Identity Proofs

**Higg** · November 15th 2012, 12:05 AM

Hi everyone!

I just have an identity proof that I am stuggling with.

Identity Proofs - Simple Question Help-lesson3hw_q2k.gif

Could anyone give me a hand?

Thanks!

**MarkFL** · November 15th 2012, 12:16 AM

That's not an identity.

**Higg** · November 15th 2012, 12:30 AM

No wonder it was so hard. Didn't even copy the question correctly haha. Thanks!

**Prove It** · November 15th 2012, 05:43 AM

Originally Posted by Higg

Hi everyone!

I just have an identity proof that I am stuggling with.

Click image for larger version.

Name: lesson3hw_Q2k.gif
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ID: 25724

Could anyone give me a hand?

Thanks!

It's not an identity, and there are not even any real solutions for x.

$\displaystyle \begin{align*} \tan^4{x} - \sec^4{x} &= 2\sin^2{x} \\ \left( \tan^2{x} - \sec^2{x} \right)\left( \tan^2{x} + \sec^2{x} \right) &= 2\sin^2{x} \\ -1\left( \tan^2{x} + \sec^2{x} \right) &= 2\sin^2{x} \\ -\left( \frac{\sin^2{x}}{\cos^2{x}} + \frac{1}{\cos^2{x}} \right) &= 2\sin^2{x} \\ -\left( \frac{\sin^2{x} + 1 }{\cos^2{x}} \right) &= 2\sin^2{x} \\ \frac{-\left( \sin^2{x} + 1 \right) }{1 - \sin^2{x}} &= 2\sin^2{x} \\ -\left( \sin^2{x} + 1 \right) &= 2\sin^2{x} \left( 1 - \sin^2{x} \right) \\ -\sin^2{x} - 1 &= 2\sin^2{x} - 2\sin^4{x} \\ 2\sin^4{x} - 3\sin^2{x} - 1 &= 0 \\ \sin^2{x} &= \frac{3 \pm \sqrt{(-3)^2 - 4(2)(-1)}}{2(2)} \\ \sin^2{x} &= \frac{3 \pm \sqrt{13}}{4} \\ \sin^2{x} &= \frac{3 + \sqrt{13}}{4} \textrm{ if we are assuming } x \textrm{ is real} \\ \sin{x} &= \pm \frac{\sqrt{{3 + \sqrt{13}} }}{2} \end{align*}$

Since $\displaystyle \begin{align*} \left| \frac{\sqrt{3 + \sqrt{13}}}{2} \right| > 1 \end{align*}$ , there are not any values of x which satisfy this equation.

Of course, complex numbers are another story...

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