Thread: Proving cofunction identity using subtraction formula.

1. Proving cofunction identity using subtraction formula.

Got this question in textbook assignments:

Prove the cofunction identity using the addition and subtraction formulas.

20. cot (pi/2 - u) = tan u

so to my knowledge, I turn it into:

1 / tan (pi/2 - u) and try to apply subtraction formula but then tan pi/2 isn't defined is it? because 1/0 isn't defined. Is this the case?

[EDIT] I've also run into this now and don't want to spam the forums so I will post here, prove:

sin(pi/2 - x) = sin(pi/2 + x)

I start with RHS = sin(pi/2) cos x + cos (pi/2) sin x
= (1) cos x + (0) sin x
= cos x

Haha this is probly also simple but alas it eludes me...

2. Hello, DannyMath!

Your reasoning is correct ... We can't work it that way.

Prove the cofunction identity using the addition and subtraction formulas.

. . $20.\;\;\cot\left(\tfrac{\pi}{2}- u\right) \:=\: \tan u$
If you don't know the addition/subtraction formula for cotangent, we can re-invent it.

$\cot(A \pm B) \;=\;\frac{1}{\tan(A+B)} \;=\;\frac{1\mp\tan A\tan B}{\tan A \pm \tan B} \;=\; \frac{1 \mp \dfrac{1}{\cot A}\,\dfrac{1}{\cot B}}{\dfrac{1}{\cot A} \pm\dfrac{1}{\cot B}}$

Multiply by $\frac{\cot A\cot B}{\cot A\cot B}\!:\quad\boxed{\cot(A \pm B) \;=\;\frac{\cot A\cot A \mp 1}{\cot B \pm \cot A}}$

We have: . $\cot\left(\tfrac{\pi}{2} - u\right) \;=\;\frac{\cot\frac{\pi}{2}\cot u + 1}{\cot u - \cot\frac{\pi}{2}} \;=\;\frac{0\cdot\cot u + 1}{\cot u - 0} \;=\;\frac{1}{\cot u} \;=\;\tan u$

3. >

4. Ok I hesitate to ask but I understand everything until after

$

\frac{1 \mp \dfrac{1}{\cot A}\,\dfrac{1}{\cot B}}{\dfrac{1}{\cot A} \pm\dfrac{1}{\cot B}}
$

The steps between the above and

$

\boxed{\cot(A \pm B) \;=\;\frac{\cot A\cot B \mp 1}{\cot B \pm \cot A}}
$

have me stumped. I tried for an hour to work it out on paper and googled it. Does it have something to do with multiplying by its reciprocal?

5. Originally Posted by DannyMath
Ok I hesitate to ask but I understand everything until after

$

\frac{1 \mp \dfrac{1}{\cot A}\,\dfrac{1}{\cot B}}{\dfrac{1}{\cot A} \pm\dfrac{1}{\cot B}}
$
Multiply both numerator and denominator by cot(A) cot(B) to get
$\frac{cot(A)cot(B)\mp 1}{cot(B)+ cot(A)}$.

The steps between the above and

$

\boxed{\cot(A \pm B) \;=\;\frac{\cot A\cot B \mp 1}{\cot B \pm \cot A}}
$

have me stumped. I tried for an hour to work it out on paper and googled it. Does it have something to do with multiplying by its reciprocal?

6. Ok I think I understand now. I was getting confused cause I know that when you multiply something by its reciprocal you get 1, but I'm starting to realize that's not what we're doing in this case haha. Thanks guys!