Possible To Do More Complex Trignometry Problems By Hand? Are there straightforward ways to do the following two problems by hand instead of using a calculator?  Any shortcuts (i.e. the denominator of question 5?  

Or is the expectation for these types of problems to just do them on the calculator ?  
 A: Hint:
For e): Use the fact that $2\pi $ is a period for $\sin $ and $\cos$.
For f):
Use addition theorem for $\sin $:
$$\sin (x+y) = \sin x\cos y +\sin y \cos x$$
and $$\cos x =\sin(90^{\circ} - x)$$ 
A: No. You should use trig  circular function formulas for simplification
HINT
e)
Note that difference in argument is $2\pi$ so that periodic trig simplification is suggested
f)
The denominator is $\sin$ addition  formula for $\sin 55^{\circ}$ Recognize with complementary angle $ =\cos 35^{\circ}$ which is same as numerator, so evaluates to unity.
A: $\cos \dfrac{23\pi}{4} 
   = \cos \left( 6\pi - \dfrac{\pi}{4} \right)
   = \cos \left(- \dfrac{\pi}{4} \right)
   = \dfrac{1}{\sqrt 2}$
$\sin \dfrac{15\pi}{4} 
   = \sin \left( 4\pi - \dfrac{\pi}{4} \right)
   = \sin\left(- \dfrac{\pi}{4} \right)
   = -\dfrac{1}{\sqrt 2}$
$$\cos \dfrac{23\pi}{4} - \sin \dfrac{15\pi}{4} = \sqrt 2$$
$$\dfrac{\cos 35^\circ}{\sin 20^\circ \cos 35^\circ + \cos 20^\circ \sin 35^\circ}
  =\dfrac{\cos 35^\circ}{\sin 55^\circ}  
  = \dfrac{\cos 35^\circ}{\cos (90^\circ - 55^\circ)}
  = 1$$
A: For all $x$ we have $\sin (-x)=-\sin x$ and $\cos (-x)=\cos x.$
For all $x$ and for all $n\in \Bbb Z$ we have $\cos (x+n\pi)=(-1)^n\cos x$ and $\sin (x+n\pi)=(-1)^n\sin x.$
For all $x$ we have $\sin (\pi/2-x)=\cos x$ and $\cos (\pi/2-x)=\sin x.$ 
For all $x,y$ we have $\sin x \cos y+\sin y\cos x=\sin (x+y)$ and $\cos x \cos y-\sin x \sin y=\cos (x+y).$
To help remember which of the LHS in the formulas in the above line has a $+$ and which has a $-,$ think of $\sin (x+y)$  increasing   and $\cos (x+y)$  decreasing, as a function of $x+y,$ when $x+y$ is small but positive.
