Jump to a New ChapterIntroduction to the SAT IIContent and Format of the SAT II Math IICStrategies for SAT II Math IICMath IIC FundamentalsAlgebraPlane GeometrySolid GeometryCoordinate GeometryTrigonometryFunctionsStatisticsMiscellaneous MathPractice Tests Are Your Best Friends
 9.1 Basic Functions 9.2 Cosecant, Secant, and Cotangent 9.3 Solving Right Triangles 9.4 Trigonometric Identities 9.5 Graphing Trigonometric Functions 9.6 The Unit Circle

 9.7 Graphing in the Entire Coordinate Plane 9.8 Inverse Trigonometric Functions 9.9 Solving Non-Right Triangles 9.10 Key Terms 9.11 Review Questions 9.12 Explanations
Explanations

1.      C

The trick to this question is knowing the identity tan2 + 1 = sec2. If you didn’t know it, you could have derived the identity from the base identity sin2 + cos2 = 1 by dividing through the equation by cos2, giving tan2 + 1 = sec2. Subtracting 1 from both sides gives tan2 = sec2 – 1.

Now, substitute into the equation from the question:

The identity tan = sin /cos allows for further simplification:

The only answer choice for that yields a cosine value of 1 is 0.

2.      B

For any x, sin–1(sin(x)) = x. So the fraction reduces to cos 60º/cos 30º. Either by calculator, memorization, or using the 30-60-90 triangle, you can evaluate the trigonometric functions: 1/2/2 = 1.

3.      C

The first step is to calculate the length of the segment RT. To do so, use the definition of the sine function, sin = opposite /hypotenuse.

Next, calculate the length of segment ST using the cosine function (cos = adjacent/hypotenuse).

Now, the trick is to see that length segment ST is actually the sum of the radii of the two circles (SU + UT). Moreover, the length of segment RT = UT because they are both radii of the larger circle. Since you know the length of segment RT (and so, segment UT), you can subtract this value from segment ST to get the length of segment SU (which is also the radius of the smaller circle and the answer):

4.      C

The figure gives a non-right triangle, so you can’t use the basic trigonometric functions to find BC. But you are given the measures of two angles and the side between them, so you can use the law of sines: BC /sin A = AC/sin B.

First, find the measure of , using the rule that the sum of angles in a triangle is 180: B = 180 – 72 – 50 = 58º. Plug the value of into the law of sines: BC /sin 50 = 16/sin 58. By solving this proportion, you arrive at BC ≈ 14.45.

5.      E

Since you have the measures of two sides and the angle between them, you can start with the law of cosines to find c. Plug the given values into the law of cosines formula: c2 = 52 + 72 + 2(5)(7) cos 110º. Working this out, you get c ≈ 7.08. Now that you have the measure of the side opposite , you can use the law of sines to find a. Substitute the values into the formula, and solve the equation:

Taking the arcsine, ≈ 68.3º.

6.      D

sin–1 2 is undefined, because no angle exists whose sine is greater than 1. tan π/2 is undefined, because cos π /2 = 0. sec π/2 is undefined, because the secant function is equal to the reciprocal of the cosine function, and cos π /2 = 0, leaving a zero in the denominator. Since tan–1 0 = 0, it is defined. cot /2 is undefined because the cotangent function is the reciprocal of the tangent function, and tan / 2 is undefined.

7.      C

There are two easy ways to simplify the given expression. Both involve using identities. The first way is to immediately factor the expression into (1 – cos2x)(1 + tan2x). Then, using the identities 1 – cos2x = sin2x and 1 + tan2x = sec2x, the expression is simplified to sin2(x)sec2(x), which can be furthered simplified:

The other way takes slightly longer but requires the use of fewer identities.

8.      A

Without knowing the double-angle identity for sine, you could have found the arcsine of 1 /2, divided that angle by 2, found the sine and cosine of that angle, and then squared their product, which gives a value of 1/16.

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