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 $sin x=frac13cos x$ and $cos x=frac13sin x$ .
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 wrong? Thanks.
Bạn đang xem: Prove that sin^6 + cos^6 = 1
asked Jun 20, 2013 at 19:31
22.3k1111 gold badges7676 silver badges151151 bronze badges
showroom a bình luận |
3 Answers 3
Sorted by: Reset to mặc định
Highest score (default) Date modified (newest first) Date created (oldest first)
$(sin^2 x + cos^2 x)^3=sin^6 x + cos^6 x + 3sin^2 x cos^2 x$
answered Jun 20, 2013 at 19:36
địa chỉ a comment |
Should be $(sin^2 x+cos^2 x)^3=1=sin^6 x+cos^6 x+3sin^4 x cos^2 x+3cos^4 x sin^2 x$
answered Jun 20, 2013 at 19:35
2,4581515 silver badges2626 bronze badges
địa chỉ a comment |
$sin^6x + cos^6x = (sin^2x)^3 + (cos^2x)^3 =(sin^2x + cos^2x)(sin^4x + cos^4x -sin^2xcos^2x)$
$sin^4x+cos^4x -sin^2xcos^2x = (sin^2x + cos^2x)^2 - 2sin^2xcos^2x -sin^2xcos^2x$
or $1-3sin^2xcos^2x = 1-3left(dfrac13 ight)^2 = dfrac23$.
edited Oct 2, 2013 at 17:43
answered Jul 21, 2013 at 8:31
6,64233 gold badges3232 silver badges5858 bronze badges
$egingroup$ It has lớn be $1 - 1 / 3$, not $1 - (1 / 3) ^ 2$, the answer is $2 / 3$. $endgroup$
Oct 2, 2013 at 10:54
địa chỉ a bình luận |
Thanks for contributing an answer khổng lồ slovenija-expo2000.comematics Stack Exchange!Please be sure to answer the question. Provide details & share your research!
But avoid …Asking for help, clarification, or responding lớn other answers.Making statements based on opinion; back them up with references or personal experience.
Use slovenija-expo2000.comJax lớn format equations. slovenija-expo2000.comJax reference.
To learn more, see our tips on writing great answers.
Xem thêm: 5 Cách Đổi Font Chữ Mặc Định Của Facebook Là Gì, “Thay Da Đổi Thịt” Cho Font Chữ Facebook
Sign up or log in
Sign up using Google
Sign up using Facebook
Sign up using e-mail and Password
Post as a guest
thư điện tử Required, but never shown
Post as a guest
thư điện tử
Required, but never shown
Post Your Answer Discard
Not the answer you're looking for? Browse other questions tagged algebra-precalculus trigonometry or ask your own question.
Featured on Meta
How can we show $cos^6x+sin^6x=1-3sin^2x cos^2x$?
Solve for $sin^2(x) = 3cos^2(x)$
What does $sin( heta) > 0$ mean here? If $ an( heta) = -frac815$, & $sin( heta) > 0$, then find $cos( heta)$.
Proving trigonometric identity $1+cot x an y=fracsin(x+y)sin xcos y$
"If $|sin x + cos x |=|sin x|+|cos x| (sin x, cos x eq 0)$, in which quadrant does $x$ lie?"
What is wrong in my answer? Subject: finding the integral of $cot x$
Maximizing $3sin^2 x + 8sin xcos x + 9cos^2 x$. What went wrong?
Using de Moivre to solve $z^3=-1$, one solution is $cos pi+i sin pi$. What am I doing wrong?
Finding $sin 2x$ from transforming $sin^4 x+ cos^4 x = frac79$ using trigonometric identities
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
Subscribe to lớn RSS
Question feed to subscribe khổng lồ this RSS feed, copy và paste this URL into your RSS reader.
Stack Exchange Network
Site thiết kế / hình ảnh © 2022 Stack Exchange Inc; user contributions licensed under cc by-sa. Rev2022.4.21.42004