# 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 ?

• I think both can be solved through the use of the sum/difference angle formulas and you’ll also need to use the half angle formula in $(f)$ (at least, the method I used required the half angle formula). – Clayton Jan 18 at 18:42
• Any time you have the sine or cosine of an angle, that can be expressed as sine of something in [0,pi/4). – Acccumulation Jan 19 at 7:23

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.

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 $$\cos x =\sin(90^{\circ} - x)$$

$$\cos \dfrac{23\pi}{4} = \cos \left( 6\pi - \dfrac{\pi}{4} \right) = \cos \left(- \dfrac{\pi}{4} \right) = \dfrac{1}{\sqrt 2}$$

$$\sin \dfrac{15\pi}{4} = \sin \left( 4\pi - \dfrac{\pi}{4} \right) = \sin\left(- \dfrac{\pi}{4} \right) = -\dfrac{1}{\sqrt 2}$$

$$\cos \dfrac{23\pi}{4} - \sin \dfrac{15\pi}{4} = \sqrt 2$$

$$\dfrac{\cos 35^\circ}{\sin 20^\circ \cos 35^\circ + \cos 20^\circ \sin 35^\circ} =\dfrac{\cos 35^\circ}{\sin 55^\circ} = \dfrac{\cos 35^\circ}{\cos (90^\circ - 55^\circ)} = 1$$

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.