Trigonometric Functions and Equations Questions and Answers

Answer: E

The value is \(\sec\theta = \frac{-5}{2\sqrt{6}}\).

Find the cosine from \(\sin^{2}\theta + \cos^{2}\theta = 1\), and choose its sign from the quadrant.

\[ \begin{aligned} \cos^{2}\theta &= 1 - \tfrac{1}{25} = \tfrac{24}{25} \\ \cos\theta &= -\tfrac{2\sqrt{6}}{5} \Rightarrow \sec\theta = -\tfrac{5}{2\sqrt{6}} \end{aligned} \]

In the second quadrant the cosine, and so the secant, is negative, which rules out \(\frac{5}{2\sqrt{6}}\). \(\frac{-2\sqrt{6}}{5}\) is the cosine itself rather than its reciprocal.

Answer: B

The expression equals sin x.

Apply \(2\sin A\cos A = \sin 2A\) twice, using one factor of 2 each time.

\[ \begin{aligned} 2\sin\tfrac{x}{4}\cos\tfrac{x}{4} &= \sin\tfrac{x}{2} \\ 2\sin\tfrac{x}{2}\cos\tfrac{x}{2} &= \sin x \end{aligned} \]

Stopping after the first step gives \(\sin\frac{x}{2}\), the last option; using a third factor of 2 that is not there gives sin 2x. Cosine answers cannot arise because the sine factor survives every step.

Answer: C

The expression equals 1.

The denominator is a difference of squares, \((1 + \sin\alpha)^{2} - \cos^{2}\alpha\).

\[ \begin{aligned} (1 + \sin\alpha)^{2} - \cos^{2}\alpha &= 1 + 2\sin\alpha + \sin^{2}\alpha - \cos^{2}\alpha \\ &= 2\sin\alpha + 2\sin^{2}\alpha \\ &= 2\sin\alpha(1 + \sin\alpha) \end{aligned} \]

Here \(1 - \cos^{2}\alpha = \sin^{2}\alpha\) was used. The denominator equals the numerator, so the value is 1 for every admissible \(\alpha\). tan \(\alpha\) and the two quotients are what partial simplifications leave behind.

Answer: D

The value is x = 1.

Evaluate the right-hand side first: \(\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}\), so it equals \(\frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}\).

\[ \begin{aligned} \frac{x}{1 + x} &= \sin\frac{\pi}{6} = \frac{1}{2} \\ 2x &= 1 + x \Rightarrow x = 1 \end{aligned} \]

The value 1/2 is \(\frac{x}{1 + x}\), not x itself. x = 2 or 3 would give \(\frac{2}{3}\) or \(\frac{3}{4}\), which are not \(\sin\frac{\pi}{6}\).

Answer: A

The value is \(x = \frac{1}{2}\).

Write \(\frac{\pi}{4} = \tan^{-1}1\) and use \(\tan^{-1}a - \tan^{-1}b = \tan^{-1}\frac{a - b}{1 + ab}\).

\[ \begin{aligned} \tan^{-1}3 - \tan^{-1}1 &= \tan^{-1}\frac{3 - 1}{1 + 3} \\ &= \tan^{-1}\frac{1}{2} \end{aligned} \]

So \(x = \frac{1}{2}\). Subtracting 1 from 3 directly gives 2, which ignores the denominator \(1 + ab\); x = 1 or 3 would make the right side \(\frac{\pi}{4}\) or \(\tan^{-1}3\) itself.

Answer: C

The value is \(\frac{1}{4\sqrt{2}}\).

Use \(\cos 75^{\circ} = \sin 15^{\circ}\):

\[ \begin{aligned} \cos 75^{\circ}\cos 15^{\circ} &= \sin 15^{\circ}\cos 15^{\circ} \\ &= \tfrac{1}{2}\sin 30^{\circ} = \tfrac{1}{4} \end{aligned} \]

Multiplying by \(\cos 45^{\circ} = \frac{1}{\sqrt{2}}\) gives \(\frac{1}{4\sqrt{2}}\). \(\frac{1}{\sqrt{2}}\) is \(\cos 45^{\circ}\) alone, and the other options use the wrong product.

Answer: A

The sum equals \(\tan\frac{\alpha}{2}\tan\frac{\beta}{2}\tan\frac{\nu}{2}\).

If \(\alpha + \beta + \nu = 2\pi\), the half-angles add to \(\pi\). For three angles A, B, C with \(A + B + C = \pi\), \(\tan(A + B) = \tan(\pi - C) = -\tan C\), which rearranges to

\[ \tan A + \tan B + \tan C = \tan A\tan B\tan C \]

Applying it with \(A = \frac{\alpha}{2}\) and so on gives option A. The negative and multiplied versions do not follow from the identity.

Check with an example: take \(\alpha = \beta = \nu = \frac{2\pi}{3}\); each half-angle is \(\frac{\pi}{3}\) with tangent \(\sqrt{3}\), and \(3\sqrt{3} = (\sqrt{3})^{3}\).

Answer: B

The value is 1.

\(\tan 315^{\circ} = \tan(360^{\circ} - 45^{\circ}) = -1\). Cotangent is odd, so \(\cot(-405^{\circ}) = -\cot 405^{\circ} = -\cot 45^{\circ} = -1\). The product is \((-1)(-1) = 1\).

−1 comes from losing one of the two signs, and the fractional values come from using sines or cosines of 30° or 45° instead.

Answer: E

The expression equals \(\tan\frac{\pi}{7}\).

Let \(a = \frac{\pi}{7}\). Use \(\sin 2a = 2\sin a\cos a\) and \(1 + \cos 2a = 2\cos^{2}a\):

\[ \begin{aligned} \frac{\sin a + 2\sin a\cos a}{\cos a + 2\cos^{2}a} &= \frac{\sin a(1 + 2\cos a)}{\cos a(1 + 2\cos a)} \\ &= \tan a \end{aligned} \]

\(\cot\frac{\pi}{7}\) is the reciprocal, and the forms with \(\frac{\pi}{14}\) would come from half-angle formulas applied to the wrong angle.

The common factor \(1 + 2\cos a\) is never zero for \(a = \frac{\pi}{7}\), so the cancellation is valid.

Answer: B

The value is \(\frac{2025}{2024}\).

Let \(\theta = \cos^{-1}\frac{2024}{2025}\), so \(\cos\theta = \frac{2024}{2025}\) and \(\sec\theta = \frac{1}{\cos\theta} = \frac{2025}{2024}\).

\(\frac{2024}{2025}\) is \(\cos\theta\) itself, and the negative option would need an angle in the second quadrant, but \(\cos^{-1}\) of a positive number lies between 0 and \(\frac{\pi}{2}\).