Prove $\sin(\pi/2-x)\cot(x+\pi/2) = -\sin(x)$

Question

I've managed to use cofunction identities to get the left side of the equation to (pi/2-x)-tanx. From here, I keep ending up at -cotx? I'm pretty sure that's wrong, and I have no idea how to to go about proving this in another way.

Answer 1

Hint: Show that $$\sin\left(\frac{\pi}{2} -x\right) = \cos(x)$$ and $$\cot\left(x +\frac{\pi}{2}\right) = -\tan(x)$$

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Both of the above can be shown using $$ \sin(x +y) = \sin(x) \cos(y) + \cos(x) \sin(y) $$ and $$ \cos(x +y) = \cos(x) \cos(y) - \sin(x) \sin(y) $$ combined with $\cot(x) = \frac{\cos(x)}{\sin(x)}$ and $\tan(x) = \frac{\sin(x)}{\cos(x)}$.

Answer 2

Take LHS First , Sin(π/2 - x) = cosx Second , cot(π/2+ x) = -tanx because in 2nd quadrant cot is negative so when cot changes to tanx we add extra negative sign.

Now Put -tanx = - sinx / cosx and cancel out cos x from numerator and denominator hence you got -six in LHS which is equal to RHS.

Answer 3

Here's a hint,

Try using the following identities \begin{align*} \cot(x) &= \frac{\cos(x)}{\sin(x)}\\ \sin(-x) &= -\sin(x)\\ \cos(-x) &= \cos(x)\\ \sin(x+\pi) &= -\sin(x)\\ \cos(x+\pi) &= -\cos(x) \end{align*}

Expand the LHS with the first one, and then try to change some terms (and maybe get them to cancel) by using the phase shift identities

Answer 4

Just muck it out.

The following are basic identities.

$\sin(\frac \pi 2-x) =\cos x$

$\sin(x+\frac \pi 2) = \cos x$

$\cos(x+\frac \pi 2) = \sin x$.

We can know this by:

1. Rote memory

2. Considering that by considering angles $\frac \pi 2 -x$ or $x +\frac \pi 2$ is simply a matter of rotating a unit circle a quarter turn and possibly reflecting the circle on an axis and thus the $x,y$ cooridinates will "switch". (It helps to draw a picture.

3. Considering $\sin (A+ B) = \sin A\cos B + \cos A \sin B$ and $\cos (A+B) =\cos A\cos B -\sin A\sin B$ and so

• $\sin (\frac \pi 2 - x) = \sin \frac \pi 2\cos x + \cos \frac \pi 2 \sin(-x)=1\cdot \cos x + 0\cdot (-\sin x) = \cos x$.
• $\sin ( x+ \frac \pi 2) = \sin x\cos \frac \pi 2 + \cos x \sin(\frac \pi 2)=\sin x \cdot 0 + \cos x\cdot 1 = \cos x$
• $\cos(x+ \frac \pi 2) = \cos x\cos\frac \pi 2 - \sin x \sin \frac \pi 2=\cos x \cdot 0 -\sin x \cdot 1=-\sin x$.

ANd as a result of these three identities.

$\sin(\frac \pi -x)\cot (x+\frac \pi 2) = $

$\sin(\frac \pi -x)\frac {\cos (x+\frac \pi 2)}{\sin (x+\frac \pi 2)}=$

$\cos x\frac {-\sin x}{\cos x}= $

$-\sin x$.

That is ..... IF $\cos x\ne 0$. Or in otherwords if $x \ne \pm\frac \pi 2$.

If $x =\pm \frac \pi 2$ then $\cot (x+\frac \pi 2)$ is not defined.