1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1

\includegraphics{figure_6} The diagram shows a triangle \(ABC\) in which \(BC = 20\) cm and angle \(ABC = 90°\). The perpendicular from \(B\) to \(AC\) meets \(AC\) at \(D\) and \(AD = 9\) cm. Angle \(BCA = \theta°\).

1. By expressing the length of \(BD\) in terms of \(\theta\) in each of the triangles \(ABD\) and \(DBC\), show that \(20\sin^2 \theta = 9\cos \theta\). [4]
2. Hence, showing all necessary working, calculate \(\theta\). [3]

1. By differentiating \(\frac{\cos x}{\sin x}\), show that if \(y = \cot x\) then \(\frac{dy}{dx} = -\cosec^2 x\). [3]
2. Hence show that \(\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \cosec^2 x \, dx = \sqrt{3}\). [2]

By using appropriate trigonometrical identities, find the exact value of

1. \(\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \cot^2 x \, dx\), [3]
2. \(\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{1}{1 - \cos 2x} \, dx\). [3]

Solve the equation \(\sec^2 \theta = 3 \cosec \theta\) for \(0° < \theta < 180°\). [5]

The parametric equations of a curve are $$x = e^{-t}\cos t, \quad y = e^{-t}\sin t.$$ Show that \(\frac{dy}{dx} = \tan(t - \frac{1}{4}\pi)\). [6]

A particle is projected from a point \(O\) on horizontal ground. The velocity of projection has magnitude \(20 \text{ m s}^{-1}\) and direction upwards at an angle \(\theta\) to the horizontal. The particle passes through the point which is 7 m above the ground and 16 m horizontally from \(O\), and hits the ground at the point \(A\).

1. Using the equation of the particle's trajectory and the identity \(\sec^2 \theta = 1 + \tan^2 \theta\), show that the possible values of \(\tan \theta\) are \(\frac{4}{3}\) and \(\frac{1}{4}\). [4]
2. Find the distance \(OA\) for each of the two possible values of \(\tan \theta\). [3]
3. Sketch in the same diagram the two possible trajectories. [2]

A particle \(P\) is projected with speed \(u\) at an angle \(\alpha\) above the horizontal from a point \(O\) on a horizontal plane and moves freely under gravity. The horizontal and vertical displacements of \(P\) from \(O\) at a subsequent time \(t\) are denoted by \(x\) and \(y\) respectively.

1. Derive the equation of the trajectory of \(P\) in the form $$y = x \tan \alpha - \frac{gx^2}{2u^2} \sec^2 \alpha.$$ [3]

During its flight, \(P\) must clear an obstacle of height \(h\) m that is at a horizontal distance of \(32\) m from the point of projection. When \(u = 40\sqrt{2}\) m s\(^{-1}\), \(P\) just clears the obstacle. When \(u = 40\) m s\(^{-1}\), \(P\) only achieves \(80\%\) of the height required to clear the obstacle.

1. Find the two possible values of \(h\). [6]

A particle \(P\) is projected from a point \(O\) on horizontal ground with speed \(u\) at an angle \(\theta\) above the horizontal, where \(\tan \theta = \frac{1}{3}\). The particle \(P\) moves freely under gravity and passes through the point with coordinates \((3a, \frac{4}{5}a)\) relative to horizontal and vertical axes through \(O\) in the plane of the motion.

1. Use the equation of the trajectory to show that \(u^2 = 25ag\). [2]
2. Express \(V^2\) in the form \(kag\), where \(k\) is a rational number. [6]

At the instant when \(P\) is moving horizontally, a particle \(Q\) is projected from \(O\) with speed \(V\) at an angle \(\alpha\) above the horizontal. The particles \(P\) and \(Q\) reach the ground at the same point and at the same time.

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. Solve, for \(-180° < \theta \leq 180°\), the equation $$3\tan(\theta + 43°) = 2\cos(\theta + 43°)$$ [6]

\includegraphics{figure_1} Figure 1 shows the triangle \(ABC\), with \(AB = 6\) cm, \(BC = 4\) cm and \(CA = 5\) cm.

1. Show that \(\cos A = \frac{3}{4}\). [3]
2. Hence, or otherwise, find the exact value of \(\sin A\). [2]

Given that \(2 \sin 2\theta = \cos 2\theta\),

1. show that \(\tan 2\theta = 0.5\). [1]
2. Hence find the values of \(\theta\), to one decimal place, in the interval \(0 \leq \theta < 360\) for which \(2 \sin 2\theta ° = \cos 2\theta °\). [5]

Find the values of \(\theta\), to 1 decimal place, in the interval \(-180 \leq \theta < 180\) for which $$2 \sin^2 \theta ° - 2 \sin \theta ° = \cos^2 \theta °.$$ [8]