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using trigonometric identities

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2

$begingroup$

For proving $$frac 16cos (4x)+7 =frac1sin^4x +cos^2x +frac1sin^2x +cos^4x $$

I tried to use that:

beginalign
sin^4 x +cos^4 x&=sin^4 x +2sin^2xcos^2 x+cos^4 x - 2sin^2xcos^2 x\
&=(sin^2x+cos^2 x)^2-2sin^2xcos^2 x\
&=1^2-frac12(2sin xcos x)^2\
&=1-frac12sin^2 (2x)\
&=1-frac12left(frac1-cos 4x2right)\
&=frac34+frac14cos 4x
endalign

but i can't try more

trigonometry

share|cite|improve this question

asked May 17 at 17:58

GeorgeGeorge

3246

$endgroup$

add a comment | 

2

$begingroup$

For proving $$frac 16cos (4x)+7 =frac1sin^4x +cos^2x +frac1sin^2x +cos^4x $$

I tried to use that:

beginalign
sin^4 x +cos^4 x&=sin^4 x +2sin^2xcos^2 x+cos^4 x - 2sin^2xcos^2 x\
&=(sin^2x+cos^2 x)^2-2sin^2xcos^2 x\
&=1^2-frac12(2sin xcos x)^2\
&=1-frac12sin^2 (2x)\
&=1-frac12left(frac1-cos 4x2right)\
&=frac34+frac14cos 4x
endalign

but i can't try more

trigonometry

share|cite|improve this question

asked May 17 at 17:58

GeorgeGeorge

3246

$endgroup$

add a comment | 

$begingroup$

For proving $$frac 16cos (4x)+7 =frac1sin^4x +cos^2x +frac1sin^2x +cos^4x $$

I tried to use that:

beginalign
sin^4 x +cos^4 x&=sin^4 x +2sin^2xcos^2 x+cos^4 x - 2sin^2xcos^2 x\
&=(sin^2x+cos^2 x)^2-2sin^2xcos^2 x\
&=1^2-frac12(2sin xcos x)^2\
&=1-frac12sin^2 (2x)\
&=1-frac12left(frac1-cos 4x2right)\
&=frac34+frac14cos 4x
endalign

but i can't try more

trigonometry

share|cite|improve this question

asked May 17 at 17:58

GeorgeGeorge

3246

$endgroup$

share|cite|improve this question

asked May 17 at 17:58

GeorgeGeorge

3246

share|cite|improve this question

asked May 17 at 17:58

GeorgeGeorge

3246

asked May 17 at 17:58

GeorgeGeorge

3246

asked May 17 at 17:58

GeorgeGeorge

3246

3 Answers
3

active

oldest

votes

7

$begingroup$

$displaystyle sin^4x+cos^2x=frac(1-cos2x)^24+frac1+cos 2x2=frac3+cos^22x4=frac3+frac1+cos 4x24=frac7+cos4x8$

$displaystyle sin^2x+cos^4x=sin^4left(frac pi2-xright)+cos^2left(frac pi2-xright)=frac7+cos4left(frac pi2-xright)8=frac7+cos4x8$

share|cite|improve this answer

answered May 17 at 18:03

CY AriesCY Aries

19.3k11844

$endgroup$

add a comment | 

5

$begingroup$

Note that
$$beginalignsin^4x +cos^2x &=
sin^2x(1-cos^2x) +cos^2x\
&=sin^2x+cos^2x(1-sin^2x)=sin^2x +cos^4x.
endalign$$
Hence, according to your work,
$$beginalign
2(sin^4x +cos^2x)&=(sin^4x +cos^2x) +(sin^2x +cos^4x)\
&=sin^4x +cos^4x+1=frac7+cos 4x4.endalign$$
Can you take it from here?

share|cite|improve this answer

edited May 18 at 7:41

answered May 17 at 18:04

Robert ZRobert Z

103k1073146

$endgroup$

add a comment | 

0

$begingroup$

$c^4-s^4=(c^2+s^2)(c^2-s^2)=c^2-s^2$

$iff c^4+s^2=c^2+s^4=P$(say) where $c=cos x,s=sin x$

$P+P=1+c^4+s^4=1+(c^2+s^2)^2-2c^2s^2=2-dfracsin^22x2=2-dfrac1-cos4x4=?$

using $sin2x=2sin xcos x,cos2y=2-2sin^2y$

We need $$dfrac1P+dfrac2P=?$$

share|cite|improve this answer

answered May 17 at 18:20

lab bhattacharjeelab bhattacharjee

231k15161283

$endgroup$

add a comment | 

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7

$begingroup$

$displaystyle sin^4x+cos^2x=frac(1-cos2x)^24+frac1+cos 2x2=frac3+cos^22x4=frac3+frac1+cos 4x24=frac7+cos4x8$

$displaystyle sin^2x+cos^4x=sin^4left(frac pi2-xright)+cos^2left(frac pi2-xright)=frac7+cos4left(frac pi2-xright)8=frac7+cos4x8$

share|cite|improve this answer

answered May 17 at 18:03

CY AriesCY Aries

19.3k11844

$endgroup$

add a comment | 

7

$begingroup$

$displaystyle sin^4x+cos^2x=frac(1-cos2x)^24+frac1+cos 2x2=frac3+cos^22x4=frac3+frac1+cos 4x24=frac7+cos4x8$

