Trigonometry Questions and Answers

Answer: B

Both functions decrease on \((\frac{\pi}{2}, \pi)\).

On \([0, 2\pi]\), \(\sin x\) increases up to \(\pi/2\), decreases from \(\pi/2\) to \(3\pi/2\), then increases again. \(\cos x\) decreases from 0 to \(\pi\) and increases from \(\pi\) to \(2\pi\). The derivatives say the same: \((\sin x)' = \cos x\) is negative on \((\pi/2, 3\pi/2)\), and \((\cos x)' = -\sin x\) is negative on \((0, \pi)\).

The two decreasing intervals overlap only on \((\pi/2, \pi)\). On \((0, \pi/2)\) sine is increasing; on \((\pi, 3\pi/2)\) cosine is increasing; on \((3\pi/2, 2\pi)\) both are increasing.

Answer: A

The value is \(\frac{a^{2}}{b\sqrt{b^{2}-a^{2}}}\).

Since \(\sin 73^{\circ} = \cos 17^{\circ}\), the expression is \(\sec 17^{\circ} - \cos 17^{\circ}\). With \(\sin 17^{\circ} = a/b\), \(\cos 17^{\circ} = \sqrt{b^{2}-a^{2}}/b\):

\[ \begin{aligned} \sec\theta - \cos\theta &= \frac{1 - \cos^{2}\theta}{\cos\theta} = \frac{\sin^{2}\theta}{\cos\theta} \\ &= \frac{a^{2}/b^{2}}{\sqrt{b^{2}-a^{2}}/b} \\ &= \frac{a^{2}}{b\sqrt{b^{2}-a^{2}}} \end{aligned} \]

The options with \(\sqrt{a^{2}-b^{2}}\) cannot be right, because \(a < b\) for a sine and that root would not be real. The option with \(b^{2}\) on top inverts the sine.

Answer: B

The value of θ is 60°.

Check the options in the left-hand side. At 60°:

\[ \begin{aligned} \sin 60^{\circ} - \cos 60^{\circ} &= \frac{\sqrt{3}}{2} - \frac{1}{2} \\ &= \frac{\sqrt{3} - 1}{2} \end{aligned} \]

That matches. At 45° the difference is 0, at 90° it is 1, and at 75° it is \(\sin 75^{\circ} - \cos 75^{\circ} = \tfrac{\sqrt{2}}{2} \approx 0.71\), whereas \(\tfrac{\sqrt{3}-1}{2} \approx 0.37\).

Answer: C

The value is \(\dfrac{2x^{2}}{y^{2} - x^{2}}\).

From \(\sin\theta = x/y\), treat \(x\) as the opposite side and \(y\) as the hypotenuse. The adjacent side is then \(\sqrt{y^{2} - x^{2}}\):

\[ \begin{aligned} \tan\theta &= \frac{x}{\sqrt{y^{2} - x^{2}}} \\ 2\tan^{2}\theta &= \frac{2x^{2}}{y^{2} - x^{2}} \end{aligned} \]

The stem's '2.' is a multiplication dot. Options A and B are \(\csc\theta\) and \(\sin\theta\) themselves, and option D has the wrong power and the wrong sign in the denominator.

Answer: B

sec A is 17/15.

With the right angle at C, the hypotenuse is AB:

\[ AB = \sqrt{15^{2} + 8^{2}} = \sqrt{289} = 17 \]

For angle A, the adjacent side is AC = 15, so \(\cos A = 15/17\) and \(\sec A = 17/15\). 15/17 is cos A, the reciprocal of the answer, and 17/5 and 5/17 do not use the sides of this triangle at all.

Answer: B

The false option is \(\cos A = \sqrt{2}\).

\(\cot A = 1\) means \(A = 45^{\circ}\), so \(\sin A = \cos A = \frac{1}{\sqrt{2}}\). The cosine of an angle can never exceed 1, so \(\sqrt{2} \approx 1.414\) is impossible.

The other three are true: \(\sin A = \frac{1}{\sqrt{2}}\); \(\tan A + \cot A = 1 + 1 = 2\); and \(\cos A\,\text{cosec}\,A = \cos A/\sin A = \cot A = 1\).

Answer: D

The statement that is not true is \(\sec C = \sqrt{29}/2\).

With the right angle at B, \(\sin A = BC/AC = 5/\sqrt{29}\). Take BC = 5 and AC = \(\sqrt{29}\); then

\[ AB = \sqrt{29 - 25} = 2 \]

So \(\cos A = 2/\sqrt{29}\), \(\tan C = AB/BC = 2/5\) and \(\cos C = BC/AC = 5/\sqrt{29}\), all true. But \(\sec C = AC/BC = \sqrt{29}/5\), not \(\sqrt{29}/2\); the false option uses AB in place of BC.

A quick check: \(\sin^2 A + \cos^2 A = 25/29 + 4/29 = 1\), confirming sides 2, 5 and \(\sqrt{29}\). Since A and C are complementary, \(\cos C = \sin A\), which is why option C is true.

Answer: D

The value is 1.

tan P = 1 means P = 45°. Then sin P = cos P = \(\frac{1}{\sqrt{2}}\), so

\[ \begin{aligned} 2\sin P\cos P &= 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \\ &= 1 \end{aligned} \]

The same follows from the identity 2 sin P cos P = sin 2P = sin 90° = 1. The value \(1/\sqrt{2}\) is sin P alone, \(\sqrt{2}\) is sin P + cos P, and 2 doubles the correct answer.

Answer: B

The value of \(\sec\theta\) is \(\frac{5}{3}\).

From \(3\sin\theta = 4\cos\theta\), \(\tan\theta = \frac{4}{3}\). Take a right triangle with opposite side 4 and adjacent side 3; its hypotenuse is 5:

\[ \begin{aligned} \sec\theta &= \frac{\text{hypotenuse}}{\text{adjacent}} \\ &= \frac{5}{3} \end{aligned} \]

\(\frac{5}{4}\) is \(\csc\theta\), \(\frac{4}{3}\) is \(\tan\theta\) and \(\frac{3}{4}\) is \(\cot\theta\), the usual ways of mixing up the ratios.

Answer: C

The value is 49/64.

Multiply out the brackets and use sin²θ + cos²θ = 1:

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

\[ \cot^{2}\theta = (7/8)^{2} = \frac{49}{64} \]

49/58 has the right numerator but a wrong denominator, and 21/24 and 35/48 do not come from squaring 7/8.