## SAT SUBJECT TEST MATH LEVEL 2

## PART 2

## REVIEW OF MAJOR TOPICS

## CHAPTER 1

Functions

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1.3 Trigonometric Functions and Their Inverses

### Answers and Explanations

Definitions

1. **(A)** Reference angle is 40°. Cosine in quadrant IV is positive.

2. **(D)** See corresponding figure. Therefore, sin .

3. **(D)** Angle is in quadrant II since sec < 0 and sin > 0. Therefore, tan .

4. **(A)** Cofunctions of complementary angles are equal. *x *– 30 + *x *= 90 finds a reference angle of 60° for *x*. The angle in quadrant III that has a reference angle of 60° is 240°.

5. **(A)** Angle is in quadrant II, and sin is positive. Angle is in quadrant IV, and sin is negative.

* 6. **(E)** Put your calculator in degree mode, cos 310° + cos 190° 0.643 + (–0.985) – 0.342. Checking the answer choices shows that –cos 70° –0.342.

Arcs and Angles

1. **(E)** .

2. **(D)**

* 3. **(C)**

* 4. **(A)** ≈ 0.7

*s* = *r* ≈ 1(0.7) ≈ 0.7.

* 5. **(C)** Change 25° to 0.436 radian .

*s *= *r*, and so 12 = *r*(0.436) and *r *= 27.5 inches.

Special Angles

1. **(A)** Sketch a –60° angle in standard position as shown in the figure below.

The tangent ratio is

2. **(C)** Sketch an angle of radians in standard position, as shown in the figure below.

The cosine ratio is .

3. **(E)** First, determine an angle between 0° and 360° that is coterminal with 540° by subtracting 360° from 540° repeatedly until the result is in this interval. In this case, one subtraction suffices. Since coterminal angles have the same trig values, csc 540° = csc 180°. Sketch the figure below

In a quadrantal angle *r *= 1, and the cosecant ratio is , which is undefined.

Graphs

1. **(C)** Period . Point *P *is of the way through the period. Amplitude is 1 because the coefficient of sin is 1. Therefore, point *P *is at .

2. **(E)** Amplitude = . Period = . Graph translated unit up. Graph looks like a cosine graph reflected about *x*-axis and shifted up unit.

* 3. **(C)** Graph 4cos using ZOOM/ZTRIG and observe that the portion of the graph between is decreasing.

* 4. **(C)** Graph the function and determine its maximum (2) and minimum (–2). Subtract and then divide by 2.

5. **(D)** Period = .

* 6. **(C)** Graph the function using 0 for Xmin and for Xmax. Observe that the maximum occurs when *x *=. Then .

7. **(D)** Period (from the figure), so *M *= . Phase shift for a sine curve in the figure is –. Therefore, *x + N* = 0 when *x = –*. Therefore,

Identities, Equations, and Inequalities

* 1. **(A)** sin 2*x* = 2 sin *x* cos .

2. **(D)** Since tangent and cotangent are cofunctions, tan *A *= cot(90° – *A*), so *B *= 90° – *A, *and *A *+ *B *= 90°.

* 3. **(D)**

4. **(A)** sin 74° = 2 sin 37° cos 37°. Since . Since 74° is in the first quadrant, the positive square root applies, so cos .

* 5. **(B)** .

* 6. **(A)** Graph *y *= sin *x *and *y *= cos in radian mode using the Xmin = 0 and Xmax = . Observe that the first graph is beneath the second on [0,0.79].

7. **(D)** Remember that the range of the sine function is [–1,1], so the second term ranges from 6 to –6.

Inverse Trig Functions

* 1. **(E)** Set your calculator to degree mode, and enter 2nd sin^{–1} .

* 2. **(D)** Set your calculator to radian mode, and enter 2nd cos^{–1}(–0.5624).

* 3. **(B)** Set your calculator to degree mode, and enter 2nd tan^{–1}(tan 128°).

* 4. **(E)** The range of inverse cotangent functions consists of only positive numbers.

5. **(A)** Since , I is true. Since cos^{–1}1 = 0 and cos^{–1}(–1) = , II is not true. Since the range of cos^{–1} is [0,], III is not true because cos^{–1} can never be negative.

6. **(E)** 3*x* = arccos , and so *x *=arccos.

Triangles

1. **(D)** Law of Sines: . The figure shows two possible locations for *B*, labeled *B*_{1} and *B*_{2}, where m *AB*_{1}*C *= 45° and m *AB*_{2}*C *= 135°. Corresponding to these, m *ACB*_{1} = 105° and m *ACB*_{2} = 15°. Of these, only 15° is an answer choice.

* 2. **(E)** By the Law of Sines: . The figure below shows this to be an ambiguous case (an angle, the side opposite, and another side), so or *C *= 180° – 41.81° = 138.19°.

3. **(A)** The angles are 15°, 45°, and 120°. Let *c *be the longest side and *b *the next longest.

* 4. **(D)** Use the Law of Cosines. Let the sides be 4, 5, and 6. 16 = 25 + 36 – 60 cos *A*. Cos , which implies that

* 5. **(C)** Law of Cosines: *d*^{2} = 36 + 64 – 96 cos 120°. *d*^{2} = 148. Therefore, *d * 12.

6. **(D)** Altitude to base= 8 sin 60° = 4. Therefore, 4 < *a* < 8.

* 7. **(D)** *A *= 45°. Law of Sines: . Therefore, *a* = 10 14.

8. **(D)** Area . Therefore, *C *= 60° or 120°.

9. **(C)** Area . *C *= 60° or 120°. Use Law of Cosines with 60° and then with 120°.

**Note:** At this point in the solution you know there have to be two values for *C*. Therefore, the answer must be Choice C or E. If *C *= 10 (from Choice E), *ABC *is a right triangle with area = · 6 · 8 = 24. Therefore, Choice E is not the answer, and so Choice C is the correct answer.

10. **(A)** In I the altitude and so 2 triangles. so only 1 triangle. In III the altitude so no triangle.