$displaystyle sin^2x+cos^4x=sin^4left(frac pi2-xright)+cos^2left(frac pi2-xright)=frac7+cos4left(frac pi2-xright)8=frac7+cos4x8$

share|cite|improve this answer

answered May 17 at 18:03

CY AriesCY Aries

19.3k11844

$endgroup$

add a comment | 

$begingroup$

$displaystyle sin^4x+cos^2x=frac(1-cos2x)^24+frac1+cos 2x2=frac3+cos^22x4=frac3+frac1+cos 4x24=frac7+cos4x8$

$displaystyle sin^2x+cos^4x=sin^4left(frac pi2-xright)+cos^2left(frac pi2-xright)=frac7+cos4left(frac pi2-xright)8=frac7+cos4x8$

share|cite|improve this answer

answered May 17 at 18:03

CY AriesCY Aries

19.3k11844

$endgroup$

share|cite|improve this answer

answered May 17 at 18:03

CY AriesCY Aries

19.3k11844

answered May 17 at 18:03

CY AriesCY Aries

19.3k11844

answered May 17 at 18:03

CY AriesCY Aries

19.3k11844

5

$begingroup$

Note that
$$beginalignsin^4x +cos^2x &=
sin^2x(1-cos^2x) +cos^2x\
&=sin^2x+cos^2x(1-sin^2x)=sin^2x +cos^4x.
endalign$$
Hence, according to your work,
$$beginalign
2(sin^4x +cos^2x)&=(sin^4x +cos^2x) +(sin^2x +cos^4x)\
&=sin^4x +cos^4x+1=frac7+cos 4x4.endalign$$
Can you take it from here?

share|cite|improve this answer

edited May 18 at 7:41

answered May 17 at 18:04

Robert ZRobert Z

103k1073146

$endgroup$

add a comment | 

5

$begingroup$

Note that
$$beginalignsin^4x +cos^2x &=
sin^2x(1-cos^2x) +cos^2x\
&=sin^2x+cos^2x(1-sin^2x)=sin^2x +cos^4x.
endalign$$
Hence, according to your work,
$$beginalign
2(sin^4x +cos^2x)&=(sin^4x +cos^2x) +(sin^2x +cos^4x)\
&=sin^4x +cos^4x+1=frac7+cos 4x4.endalign$$
Can you take it from here?

share|cite|improve this answer

edited May 18 at 7:41

answered May 17 at 18:04

Robert ZRobert Z

103k1073146

$endgroup$

add a comment | 

$begingroup$

Note that
$$beginalignsin^4x +cos^2x &=
sin^2x(1-cos^2x) +cos^2x\
&=sin^2x+cos^2x(1-sin^2x)=sin^2x +cos^4x.
endalign$$
Hence, according to your work,
$$beginalign
2(sin^4x +cos^2x)&=(sin^4x +cos^2x) +(sin^2x +cos^4x)\
&=sin^4x +cos^4x+1=frac7+cos 4x4.endalign$$
Can you take it from here?

share|cite|improve this answer

edited May 18 at 7:41

answered May 17 at 18:04

Robert ZRobert Z

103k1073146

$endgroup$

share|cite|improve this answer

edited May 18 at 7:41

answered May 17 at 18:04

Robert ZRobert Z

103k1073146

answered May 17 at 18:04

Robert ZRobert Z

103k1073146

answered May 17 at 18:04

Robert ZRobert Z

103k1073146

0

$begingroup$

$c^4-s^4=(c^2+s^2)(c^2-s^2)=c^2-s^2$

$iff c^4+s^2=c^2+s^4=P$(say) where $c=cos x,s=sin x$

$P+P=1+c^4+s^4=1+(c^2+s^2)^2-2c^2s^2=2-dfracsin^22x2=2-dfrac1-cos4x4=?$

using $sin2x=2sin xcos x,cos2y=2-2sin^2y$

We need $$dfrac1P+dfrac2P=?$$

share|cite|improve this answer

answered May 17 at 18:20

lab bhattacharjeelab bhattacharjee

231k15161283

$endgroup$

add a comment | 

0

$begingroup$

$c^4-s^4=(c^2+s^2)(c^2-s^2)=c^2-s^2$

$iff c^4+s^2=c^2+s^4=P$(say) where $c=cos x,s=sin x$

$P+P=1+c^4+s^4=1+(c^2+s^2)^2-2c^2s^2=2-dfracsin^22x2=2-dfrac1-cos4x4=?$

using $sin2x=2sin xcos x,cos2y=2-2sin^2y$

We need $$dfrac1P+dfrac2P=?$$

share|cite|improve this answer

answered May 17 at 18:20

lab bhattacharjeelab bhattacharjee

231k15161283

$endgroup$

add a comment | 

$begingroup$

$c^4-s^4=(c^2+s^2)(c^2-s^2)=c^2-s^2$

$iff c^4+s^2=c^2+s^4=P$(say) where $c=cos x,s=sin x$

$P+P=1+c^4+s^4=1+(c^2+s^2)^2-2c^2s^2=2-dfracsin^22x2=2-dfrac1-cos4x4=?$

using $sin2x=2sin xcos x,cos2y=2-2sin^2y$

We need $$dfrac1P+dfrac2P=?$$

share|cite|improve this answer

answered May 17 at 18:20

lab bhattacharjeelab bhattacharjee

231k15161283

$endgroup$

share|cite|improve this answer

answered May 17 at 18:20

lab bhattacharjeelab bhattacharjee

231k15161283

answered May 17 at 18:20

lab bhattacharjeelab bhattacharjee

231k15161283

answered May 17 at 18:20

lab bhattacharjeelab bhattacharjee

231k15161283