IB Maths AA HL Reciprocal and Inverse Trig Functions Questions

1. Rewrite the reciprocal trigonometric functions in terms of the primary trigonometric functions: $\frac{1}{\cos (\pi /3)}+\frac{1}{\tan (\pi /4)}$.

2. Evaluate the exact value of the primary functions using the unit circle: $\cos \left(\frac{\pi }{3}\right)=\frac{1}{2}$ and $\tan \left(\frac{\pi }{4}\right)=1$.

3. Substitute these values back into the fractions: $\frac{1}{1/2}+\frac{1}{1}$.

4. Simplify the fractions: $2+1$.

5. State the final exact sum: $\mathbf{3}$.

1. Identify the relevant Pythagorean identity linking secant and tangent from the formula booklet: $1+{\tan }^{2}x={\sec }^{2}x$.

2. Substitute this identity into the equation to create a uniform trigonometric function: $(1+{\tan }^{2}x)+2\tan x=4$.

3. Rearrange the terms to form a quadratic equation in terms of $\tan x$: ${\tan }^{2}x+2\tan x-3=0$.

4. Factorise the quadratic expression: $(\tan x+3)(\tan x-1)=0$, giving $\tan x=1$ or $\tan x=-3$.

5. Solve the first case: $\tan x=1⟹x=\frac{\pi }{4}\approx 0.785$ radians.

6. Solve the second case using your GDC: $\arctan (-3)=-1.249...$ Since this is outside the domain $[0,\pi ]$, add the period of tangent ($\pi$) to find the valid second-quadrant solution: $x=-1.249...+\pi =1.892...$ Final answers: $\mathbf{x}\mathbf{=}\mathbf{0.785}\mathbf{,}\mathbf{x}\mathbf{=}\mathbf{1.89}$.

1. Assign a variable to the inner inverse trigonometric function to simplify the expression: Let $\theta =\arccos \left(\frac{2}{3}\right)$.

2. Rewrite this relationship using primary trigonometry: $\cos \theta =\frac{2}{3}$. Because the range of arccosine is $[0,\pi ]$ and the value is positive, $\theta$ must be an acute angle in Quadrant 1.

3. Use the Pythagorean identity to find $\sin \theta$: $\sin \theta =\sqrt{1-{\cos }^2\theta }=\sqrt{1-{\left(\frac{2}{3}\right)}^2}=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3}$.

4. Substitute $\theta$ back into the original expression, which now reads $\sin (2\theta )$.

5. Apply the double angle identity for sine: $\sin (2\theta )=2\sin \theta \cos \theta$.

6. Calculate the final exact value by substituting the known fractions: $2\left(\frac{\sqrt{5}}{3}\right)\left(\frac{2}{3}\right)=\frac{\mathbf{4}\sqrt{\mathbf{5}}}{\mathbf{9}}$.