Trigonometry Questions and Answers

Answer: D

The value is \((2 - e^{2})^{3/2}\).

Reading the expression as \(\sec\alpha + \tan^{3}\alpha \cdot \csc\alpha\), first simplify the second term:

\[ \begin{aligned} \tan^{3}\alpha \csc\alpha &= \tan^{2}\alpha \cdot \frac{\sin\alpha}{\cos\alpha} \cdot \frac{1}{\sin\alpha} \\ &= \tan^{2}\alpha \sec\alpha \end{aligned} \]

so the whole expression is

\[ \begin{aligned} \sec\alpha(1 + \tan^{2}\alpha) &= \sec^{3}\alpha \\ &= (1 + \tan^{2}\alpha)^{3/2} \\ &= (2 - e^{2})^{3/2} \end{aligned} \]

\((2 - e^{2})^{1/2}\) is \(\sec\alpha\) alone, the first term without the second. The options with \(2 + e^{2}\) get the sign of \(e^{2}\) wrong when adding 1 to \(1 - e^{2}\).

Answer: A

45 degrees is 0.785 radian.

Since \(180^\circ = \pi\) radians:

\[ \begin{aligned} 45^\circ &= 45 \times \frac{\pi}{180} \\ &= \frac{\pi}{4} \approx 0.785 \end{aligned} \]

The other options are the radian values of other common angles: 1.57 is \(\frac{\pi}{2}\) (90°), 1.05 is \(\frac{\pi}{3}\) (60°) and 6.28 is \(2\pi\) (360°).

Answer: A

\(\tan\theta = -\frac{4}{3}\).

From \(\cos\theta = -\frac{3}{5}\), use \(\sin^{2}\theta + \cos^{2}\theta = 1\):

\[ \begin{aligned} \sin^{2}\theta &= 1 - \frac{9}{25} = \frac{16}{25} \\ \sin\theta &= \frac{4}{5} \\ \tan\theta &= \frac{4/5}{-3/5} = -\frac{4}{3} \end{aligned} \]

Sine is positive in the second quadrant, which is why \(\sin\theta\) is \(+\frac{4}{5}\), and tangent, sine over cosine, is negative there. That rules out \(\frac{4}{3}\). \(-\frac{4}{5}\) is a value of sine with the wrong sign, and \(\frac{3}{5}\) is cosine without its sign.

Answer: B

The value is 2.

Expand both squares; the cross terms cancel:

\[ \begin{aligned} (s + c)^{2} + (s - c)^{2} &= 2s^{2} + 2c^{2} \\ &= 2(\sin^{2}15^\circ + \cos^{2}15^\circ) \\ &= 2 \times 1 = 2 \end{aligned} \]

Here \(s = \sin 15^\circ\) and \(c = \cos 15^\circ\). The \(+2sc\) from the first square and the \(-2sc\) from the second cancel, so the angle does not matter at all. 1 forgets the factor 2: it is the value of \(\sin^{2} + \cos^{2}\) alone. 0 and \(-1\) are impossible, since a sum of two squares cannot be zero or negative here.

Answer: B

We use the fundamental trigonometric identity \(\sin^2 \theta + \cos^2 \theta = 1\).

\((\frac{3}{5})^2 + \cos^2 \theta = 1\)

\(\frac{9}{25} + \cos^2 \theta = 1\)

\(\cos^2 \theta = 1 - \frac{9}{25} = \frac{16}{25}\)

\(\cos \theta = \sqrt{\frac{16}{25}} = \frac{4}{5}\) (Assuming θ is in the first quadrant).

Answer: B

\[ \tan 45^{\circ} = \frac{\sin 45^{\circ}}{\cos 45^{\circ}} = \frac{1/\sqrt{2}}{1/\sqrt{2}} = 1 \]

Geometrically, a 45° right triangle is isosceles, so the opposite and adjacent sides are equal.

Answer: B

This is the fundamental Pythagorean identity.

\[ \sin^{2}A + \cos^{2}A = 1 \]

It follows from Pythagoras applied to a right triangle of hypotenuse 1, and it holds for every angle \( A \).

Answer: B

Let 'h' be the height of the tree and 's' be the length of the shadow. The angle of elevation is θ.

We have \(\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{s}\).

\(\tan 60° = \frac{h}{10}\)

Since \(\tan 60° = \sqrt{3}\), we have \(\sqrt{3} = \frac{h}{10}\).

h = 10√3 meters.

Answer: B

This is another key Pythagorean identity. It is derived from \(\sin^2 \theta + \cos^2 \theta = 1\) by dividing the entire equation by \(\cos^2 \theta\).

\(\frac{\sin^2 \theta}{\cos^2 \theta} + \frac{\cos^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta}\)

\(\tan^2 \theta + 1 = \sec^2 \theta\)

Rearranging gives \(\sec^2 \theta - \tan^2 \theta = 1\).

Answer: A

If \(\tan \theta = 1\), then \(\theta = 45°\).

We need to find the value of \(\sin 45° \cos 45°\).

\(\sin 45° = \frac{1}{\sqrt{2}}\) and \(\cos 45° = \frac{1}{\sqrt{2}}\).

So, \(\sin \theta \cos \theta = \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = \frac{1}{2}\).