# Another question... the value of sin x

• Feb 4th 2014, 08:04 AM
Deci
Another question... the value of sin x
Another question...

If $cosec^2 x = -\sqrt{7} sec^2 x$

then the value of sin x is...

the correct ans is $\frac{\sqrt{7}-2 \sqrt{2}}{2}$

But why I calculate and found $\frac{\sqrt{7}- \sqrt{11}}{2}$ instead?
• Feb 4th 2014, 08:38 AM
romsek
Re: Another question... the value of sin x
Quote:

Originally Posted by Deci
Another question...

If $cosec^2 x = -\sqrt{7} sec^2 x$

then the value of sin x is...

the correct ans is $\frac{\sqrt{7}-2 \sqrt{2}}{2}$

But why I calculate and found $\frac{\sqrt{7}- \sqrt{11}}{2}$ instead?

Attachment 30120

Neither of those answers is correct. Assuming that by cosec(x) you mean cosecant(x) = 1/sin(x). It's usually abbreviated csc(x).

What I get is

$\csc^2(x)=-\sqrt{7}\sec^2(x)$

$\sin^2(x)=-\frac{\sqrt{7}}{7}\cos^2(x)$

$\sin^2(x)=-\frac{\sqrt{7}}{7}\left(1-\sin^2(x)\right)$

$\sin^2(x)\left(1-\frac{\sqrt{7}}{7}\right)=-\frac{\sqrt{7}}{7}$

$\sin^2(x)\left(\frac{7-\sqrt{7}}{7}\right)=-\frac{\sqrt{7}}{7}$

$\sin^2(x)=\frac{\sqrt{7}}{\sqrt{7}-7}}=-\frac{1+\sqrt{7}}{6}$

$\sin(x)=\pm \sqrt{-\frac{1+\sqrt{7}}{6}}$

Now look at that last term on the right. It's the square root of a negative number. Over the Real numbers this is not defined. Is this equation meant to be over the Complex numbers? If so the last line is your solution. If not, there is no solution.
• Feb 4th 2014, 10:01 AM
Soroban
Re: Another question... the value of sin x
Hello, Deci!

There must be a typo.
The problem has no real solution.

Quote:

$\text{If }\,\csc^2\!x \:=\:-\sqrt{7}\sec^2\!x,\,\text{ find the value of }\sin x.$

$\csc^2\!x\:=\:-\sqrt{7}\sec^2\!x \quad\Rightarrow\quad \frac{1}{\sin^2\!x} \:=\:-\frac{\sqrt{7}}{\cos^2\!x} \quad\Rightarrow\quad -\frac{1}{\sqrt{7}} \:=\:\frac{\sin^2\!x}{\cos^2\!x}$

$\text{But }\,\tan^2\!x \:=\:-\frac{1}{\sqrt{7}}\,\text{ has no real roots.}$