Question:
Use the symmetric and periodic properties of the sine, cosine and tangent functions to establish the following results.

$\displaystyle (a) \sin (90-\theta)^o = \cos \theta^o$
$\displaystyle (b) \sin (270 + \theta)^o = -\cos \theta^o$
$\displaystyle (c) \tan (\theta - 180)^o = \tan \theta^o$

Attempt:
I don't know how to start

Hello,

For the first one, there are two ways imho :
- sketch a trigonometrical circle and you will see that this relation is true
- use the formula sin(a-b)=sin(a)cos(b)-sin(b)cos(a)

For the second one, use the periodicity :
- the sine function is 360-periodic. Hence $\displaystyle sin(270+\theta)=sin((270+\theta)-360)=sin(\theta-90)$

- As the sine function is an odd function, sin(-x)=-sin(x)
- $\displaystyle sin(\theta-90)=-sin(90-\theta)$ and you retrieve the first relation