trig

**tangerine_tomato** · June 13th 2008, 10:52 PM

given tanΘ = 2, use trigometric identities to find the exact value of cscΘ(90°-Θ)

thank you

**Chris L T521** · June 13th 2008, 11:28 PM

Originally Posted by tangerine_tomato

given tanΘ = 2, use trigometric identities to find the exact value of cscΘ(90°-Θ)

thank you

You can observe that $\csc(90^o-\vartheta)=\sec(\vartheta)$ . But if you didn't know that, how would you go about solving this problem?

Note 3 things:

$\sec(\varphi)=\frac{1}{\cos(\varphi)}$
$\csc(\varphi)=\frac{1}{\sin(\varphi)}$ and $\sin(\varphi-\vartheta)=\sin(\varphi)\cos(\vartheta)-\cos(\varphi)\sin(\vartheta)$

Thus,

$\csc(90^o-\vartheta)=\frac{1}{\sin(90^o-\vartheta)}=\frac{1}{\sin(90^o)\cos(\vartheta)-\cos(90^o)\sin(\vartheta)}$

Simplifying, we get:

$\csc(90^o-\vartheta)=\frac{1}{\cos(\vartheta)}=\sec(\varthet a)$

Since $\tan(\vartheta)=2$ , we can conclude that $\cos(\vartheta)=\frac{1}{\sqrt{5}}$

Thus, $\csc(90^o-\vartheta)=\sec(\vartheta)=\color{red}\boxed{\sqrt {5}}$

Does this make sense?

--Chris

**Mathstud28** · June 13th 2008, 11:30 PM

Originally Posted by Chris L T521

You can observe that $\csc(90^o-\vartheta)=\sec(\vartheta)$ . But if you didn't know that, how would you go about solving this problem?

Note 3 things:

$\sec(\varphi)=\frac{1}{\cos(\varphi)}$
$\csc(\varphi)=\frac{1}{\sin(\varphi)}$ and $\sin(\varphi-\vartheta)=\sin(\varphi)\cos(\vartheta)-\cos(\varphi)\sin(\vartheta)$

Thus,

$\csc(90^o-\vartheta)=\frac{1}{\sin(90^o-\vartheta)}=\frac{1}{\sin(90^o)\cos(\vartheta)-\cos(90^o)\sin(\vartheta)}$

Simplifying, we get:

$\csc(90^o-\vartheta)=\frac{1}{\cos(\vartheta)}=\sec(\varthet a)$

Since $\tan(\vartheta)=2$ , we can conclude that $\cos(\vartheta)=\frac{1}{\sqrt{5}}$

Thus, $\csc(90^o-\vartheta)=\sec(\vartheta)=\color{red}\boxed{\sqrt {5}}$

Does this make sense?

--Chris

Two "var"s in one post...you..you've made me proud

**tangerine_tomato** · June 14th 2008, 02:57 AM

Originally Posted by Chris L T521

You can observe that $\csc(90^o-\vartheta)=\sec(\vartheta)$ . But if you didn't know that, how would you go about solving this problem?

Note 3 things:

$\sec(\varphi)=\frac{1}{\cos(\varphi)}$
$\csc(\varphi)=\frac{1}{\sin(\varphi)}$ and $\sin(\varphi-\vartheta)=\sin(\varphi)\cos(\vartheta)-\cos(\varphi)\sin(\vartheta)$

Thus,

$\csc(90^o-\vartheta)=\frac{1}{\sin(90^o-\vartheta)}=\frac{1}{\sin(90^o)\cos(\vartheta)-\cos(90^o)\sin(\vartheta)}$

Simplifying, we get:

$\csc(90^o-\vartheta)=\frac{1}{\cos(\vartheta)}=\sec(\varthet a)$

Since $\tan(\vartheta)=2$ , we can conclude that $\cos(\vartheta)=\frac{1}{\sqrt{5}}$

Thus, $\csc(90^o-\vartheta)=\sec(\vartheta)=\color{red}\boxed{\sqrt {5}}$

Does this make sense?

--Chris

i dont understand that stuff in the 3rd line after the word thus

is that cause of the complementary angle theorem

and i dont understand anything after that

but im too tired ATM to make much effort to understand it

**mr fantastic** · June 14th 2008, 03:22 AM

Originally Posted by tangerine_tomato

i dont understand that stuff in the 3rd line after the word thus

is that cause of the complementary angle theorem

and i dont understand anything after that

but im too tired ATM to make much effort to understand it

It all boils down to finding $\cos \theta$ from $\tan \theta = 2$ .

Half the solution is given above. The other half is ${\color{red}-}\sqrt{5}$ .

**Chris L T521** · June 14th 2008, 10:08 AM

Originally Posted by mr fantastic

It all boils down to finding $\cos \theta$ from $\tan \theta = 2$ .

Half the solution is given above. The other half is ${\color{red}-}\sqrt{5}$ .

Shoot...darn negative sign...

...wait...how would we know for sure it was in the first quadrant instead of the third quadrant or vice versa? I didn't see any restriction on the angle...

EDIT: Just realized what he meant...never mind...

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