SAT 2016

CHAPTER 10

THE SAT MATH: ADDITIONAL TOPICS

Skill 2: Understanding Basic Trigonometry

Lesson 9: The basic trigonometric functions

Which of these is equivalent to Images? (No calculator)

A)   Images

B)   Images

C)   Images

D)   Images

The Basic Trigonometric Functions

Any of the three basic trigonometric functions, like all functions, takes an input number and transforms it into an output number. A trigonometric function takes an angle, θ, as an input and constructs a right triangle with θ as one of its acute angles. The output is then the ratio of two sides of that triangle as defined by the mnemonic SOH-CAH-TOA.

image

But these definitions are limited, because they only work when θ is an acute angle. What if it’s a larger angle, like 135°, or even a negative angle, like –20°? To find the trigonometric ratios for these angles, we use the unit circle.

The Unit Circle

The unit circle is just a circle with radius 1 centered at the origin of the xy-plane. When using the unit circle, all angles must be in standard position, that is, with vertex at the origin and measured counterclockwise from the positive x-axis (just like the angle θ in the following figure).

image

When an angle, θ, is in standard position, its terminal ray intersects the unit circle in the point (x1y1). If we drop a vertical line segment from this point to the x-axis, we form a right triangle with legs of length x1 and y1 and a hypotenuse of length 1 (as shown above).

So now let’s go back to the definitions of the basic trigonometric functions. In terms of this right triangle, what are the sine, cosine, and tangent of θ?

image

This suggests three important theorems:

•  The sine of any angle is the y-coordinate of its corresponding point on the unit circle.

•  The cosine of any angle is the x-coordinate of its corresponding point on the unit circle.

•  The tangent of any angle is the ratio of the y-coordinate to the x-coordinate of its corresponding point on the unit circle.

(Medium-hard) First, you may find it useful to convert the angles to degree measures using the conversion factor (180°/π radians). This gives us π/4 radians = 45° and π/6 radians = 30°. We should recognize these as angles in two Special Right Triangles:

image

Using the definitions for sine and cosine above, these triangles show us that Images and Images. Therefore, Images Imagesand the correct answer is (C).

If sin x = a and sin z = −a, where x and z are in radians, and Images, which of the following could be the value of z in terms of x?

A)   π – x

B)   x – π

C)   2π + x

D)   Images

(Hard) The statement Images indicates that x is an angle in quadrant II, where the sine (the y-value of the points on the unit circle) is positive. Let’s draw this situation on the unit circle so we can visualize it. (We don’t want to confuse the angles called x and y in the problem with thex-coordinates and y-coordinates in the xy-plane. For this reason, let’s label the terminal rays for the angles “angle x” and “angle z.”) Recall that the sine of any angle is the y-coordinate of the point on the unit circle that corresponds to that angle. If sin x = a, then a is the y-coordinate of the point on the unit circle that corresponds to “angle x,” as shown in the diagram. If sin z = –a, then –a is the y-coordinate of the point on the unit circle that corresponds to “angle z.” There are two possible locations for “angle z” as shown in the diagram.

image

At this point, it may be easiest to simply pick a value for “angle x” that is between π/2 (≈ 1.57) and π (≈ 3.14), such as x = 2. Since sin 2 ≈ 0.909 (remember to put your calculator into “radian mode”), a = 0.909. Now we just need to find which angle among the choices has a sine of − 0.909

A)   sin(π –2) = 0.0909

B)   sin(2 – π) = –0.909

C)   sin(2π + π) = 0.0909

D)   sin(2 – π/2) = 0.416

Therefore, the correct answer is (B).

The Pythagorean Identity

sin2 x + cos2 x = 1 for all values of x

An identity is an algebraic equation that is true for all values of the unknown, and not just for particular values. We can prove the Pythagorean Identity by just applying the Pythagorean Theorem to our right triangle and thinking about the trigonometric ratios.

image

Apply Pythagorean Theorem:

(opposite)2 + (adjacent)2 = (hypotenuse)2

Divide by (hypotenuse)2:

Images

Simplify using trig definitions:

sin2 θ + cos2 θ = 1

If b is an angle measure such that sin Images, what is the value of (sin b – cos b)2?

A)   Images

B)   Images

C)   Images

D)   Images

(Medium) The expression we are trying to evaluate includes squared trigonometric ratios, so we will probably have to take advantage of the Pythagorean Identity.

Expression to evaluate:

(sin b – cos b)2

FOIL:

sin2 b – 2sin b cos b + cos2 b

Rearrange using Commutative Law of Addition:

sin2 b + cos2 b – 2sin b cos b

Substitute sin2 b + cos2 b = 1:

1 – 2sin b cos b

Now we’ll have to find the value of sin b cos b, which we can find with the given equation.

Given equation:

Images

Multiply by cos b:

Images

Substitute sin b cos b =Images

into original expression:

Images

So the correct answer is (A).

