Trigonometry Questions and Answers

Answer: C

This is in the form of the cosine double angle identity: \(\cos(2A) = 1 - 2\sin^2 A\).

Here, A = 22.5°.

The expression is equal to \(\cos(2 \times 22.5°) = \cos(45°)\).

The value of \(\cos 45°\) is \(1/\sqrt{2}\).

Answer: B

We are given \(\tan \theta + \cot \theta = 2\). Squaring both sides:

\((\tan \theta + \cot \theta)^2 = 2^2\)

\(\tan^2 \theta + \cot^2 \theta + 2 \tan \theta \cot \theta = 4\)

Since \(\tan \theta \cot \theta = 1\), the equation becomes:

\(\tan^2 \theta + \cot^2 \theta + 2(1) = 4\)

\(\tan^2 \theta + \cot^2 \theta = 2\).

Answer: B

We use the complementary angle identity \(\sin(90° - \theta) = \cos \theta\).

\(\sin 65° = \sin(90° - 25°) = \cos 25°\).

The expression becomes \(\sin^2 25° + \cos^2 25°\).

Using the Pythagorean identity \(\sin^2 A + \cos^2 A = 1\), the value is 1.

Answer: A

Let L be the length of the string (hypotenuse) and h be the height (opposite side).

\(\sin 60° = \frac{\text{height}}{\text{length}} = \frac{75}{L}\)

\(\frac{\sqrt{3}}{2} = \frac{75}{L}\)

\(L = \frac{75 \times 2}{\sqrt{3}} = \frac{150}{\sqrt{3}} = \frac{150\sqrt{3}}{3} = 50\sqrt{3}\) m.

Answer: A

This is the double-angle formula for the tangent.

\[ \tan 2A = \frac{2\tan A}{1 - \tan^{2}A} \]

With \( A = 30^{\circ} \):

\[ \frac{2\tan 30^{\circ}}{1 - \tan^{2}30^{\circ}} = \tan 60^{\circ} \]

Checking numerically: \( \tfrac{2/\sqrt{3}}{1 - 1/3} = \tfrac{2/\sqrt{3}}{2/3} = \sqrt{3} = \tan 60^{\circ} \).

Answer: B

Given \(\sin A = \frac{12}{13}\) (Opposite/Hypotenuse). We can find the adjacent side using the Pythagorean theorem: \(Adj = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5\).

\(\cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{5}{13}\).

\(\sec A = \frac{1}{\cos A} = \frac{13}{5}\).

Answer: A

Let the mountain be at point P, with height PQ = 500m. Let the boats be A and B.

For the farther boat (B), angle of elevation is 30°. Let its distance be d₂. \(\tan 30° = \frac{500}{d_2} \Rightarrow d_2 = 500\sqrt{3}\).

For the nearer boat (A), angle of elevation is 45°. Let its distance be d₁. \(\tan 45° = \frac{500}{d_1} \Rightarrow d_1 = 500\).

The distance between the boats is \(d_2 - d_1 = 500\sqrt{3} - 500 = 500(\sqrt{3}-1)\) m.

Answer: B

This is the cosine double angle formula, \(\cos(2A) = 1 - 2\sin^2 A\). With A = 45°, the expression is \(\cos(2 \times 45°) = \cos(90°) = 0\).

Alternatively, \(\sin 45° = \frac{1}{\sqrt{2}}\). So, \(1 - 2(\frac{1}{\sqrt{2}})^2 = 1 - 2(\frac{1}{2}) = 1 - 1 = 0\).

Answer: C

Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):

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

\(\cos^2 \theta = 1 - \frac{a^2}{b^2} = \frac{b^2-a^2}{b^2}\)

\(\cos \theta = \sqrt{\frac{b^2-a^2}{b^2}} = \frac{\sqrt{b^2-a^2}}{b}\).

Answer: A

\[ \sin 30^{\circ} = \frac{1}{2}, \qquad \cos 60^{\circ} = \frac{1}{2} \]

\[ \left(\frac{1}{2}\right)^{2} + \left(\frac{1}{2}\right)^{2} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \]

The Pythagorean identity does not apply here: it needs the same angle in both terms, and 30° and 60° are different.