Trigonometric functions and their properties

Correct answer: A. (\frac{12}{13})

( \cos^2 \theta=1-\sin^2 \theta=\frac{144}{169}), and in the first quadrant it is positive. In exams do not forget quadrant sign.

Correct answer: B. (\frac{4}{5})

( \sin^2 \theta=1-\cos^2 \theta=\frac{16}{25}), and (\sin \theta) is positive in the second quadrant. In exams ASTC rule helps.

Correct answer: B. (-\tan \theta)

( \tan \theta) is an odd function, so ( \tan(-\theta)=-\tan \theta). In exams focus on sign change.

Correct answer: C. (\sin \theta)

( \pi-\theta) lies in the second quadrant and sine remains positive. In exams remember allied angle formulas.

Correct answer: A. (-\cos \theta)

( \pi-\theta) is in the second quadrant where cosine is negative. In exams always apply quadrant sign.

Correct answer: B. (-\tan \theta)

( \pi-\theta) is in the second quadrant and tangent is negative. In exams pay special attention to the sign of tangent.

Correct answer: B. (-\sin \theta)

( \pi+\theta) is in the third quadrant where sine is negative. In exams identifying allied angles is important.

Correct answer: C. (-\cos \theta)

( \pi+\theta) is in the third quadrant and cosine is negative. In exams recall the sign table quickly.

Correct answer: B. (\tan \theta)

The period of ( \tan \theta) is ( \pi), so ( \tan(\pi+\theta)=\tan \theta). In exams remember the periodicity of tangent.

Correct answer: B. (-\tan \theta)

(2\pi-\theta) is in the fourth quadrant where tangent is negative. In exams derive tangent from sine and cosine signs.

Correct answer: A. (\theta=n\pi)

( \sin \theta) is zero at points like (0), (\pi), and (2\pi). In exams write the general solution with (n\in\mathbb{Z}).

Correct answer: B. (\theta=\frac{(2n+1)\pi}{2})

( \cos \theta) is zero at odd multiples of ( \frac{\pi}{2}). In exams remember the pattern ( \frac{(2n+1)\pi}{2}).

Correct answer: A. (\theta=n\pi)

( \tan \theta=0) occurs when ( \sin \theta=0) and ( \cos \theta\neq0). In exams use period ( \pi) for tangent.

Correct answer: A. (\theta=n\pi+(-1)^n\alpha)

The combined solution of ( \sin \theta=\sin \alpha) is ( \theta=n\pi+(-1)^n\alpha). In exams learn to write two cases in one formula.

Correct answer: B. (\theta=2n\pi\pm\alpha)

For ( \cos \theta=\cos \alpha), angles occur in both directions. In exams do not forget ( \pm\alpha).

Correct answer: B. (\theta=n\pi+\alpha)

The period of ( \tan \theta) is ( \pi), so the solution is ( \theta=n\pi+\alpha). In exams use period ( \pi) for tangent equations.

Correct answer: C. \(\cot \theta\)

At \( \frac{\pi}{2}-\theta\), tangent changes to its co-function cotangent. In exams remember complementary angle property.

Correct answer: D. \(\tan \theta\)

At \( \frac{\pi}{2}-\theta\), cotangent changes to the co-function tangent. In exams remember reciprocal function pairs.

Correct answer: C. (\frac{\pi}{4})

From ( \sin \theta=\cos \theta), ( \tan \theta=1), and in the first quadrant ( \theta=\frac{\pi}{4}). In exams divide by ( \cos \theta) only when it is nonzero.

Correct answer: A. (\frac{\pi}{6},\frac{5\pi}{6})

The reference angle for ( \sin \theta=\frac{1}{2}) is ( \frac{\pi}{6}), and sine is positive in the first and second quadrants. In exams choose only answers in the given interval.

Correct answer: A. (\frac{\pi}{3},\frac{5\pi}{3})

For ( \cos \theta=\frac{1}{2}), the reference angle is ( \frac{\pi}{3}), and cosine is positive in the first and fourth quadrants. In exams write both solutions up to (2\pi).

Correct answer: B. (\frac{\pi}{3})

The reference angle for ( \tan \theta=\sqrt{3}) is ( \frac{\pi}{3}), and in (0<\theta<\pi) tangent is positive in the first quadrant. In exams choose the quadrant by sign.

Correct answer: B. (1)

Because (1-\cos^2 \theta=\sin^2 \theta), the ratio is (1). In exams always check the denominator condition.

Correct answer: C. (1)

Because (1-\sin^2 \theta=\cos^2 \theta), the value is (1). In exams replace the numerator using the basic identity.

Correct answer: B. (\frac{4}{5})

In ( \tan \theta=\frac{4}{3}), opposite is (4), adjacent is (3), and hypotenuse is (5). In exams identify Pythagorean triplets.