what have I done wrong? "if [(sinx)^4]/2 + [(cosx)^4]/3 = 1/5 then prove that (tanx)^2=2/3"

[ALIAHMAD]

if [(sinx)^4]/2 + [(cosx)^4]/3 = 1/5

then
prove that (tanx)^2=2/3


I tried solving this as follows

taking LCM as 6

{3[(sinx)^4] + 2[(cosx)^4]}/6=1/5

{3[(sinx)^4] + 2[(cosx)^4]}=6/5

now using sin^n+cos^n=1 property

(sinx)^4 + 2=6/5

(sinx)^4=-4/5

but -ve should not come there as sin is raised to 4th power

what I did wrong ?

 

[topsquark]

if [(sinx)^4]/2 + [(cosx)^4]/3 = 1/5

then
prove that (tanx)^2=2/3


I tried solving this as follows

taking LCM as 6

{3[(sinx)^4] + 2[(cosx)^4]}/6=1/5

{3[(sinx)^4] + 2[(cosx)^4]}=6/5

now using sin^n+cos^n=1 property

(sinx)^4 + 2=6/5

(sinx)^4=-4/5

but -ve should not come there as sin is raised to 4th power

what I did wrong ?

[imath]sin^n(x) + cos^n(x) = 1[/imath] only when n = 2.

-Dan

 

[Steven G]

{3[(sinx)^4] + 2[(cosx)^4]}/6=1/5

{3[(sinx)^4] + 2[(cosx)^4]}=6/5

How can the same expression equal both 1/5 and 6/5??

 

[topsquark]

How can the same expression equal both 1/5 and 6/5??

The bottom equation is 6 times the top equation.

-Dan

 

[MarkFL]

Dividing through by the 4th power of the cosine function, multiplying by 30 and using a Pythagorean identity, you may write:

\(\displaystyle 15\tan^4(x)+10=6\left(\tan^2(x)+1\right)^2\)

Arrange this as a quadratic in the square of the tangent function...

 

[Dr.Peterson]

Alternatively, you can replace [imath]\cos^2(x)[/imath] with [imath]1-\sin^2(x)[/imath] and end up with a quadratic in [imath]\sin(x)[/imath], which can be nicely factored.

 

[NotJeffM]

The first major thing I do (when possible) in this kind of problem is to state things in terms of ONE trigonometric function, but here I would clear fractions as a minor clean-up step even before that. (I hate fractions.)

[math] \dfrac{\sin^4(x)}{2} + \dfrac{\cos^4(x)}{3} = \dfrac{1}{5} \implies \\ 15 \sin^4(x) + 10 \cos^4(x) = 6 \implies \\ 15 \sin^4(x) + 10 \{ \cos^2(x) \}^2 = 6 \implies \\ 15 \sin^4(x) + 10 \{ 1 - \sin^2(x) \}^2 = 6 \implies \\ 15 \sin^4(x) + 10 - 20 \sin^2(x) + 10 \sin^4(x) = 6 \implies \\ 25 \sin^4(x) - 20\sin^2(x) + 4 = 0. [/math]
Now that can be dealt with directly, but some people may find a u-substitution helpful.

[math] \text {Set } u = \sin^2(x) \implies 25u^2 - 20u + 4 = 0 \implies (5u - 2)(5u - 2) = 0 \implies \\ u = \dfrac{2}{5} \implies \sin^2(x) = \dfrac{2}{5} \implies \cos^2(x) = 1 - \dfrac{2}{5} = \dfrac{3}{5}. [/math]
Before proceeding any further, I’d check to see whether those answers are correct. Be careful.

The rest of the proof is trivial.

I do not say that all similar problems can conveniently be reduced to problems in just one trigonometric function, but those that can be so reduced are usually made much simpler. Also, u-substitutions are never necessary, but they often give clarity.

 

[MarkFL]

Dividing through by the 4th power of the cosine function, multiplying by 30 and using a Pythagorean identity, you may write:

\(\displaystyle 15\tan^4(x)+10=6\left(\tan^2(x)+1\right)^2\)

Arrange this as a quadratic in the square of the tangent function...

\(\displaystyle 15\tan^4(x)+10=6\left(\tan^4(x)+2\tan^2(x)+1\right)\)

\(\displaystyle 15\tan^4(x)+10=6\tan^4(x)+12\tan^2(x)+6\)

\(\displaystyle 9\tan^4(x)-12\tan^2(x)+4=0\)

\(\displaystyle \left(3\tan^2(x)-2\right)^2=0\)

\(\displaystyle \tan^2(x)=\frac{2}{3}\)