2. Calculate the following without the use of a calculator: (a) \( \cos 75^{\circ} \quad \) (b) \( \sin 15^{\circ} \quad \) (c) \( \cos 165^{\circ} \) Triangle ABC is given. (a) Determine angle A in terms of B and C . (b) Hence, show that: \( \sin \mathrm{A}=\sin \mathrm{B} \cdot \cos \mathrm{C}+\cos \mathrm{B} \cdot \sin \mathrm{C} \) Show that if \( \sin \left(\theta+60^{\circ}\right)=2 \sin \theta \) then \( \tan \theta=\frac{\sqrt{3}}{3} \)

### (a) Calculation of Trigonometric Values without a Calculator #### (i) \( \cos 75^\circ \) Express \( 75^\circ \) as \( 45^\circ+30^\circ \). Then use the cosine sum formula: \[ \cos(75^\circ) = \cos(45^\circ+30^\circ) = \cos45^\circ\cos30^\circ - \sin45^\circ\sin30^\circ. \] Substitute the well‐known values: \[ \cos45^\circ=\frac{\sqrt{2}}{2},\quad \cos30^\circ=\frac{\sqrt{3}}{2},\quad \sin45^\circ=\frac{\sqrt{2}}{2},\quad \sin30^\circ=\frac{1}{2}. \] Thus, \[ \cos75^\circ = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2}\cdot\frac{1}{2} = \frac{\sqrt{6}-\sqrt{2}}{4}. \] #### (ii) \( \sin 15^\circ \) Express \( 15^\circ \) as \( 45^\circ-30^\circ \) and apply the sine difference formula: \[ \sin(15^\circ) = \sin(45^\circ-30^\circ) = \sin45^\circ\cos30^\circ - \cos45^\circ\sin30^\circ. \] Substitute the values: \[ \sin15^\circ = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2}\cdot\frac{1}{2} = \frac{\sqrt{6}-\sqrt{2}}{4}. \] #### (iii) \( \cos 165^\circ \) Write \( 165^\circ \) as \( 180^\circ-15^\circ \) to use the identity for cosine in the second quadrant: \[ \cos165^\circ = \cos(180^\circ-15^\circ) = -\cos15^\circ. \] Now find \( \cos15^\circ \) using the cosine difference formula: \[ \cos15^\circ = \cos(45^\circ-30^\circ) = \cos45^\circ\cos30^\circ + \sin45^\circ\sin30^\circ. \] Substitute the values: \[ \cos15^\circ = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}\cdot\frac{1}{2} = \frac{\sqrt{6}+\sqrt{2}}{4}. \] Thus, \[ \cos165^\circ = -\frac{\sqrt{6}+\sqrt{2}}{4}. \] --- ### (b) Triangle ABC – Angle Relationships #### (i) Expressing Angle \( \mathrm{A} \) in Terms of \( \mathrm{B} \) and \( \mathrm{C} \) In any triangle, the sum of the angles is \( 180^\circ \). Therefore, \[ \mathrm{A} + \mathrm{B} + \mathrm{C} = 180^\circ. \] Solving for \( \mathrm{A} \): \[ \mathrm{A} = 180^\circ - (\mathrm{B}+\mathrm{C}). \] #### (ii) Showing \( \sin\mathrm{A}=\sin\mathrm{B}\cos\mathrm{C}+\cos\mathrm{B}\sin\mathrm{C} \) Since \( \mathrm{A} = 180^\circ - (\mathrm{B}+\mathrm{C}) \), we have: \[ \sin\mathrm{A} = \sin\big(180^\circ - (\mathrm{B}+\mathrm{C})\big). \] Using the identity \( \sin(180^\circ-x)=\sin x \) gives: \[ \sin\mathrm{A} = \sin(\mathrm{B}+\mathrm{C}). \] Apply the sine addition formula: \[ \sin(\mathrm{B}+\mathrm{C}) = \sin\mathrm{B}\cos\mathrm{C} + \cos\mathrm{B}\sin\mathrm{C}. \] Thus, \[ \sin\mathrm{A} = \sin\mathrm{B}\cos\mathrm{C}+\cos\mathrm{B}\sin\mathrm{C}. \] --- ### (c) Solving \( \sin(\theta+60^\circ)=2\sin\theta \) for \( \tan\theta \) Start with the given equation: \[ \sin(\theta+60^\circ)=2\sin\theta. \] Expand the left-hand side using the sine addition formula: \[ \sin\theta\cos60^\circ + \cos\theta\sin60^\circ = 2\sin\theta. \] Substitute the known values \( \cos60^\circ=\frac{1}{2} \) and \( \sin60^\circ=\frac{\sqrt{3}}{2} \): \[ \sin\theta\cdot\frac{1}{2} + \cos\theta\cdot\frac{\sqrt{3}}{2} = 2\sin\theta. \] Multiply the entire equation by 2 to clear the denominators: \[ \sin\theta+\sqrt{3}\cos\theta=4\sin\theta. \] Rearrange the equation: \[ \sqrt{3}\cos\theta = 4\sin\theta - \sin\theta = 3\sin\theta. \] Divide both sides by \( \cos\theta \) (assuming \( \cos\theta\ne 0 \)): \[ \sqrt{3} = 3\tan\theta. \] Solve for \( \tan\theta \): \[ \tan\theta = \frac{\sqrt{3}}{3}. \]