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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?

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Or is the expectation for these types of problems to just do them on the calculator ?

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    $\begingroup$ 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). $\endgroup$ – Clayton Jan 18 at 18:42
  • $\begingroup$ Any time you have the sine or cosine of an angle, that can be expressed as sine of something in [0,pi/4). $\endgroup$ – Acccumulation Jan 19 at 7:23
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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.

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

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

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

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

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

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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.

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