3.4.7 Trigonometry (A-level only)

i.

Show that

\frac{1}{\tan \theta} - \tan \theta \equiv \frac{\cos 2\theta}{\sin \theta \cos \theta}

for $θ≠nπ2\theta \neq \frac{n\pi}{2}$ where $\in \mathbb{Z}$ .

[3]

ii.

Solve, for $0∘≤x<90∘0^\circ \le x < 90^\circ$ , the equation

\sin^2(2x - 15^\circ) = 2

giving your answers in degrees to one decimal place. (Solutions based entirely on graphical or numerical methods are not acceptable.)

[6]

i.

Show that

\frac{1}{\tan \theta} - \tan \theta \equiv \frac{\cos 2\theta}{\sin \theta \cos \theta}

for $θ≠nπ2\theta \neq \frac{n\pi}{2}$ where $\in \mathbb{Z}$ .

[3]

ii.

Solve, for $0∘≤x<90∘0^\circ \le x < 90^\circ$ , the equation

\sin^2(2x - 15^\circ) = 2

giving your answers in degrees to one decimal place. (Solutions based entirely on graphical or numerical methods are not acceptable.)

[6]

3.4.7 Trigonometry (A-level only) Questions