000_01110000
Nov 27 at 16:00
To me it seems okay.
– Akash Roy
Nov 27 at 15:43
To me it seems okay.
– Akash Roy
Nov 27 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 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 at 15:47
@RobertZ Sorry, I have made a mistake in the question, I shall edit it.
– 01110000_01110000
Nov 27 at 15:51
@RobertZ Sorry, I have made a mistake in the question, I shall edit it.
– 01110000_01110000
Nov 27 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 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 at 15:52
@user3482749 Thanks!
– 01110000_01110000
Nov 27 at 16:00
@user3482749 Thanks!
– 01110000_01110000
Nov 27 at 16:00
|
show 1 more comment
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3015920%2fsolve-the-following-algebraic-equations%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3015920%2fsolve-the-following-algebraic-equations%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
To me it seems okay.
– Akash Roy
Nov 27 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 at 15:47
@RobertZ Sorry, I have made a mistake in the question, I shall edit it.
– 01110000_01110000
Nov 27 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 at 15:52
@user3482749 Thanks!
– 01110000_01110000
Nov 27 at 16:00