Prove that sin^6x + cos^6x = 1
First, notice that $(sin x +cos x)^2=sin^2 x+cos^2 x+sin 2x=1+ frac 23=frac53$ .
Now, from what was given we have $sin x=frac13cos x$ và $cos x=frac13sin x$ .
Next, $(sin^2 x+cos^2 x)^3=1=sin^6 x+cos^6 x+3sin^2 x cos x+3cos^2 x sin x$ .
Now we substitute what we found above from the given:
$sin^6 x+cos^6+sin x +cos x=1$
$sin^6 x+cos^6=1-(sin x +cos x)$
$sin^6 x+cos^6=1-sqrt frac 53$
Not only is this not positive, but this is not even a rational number. What did I vì chưng wrong? Thanks.
Bạn đang xem: Prove that sin^6x + cos^6x = 1
algebra-precalculus trigonometry
cốt truyện
Cite
Follow
asked Jun 20, 2013 at 19:31
OviOvi
22.4k1111 gold badges7777 silver badges151151 bronze badges
$endgroup$
5
showroom a bình luận |
3 Answers 3
Sorted by: Reset to default
Highest score (default) Date modified (newest first) Date created (oldest first)
10
$egingroup$
$(sin^2 x + cos^2 x)^3=sin^6 x + cos^6 x + 3sin^2 x cos^2 x$
share
Cite
Follow
answered Jun 20, 2013 at 19:36
user67803user67803
$endgroup$
showroom a bình luận |
5
$egingroup$
Should be $(sin^2 x+cos^2 x)^3=1=sin^6 x+cos^6 x+3sin^4 x cos^2 x+3cos^4 x sin^2 x$
tóm tắt
Cite
Follow
answered Jun 20, 2013 at 19:35
MaazulMaazul
2,4681616 silver badges2626 bronze badges
$endgroup$
địa chỉ a phản hồi |
2
$egingroup$
$sin^6x + cos^6x = (sin^2x)^3 + (cos^2x)^3 =(sin^2x + cos^2x)(sin^4x + cos^4x -sin^2xcos^2x)$
$sin^4x+cos^4x -sin^2xcos^2x = (sin^2x + cos^2x)^2 - 2sin^2xcos^2x -sin^2xcos^2x$
or $1-3sin^2xcos^2x = 1-3left(dfrac13 ight)^2 = dfrac23$.
chia sẻ
Cite
Follow
edited Oct 2, 2013 at 17:43
answered Jul 21, 2013 at 8:31
ShobhitShobhit
6,66233 gold badges3333 silver badges5858 bronze badges
$endgroup$
3
$egingroup$ It has to lớn be $1 - 1 / 3$, not $1 - (1 / 3) ^ 2$, the answer is $2 / 3$. $endgroup$
–user98213
Oct 2, 2013 at 10:54
add a phản hồi |
Your Answer
Thanks for contributing an answer khổng lồ goodsmart.com.vnematics Stack Exchange!
Please be sure to lớn answer the question. Provide details and share your research!But avoid …
Asking for help, clarification, or responding to lớn other answers.Making statements based on opinion; back them up with references or personal experience.Xem thêm: Tả Cảnh Buổi Sáng Ở Công Viên Vào Buổi Sáng Sớm (20 Mẫu), Tả Cảnh Công Viên Vào Buổi Sáng Sớm (20 Mẫu)
Use goodsmart.com.vnJax lớn format equations. goodsmart.com.vnJax reference.
To learn more, see our tips on writing great answers.
Xem thêm: Học Ngay Cách Làm Thịt Bò Ngâm Nước Mắm Thơm Ngon, Đậm Đà Dịp Tết
Draft saved
Draft discarded
Sign up or log in
Sign up using Google
Sign up using Facebook
Sign up using e-mail and Password
Submit
Post as a guest
Name
e-mail Required, but never shown
Post as a guest
Name
Required, but never shown
Post Your Answer Discard
By clicking “Post Your Answer”, you agree khổng lồ our terms of service, privacy policy và cookie policy
Not the answer you're looking for? Browse other questions tagged algebra-precalculus trigonometry or ask your own question.
Featured on Meta
Linked
13
How can we show $cos^6x+sin^6x=1-3sin^2x cos^2x$?
Related
6
Solve for $sin^2(x) = 3cos^2(x)$
1
What does $sin( heta) > 0$ mean here? If $ an( heta) = -frac815$, and $sin( heta) > 0$, then find $cos( heta)$.
2
Proving trigonometric identity $1+cot x an y=fracsin(x+y)sin xcos y$
0
Evaluate cos<(1/2)
1
"If $|sin x + cos x |=|sin x|+|cos x| (sin x, cos x eq 0)$, in which quadrant does $x$ lie?"
2
What is wrong in my answer? Subject: finding the integral of $cot x$
3
Maximizing $3sin^2 x + 8sin xcos x + 9cos^2 x$. What went wrong?
0
Using de Moivre khổng lồ solve $z^3=-1$, one solution is $cos pi+i sin pi$. What am I doing wrong?
1
Finding $sin 2x$ from transforming $sin^4 x+ cos^4 x = frac79$ using trigonometric identities
1
Why is $frac11+(frac-sin x1+cos x)^2 equiv frac(1+cos x)^2sin^2x+1+2cos x+cos^2x$?
Hot Network Questions more hot questions
Question feed
Subscribe to RSS
Question feed lớn subscribe to lớn this RSS feed, copy and paste this URL into your RSS reader.
goodsmart.com.vnematics
Company
Stack Exchange Network
Site kiến thiết / logo sản phẩm © 2022 Stack Exchange Inc; user contributions licensed under cc by-sa. Rev2022.6.10.42345
Your privacy
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
