Addition formulae questions with answers

Questions

1. Question 1Exact values

   Without using a calculator, find the exact value of $\cos 105^\circ$.

   Show the worked solution

   1. Step 1: Write the angle as a sum or difference of angles whose exact values you know$105^\circ = 60^\circ + 45^\circ$
   2. Step 2: Which addition formula do you use?$\cos(A + B) = \cos A\cos B - \sin A\sin B$
   3. Step 3: Write down the exact values you need$\sin 60^\circ = \dfrac{\sqrt{3}}{2}$$\sin 45^\circ = \dfrac{\sqrt{2}}{2}$$\cos 60^\circ = \dfrac{1}{2}$$\cos 45^\circ = \dfrac{\sqrt{2}}{2}$
   4. Step 4: Put them into the formula, then simplify$\cos 105^\circ = \cos 60^\circ\cos 45^\circ - \sin 60^\circ\sin 45^\circ$$= \left(\dfrac{1}{2}\right)\left(\dfrac{\sqrt{2}}{2}\right) - \left(\dfrac{\sqrt{3}}{2}\right)\left(\dfrac{\sqrt{2}}{2}\right)$$= -\dfrac{\sqrt{6} - \sqrt{2}}{4}$
2. Question 2From given ratios

   Given that $\sin A = \dfrac{12}{13}$ and $\cos B = -\dfrac{15}{17}$, where $A$ is acute and $B$ is obtuse, find the exact value of $\cos(A - B)$.

   Show the worked solution

   1. Step 1: Find cos A from sin²A + cos²A = 1 (or a triangle). Is cos A positive or negative?$\cos A = \dfrac{5}{13}$
   2. Step 2: Find sin B the same way. Is sin B positive or negative?$\sin B = \dfrac{8}{17}$
   3. Step 3: Use the addition formula: put the values in, then simplify$\cos(A - B) = \cos A\cos B + \sin A\sin B$$\cos(A - B) = \left(\dfrac{5}{13}\right)\left(\dfrac{-15}{17}\right) + \left(\dfrac{12}{13}\right)\left(\dfrac{8}{17}\right)$$= \dfrac{21}{221}$
3. Question 3Solving equations

   Solve $\sin(x + 45^\circ) = 0.5\cos x$ for $0^\circ \leq x \leq 360^\circ$.

   Show the worked solution

   1. Step 1: Expand the left side with the addition formula$\sin(x + 45^\circ) = \sin x\cos 45^\circ + \cos x\sin 45^\circ$$\cos 45^\circ = \dfrac{\sqrt{2}}{2}$$\sin 45^\circ = \dfrac{\sqrt{2}}{2}$
   2. Step 2: Collect the sin x and cos x terms, then divide by cos x (sin x ÷ cos x = tan x)$\dfrac{\sqrt{2}}{2}\sin x = \left(0.5 - \dfrac{\sqrt{2}}{2}\right)\cos x$$\tan x = -0.2929\ldots$
   3. Step 3: Basic angle β: use the positive value$\beta = \tan^{-1}\left(0.2929\ldots\right)$$\beta = 16.3^\circ$
   4. Step 4: ASTC (All, Sine, Tan, Cos): in which quadrants does tan x have this sign?$\tan x$ is negative in quadrants 2 and 4.
   5. Step 5: List every angle x in the range, quadrant by quadrant$x = 180^\circ - 16.3^\circ$$= 163.7^\circ$$x = 360^\circ - 16.3^\circ$$= 343.7^\circ$
4. Question 4Exact values in radians (π/12)

   Without using a calculator, find the exact value of $\cos \dfrac{11\pi}{12}$.

   Show the worked solution

   1. Step 1: Write the angle as a sum or difference of angles whose exact values you know$\dfrac{11\pi}{12} = \dfrac{2\pi}{3} + \dfrac{\pi}{4}$
   2. Step 2: Which addition formula do you use?$\cos(A + B) = \cos A\cos B - \sin A\sin B$
   3. Step 3: Write down the exact values you need$\sin \dfrac{2\pi}{3} = \dfrac{\sqrt{3}}{2}$$\sin \dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2}$$\cos \dfrac{2\pi}{3} = -\dfrac{1}{2}$$\cos \dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2}$
   4. Step 4: Put them into the formula, then simplify$\cos \dfrac{11\pi}{12} = \left(-\dfrac{1}{2}\right)\left(\dfrac{\sqrt{2}}{2}\right) - \left(\dfrac{\sqrt{3}}{2}\right)\left(\dfrac{\sqrt{2}}{2}\right)$$= -\dfrac{\sqrt{6} + \sqrt{2}}{4}$
5. Question 5sec, cosec, cot of 15°: rationalise

   Without using a calculator, find the exact value of $\cot \dfrac{\pi}{12}$, in simplest surd form with a rational denominator.

   Show the worked solution

   1. Step 1: First find the exact value of the ratio it is the reciprocal of, with an addition formula$\tan \dfrac{\pi}{12} = 2 - \sqrt{3}$
   2. Step 2: Take the reciprocal$\cot \dfrac{\pi}{12} = \dfrac{1}{2 - \sqrt{3}}$
   3. Step 3: Rationalise: multiply top and bottom by the conjugate surd$= \dfrac{(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})}$$= \dfrac{(2 + \sqrt{3})}{4 - 3}$$= 2 + \sqrt{3}$