Thread: Trig Identities - Double Angles

1. Trig Identities - Double Angles

Prove the following using double angle formulas:

$\frac{sin(\pi+x)}{cos(2\pi+x)} + \frac{sec(\pi-x)}{csc(2\pi-x)} = 0$

-----

My Attempt:

$
LS = \frac{sin\pi cosx + cos\pi sinx}{cos2\pi cosx - sin2\pi sinx} + \frac{1}{cos\pi cosx + sin\pi sinx}$

$= \frac {-sinx}{cosx} + \frac{1}{-cosx}\times(\frac{-cosx}{1})$

$= \frac {-sinx}{cosx} + \frac {-cosx}{-cosx}$

$= \frac {(-sinx)(-cosx) + (-cosx)(cosx)}{(cosx)(-cosx)}$

2. Your
[sec(pi -x) / csc(2pi -x)] = 1 / [cos(pi)cos(x) +sin(pi)sin(x)]
is not correct.

[sec(pi -x) / csc(2pi -x)]
= [1 / cos(pi -x)] / [1 / sin(2pi -x)]
= sin(2pi -x) / cos(2pi -x)

So,
[sin(pi +x) / cos(2pi +x)] +[sec(pi -x) / csc(2pi -x)] = 0

LHS =
= [sin(pi +x) / cos(2pi +x)] +[sec(pi -x) / csc(2pi -x)]

= [sin(pi +x) / cos(2pi +x)] +[sin(2pi -x) / cos(pi -x)]

= [{sin(pi)cos(x) +cos(pi)sin(x)} / {cos(2pi)cos(x) -sin(2pi)sin(x)}]
+ [{sin(2pi)cos(x) -cos(2pi)sin(x)} / {cos(pi)cos(x) +sin(pi)sin(x)]

Now,
sin(pi) = 0
cos(pi) = -1
sin(2pi) = 0
cos(2pi) = 1

= [{0*cos(x) +(-1)sin(x)} / {(1)cos(x) -0*sin(x)}]
+ [{0*cos(x) -(1)sin(x)} / {(-1)cos(x) +0*sin(x)]

= [-sin(x) / cos(x)] +[-sin(x) / -cos(x)]

= -tan(x) +tan(x)

= 0

= RHS

Therefore, proven.

3. Hello, Macleef!

Prove: . $\frac{\sin(\pi+x)}{\cos(2\pi+x)} + \frac{\sec(\pi-x)}{\csc(2\pi-x)} = 0$

The first fraction is: . $\frac{\sin(\pi+x)}{\cos(2\pi+x)} \;=\;\frac{\sin(\pi)\cos(x) + \sin(x)\cos(\pi)}{\cos(2\pi)\cos(x) - \sin(2\pi)\sin(x)}$

. . $= \;\frac{0\cdot\cos(x) + \sin(x)\cdot(-1)}{1\cdot\cos(x) - 0\cdot\sin(x)} \;=\; \frac{-\sin(x)}{\cos(x)} \;=\;-\tan(x)$

The second fraction is: . $\frac{\sin(2\pi-x)}{\cos(\pi-x)} \;=\;\frac{\sin(2\pi)\cos(x) - \sin(x)\cos(2\pi)}{\cos(\pi)\cos(x) + \sin(\pi)\sin(x)}$

. . $= \;\frac{0\cdot\cos(x) - \sin(\pi)\cdot1}{(-1)\cos(x) + 0\cdot\sin(x)} \;=\;\frac{-\sin(x)}{-\cos(x)} \;=\;\tan(x)$

The problem becomes: . $-\tan(x) + \tan(x) \;=\;0$