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 $sin x=frac13cos x$ & $cos x=frac13sin x$ .
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 bởi wrong? Thanks.
Bạn đang xem: Prove the following identities sin^6x + cos^6x = 1
asked Jun 20, 2013 at 19:31
22.2k1111 gold badges7575 silver badges150150 bronze badges
add a comment |
3 Answers 3
Sorted by: Reset to default
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
showroom a bình luận |
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 bình luận |
$sin^6x + cos^6x = (sin^2x)^3 + (cos^2x)^3 =(sin^2x + cos^2x)(sin^4x + cos^4x -sin^2xcos^2x)$
$sin^4x+cos^4x -sin^2xcos^2x = (sin^2x + cos^2x)^2 - 2sin^2xcos^2x -sin^2xcos^2x$
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
6,63233 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 phản hồi |
Thanks for contributing an answer to lớn nasaconstellation.comematics Stack Exchange!Please be sure to 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.
Use nasaconstellation.comJax lớn format equations. nasaconstellation.comJax reference.
To learn more, see our tips on writing great answers.
Xem thêm: Tài Liệu Giáo Dục Kỹ Năng Sống Cho Học Sinh Tiểu Học Sinh Tiểu Học Chọn Lọc
Sign up or log in
Sign up using Google
Sign up using Facebook
Sign up using thư điện tử and Password
Post as a guest
e-mail Required, but never shown
Post as a guest
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$, và $sin( heta) > 0$, 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?
Whats wrong in this approach khổng lồ evaluate $int_0^fracpi2 fracsin xcos xdxsin x+cos x$
Using de Moivre to lớn 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
Hot Network Questions more hot questions
Subscribe to lớn RSS
Question feed lớn subscribe khổng lồ this RSS feed, copy and paste this URL into your RSS reader.
Stack Exchange Network
site thiết kế / biệu tượng công ty © 2022 Stack Exchange Inc; user contributions licensed under cc by-sa. Rev2022.4.1.41833