Lesson 10: The trigonometry of complementary angles

If sin Images and Images, which of the following is equal to sin Images?

A)   Images

B)   Images

C)   Images

D)   Images

Trigonometry of Complementary Angles

image

The two acute angles in a right triangle are complements of one another, that is, they have a sum of 90° (or, in radians, Images). So, if one of the angles has a radian measure of x, the other has a measure of Images – x.

If we look at the trigonometric ratios for this new angle, we see that these ratios are related to the trigonometric ratios of its complement by the following rule: The trigonometric ratio of any angle equals the cofunction of its complement.

image

(Medium-hard) Let’s start by drawing a picture of this situation. Since y is the measure of an acute angle, we can imagine it as the interior angle of a right triangle. Since its sine is equal to a/b, we can say that the opposite side has measure a and the hypotenuse has measure b.

image

Now we can find the length of the remaining leg (let’s call it k) in terms of a and b using the Pythagorean

Theorem:

k2 + a2 = b2

Subtract a2:

k2 = b2 – a2

Take the square root:

Images

Also, we know that the other acute angle has a measure of Images –y, so let’s complete the picture:

image

Now, finding the value of Images is just a matter of using the definition of sine: SOH.

image

So the correct answer is (D).

Exercise Set 3: Trigonometry (No Calculator)

1

What is the greatest possible value of f if Images?

image

2

If Images, what is the value of Images?

image

3

If (sin x – cos x)2 = 0.83, what is the value of (sin x + cos x)2?

image

4

Which of the following is equivalent to Images?

A)   Images

B)   Images

C)   Images

D)   1

5

image

If sin >θ < 0 and sin θ cos θ < 0, then θ must be in which quadrant of the figure above?

A)   I

B)   II

C)   III

D)   IV

6

If Images and Images, which of the following expressions is equal to Images?

A)   Images

B)   Images

C)   1 − sin2x

D)   Images

7

If sin b = a, which of the following could be the value of cos (b + π)?

A)   Images

B)   a2

C)   Images

D)   1 – a2

8

If Images and Images, what is the value of cos x?

A)   Images

B)   Images

C)   Images

D)   Images

EXERCISE SET 3: TRIGONOMETRY ANSWER KEY

No Calculator

1.  7/2 or 3.5   The discussion in Lesson 9 about the definition of the sine function and the unit circle made it clear that the value of the sine function ranges from –1 to 1. Therefore, the maximum value of Images is images or 3.5.

2.  1/36 or .027 or .028   An radian measure of π/3 is equivalent to 60 °. If you haven’t memorized the fact that cos(60 °) = ½, you can derive it from the Reference Information at the beginning of every SAT Math section, which includes the 30°-60°-90° special right triangle. Since a = ½, (a/3)2 = (1/6)2 = 1/36.

3.  1.17

(sin x – cos x)2 = 0.83

FOIL:

sin2 x – 2sin x cos x + cos2 x = 0.83

Regroup:

sin2 x + cos2 x – 2sin x cos x = 0.83

Simplify:

1 – 2sin x cos x = 0.83

Subtract 1:

–2sin x cos x = –0.17

Multiply by –1:

2sin x cos x = 0.17

Evaluate this expression:

(sin x + cos x)2

FOIL:

sin2 x + 2sin x cos x + cos2 x

Regroup:

sin2 x + cos2 x + 2sin x cos x

Substitute:

1 + 0.17 = 1.17

4.  D   sin(π/6) = ½ and cos(π/3) = ½, so sin(π/6)/cos(π/3) = 1.

5.  D   If sin θ < 0, then θ must be either in quadrant III or in quadrant IV. (Remember that sine corresponds to the y-coordinates on the unit circle, so it is negative in those quadrants where the y-coordinates are negative.) If sin θ cos θ < 0, then cos θ must be positive (because a negative times a positive is a negative). Since cos θ is only positive in quadrants I and IV (because cosine corresponds to the x-coordinates on the unit circle), θ must be in quadrant IV

6.  B   First, notice that a/b and b/a are reciprocals. Next, we can use the identity in Lesson 10 that Images to see that choice (B) is just the reciprocal of sin x. Alternately, we can just choose a value of x, like x = 1, and evaluate sin 1 = 0.841. The correct answer is the expression that gives a value equal to the reciprocal of 0.841, which is 1/0.841 = 1.19. Plugging in x = 1 gives (A) 0.841, (B) 1.19, (C) 0.292, (D) 0.540.

7.  C   Recall from the Pythagorean Identity that cos Images. Substituting sin b = a gives cos Images. The angle b + π is the reflection of angle b through the origin, so cos(b + π) is the opposite of cos b, which means that cos Images.

8.  D   Recall from the Pythagorean Identity that cos2 x = 1 − sin2 x

Images

Substitute cos2 x = 1 – sin2 x:

Images

Cancel common factor:

Images

Reciprocate:

Images