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SAT II Math I : SAT Subject Test in Math I

Study concepts, example questions & explanations for SAT II Math I

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6 Diagnostic Tests 113 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #4 : Right Triangles And Similar Triangles

What is the length of the leg of a right triangle whose hypotenuse is 5cm and other leg is 4cm?

Possible Answers:

4cm

3cm

10cm

9cm

6.40cm

Correct answer:

3cm

Explanation:

Pythagorean Theorem: a^2+b^2=c^2

One leg is 4cm and the hypotenuse is 5cm. Plug in 4 for one of the legs and 5 for the hypotenuse (c).

a^2+4^2=5^2

a^2+16=25

Subtracting 16 from either side of the equation gives us:

a^2=9

The last step is to take the square root both sides resulting in:

a=3cm

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Example Question #1 : Trigonometric Identities

Simplify the expression $\frac{\sin x}{1 + \cos x} + \frac{1 + \cos x}{\sin x}$ .

Possible Answers:

$1 + \tan x$

$2\csc x$

$\cot x - 1$

$2\sec x$

$2\sin x + 2\cos x$

Correct answer:

$2\csc x$

Explanation:

$\frac{\sin x}{1 + \cos x} + \frac{1 + \cos x}{\sin x}$

Find a common denominator

$\frac{(\sin x) \sin x}{(1 + \cos x) \sin x} + \frac{(1 + \cos x)(1 + \cos x)}{\sin x (1 + \cos x)}$

Multiply the numerators and leave the denominators factored

$\frac{ \sin^2 x}{(1 + \cos x) \sin x} + \frac{1 + 2\cos x + \cos^2 x}{\sin x (1 + \cos x)}$

Add numerators

$\frac{1 + 2\cos x + \cos^2 x + \sin^2 x}{\sin x (1 + \cos x)}$

Pythagorean identity

$\frac{1 + 2\cos x + 1}{\sin x (1 + \cos x)}$

Combine like terms

$\frac{2 + 2\cos x}{\sin x (1 + \cos x)}$

Factor numerator

$\frac{2(1 + \cos x)}{\sin x (1 + \cos x)}$

Reduce

$\frac{2}{\sin x}$

Reciprocal identity

$2\csc x$

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Example Question #1 : Trigonometric Identities

If $\sin \beta = \frac{3}{5}$ and $0 < \beta < \frac{\pi}{2}$ , evaluate $\cos 2\beta$ .

Possible Answers:

$\frac{4}{5}$

$\frac{7}{25}$

$\frac{8}{5}$

$\frac{9}{25}$

$\frac{16}{25}$

Correct answer:

$\frac{7}{25}$

Explanation:

The easiest identity to use here is:

$\cos 2\beta = 1 - 2 \sin^2 \beta$

Substituting in the given values we get:

$= 1 - 2 \left ( \frac{3}{5} \right )^2 = 1 - \frac{18}{25} = \frac{7}{25}$

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Example Question #2 : Trigonometric Identities

Which of the following is the simplified version of $\cot x \sec x \cos x \sin x$ ?

Possible Answers:

$\sec x$

$\tan x$

$\cos x$

$\csc x$

$\cot x$

Correct answer:

$\cos x$

Explanation:

To solve this problem we need to rewrite it in terms of $\cos x$ and $\sin x$ .

Rewriting cotangent we get the following.

$\cot x= \frac{1}{\tan x}\rightarrow \tan x=\frac{\sin x}{\cos x} \rightarrow \cot x=\frac{\cos x}{\sin x}$

Rewriting secant we get the following.

$\sec x=\frac{1}{\cos x}$

Thus, we can substitute these identities into our original problem and simplify.

$\cot x \sec x \cos x \sin x = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos x} \cdot \frac{\cos x}{1} \cdot \frac{\sin x}{1} = \cos x$

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Example Question #51 : Trigonometric Identities

Using trigonometric identities prove whether the following is valid:

$\sin^{2}A + \cot^{2}A = \frac{2\sin^{4}A +2\cos^{2}A}{1 - \cos 2A}$

Possible Answers:

Only in the range of: $0 < A < 2\pi$

Only in the range of: $0 < A < \frac{\pi}{2}$

Uncertain

False

True

Correct answer:

True

Explanation:

We can work with either side of the equation as we choose. We work with the right hand side of the equation since there is an obvious double angle here. We can factor the numerator to receive the following:

$\sin^{2}A + \cot^{2}A = \frac{2\left(\sin^{4}A + \cos^{2}A\right)}{1-\cos 2A}$

Next we note the power reducing formula for sine so we can extract the necessary components as follows:

$\sin^{2}A + \cot^{2}A = \frac{2}{1-\cos 2A}\left(\sin^{4}A + \cos^{2}A\right)$

The power reducing formula must be inverted giving:

$\sin^{2}A + \cot^{2}A = \frac{1}{\sin^{2}A}\left(\sin^{4}A + \cos^{2}A\right)$

Now we can distribute and reduce:

$\sin^{2}A + \cot^{2}A = \sin^{2}A + \frac{\cos^{2}A}{\sin^{2}A}$

Finally recalling the basic identity for the cotangent:

$\sin^{2}A + \cot^{2}A = \sin^{2}A + \cot^{2}A$

This proves the equivalence.

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Example Question #52 : Trigonometric Identities

Use the power reducing formulas for trigonometric functions to reduce and simplify the following equation:

$\sin^{2}A \cos^{2} A$

Possible Answers:

$1 + \cos^{2}A$

$1 + \cos 2A$

$\frac{1 + \cos^{2}2A}{2}$

$\frac{3 + \cos 4A}{8}$

$\frac{1 + 2\cos A}{2}$

Correct answer:

$\frac{3 + \cos 4A}{8}$

Explanation:

The power reducing formulas for both sine and cosine differ in only the operation in the numerator. Applying the power reducing formulas here we get:

$\frac{1-\cos 2A}{2}\times \frac{1 + \cos 2A}{2}$

Multiplying the binomials in the numerator and multiplying the denominators:
$\frac{1-\cos 2A + \cos 2A - \cos^{2} 2A}{4}$

Reducing the numerator:

$\frac{1+\cos^{2}2A}{4}$

We again use the power reducing formula for cosine as follows:

$\frac{1 + \frac{1 + \cos 4A}{2}}{4}$

Combining the numerator by determining a common denominator:

$\frac{\frac{3 + \cos 4A}{2}}{4}$

Now simply reducing the double fraction:

$\frac{3 + \cos 4A}{8}$

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Example Question #1 : Identities Of Squared Trigonometric Functions

Using trigonometric identities, determine whether the following is valid:

$\sin^{2} A \cos^{2} A = \frac{1-\cos^{4}A-2\cos^{2}A\sin^{2}A + \sin^{4}A}{4}$

Possible Answers:

Only valid in the range of: $0 < A < \frac{\pi}{2}$

False

True

Only valid in the range of: $0 < A < 2\pi$

Uncertain

Correct answer:

False

Explanation:

In this case we choose to work with the side that appears to be simpler, the left hand side. We begin by using the power reducing formulas:

$\frac{1-\cos 2A}{2}\frac{1+\cos 2A}{2} = \frac{1-\cos^{4}A -2\cos^{2}A\sin^{2}A+\sin^{4}A}{4}$

Next we perform the multiplication on the numerator:

$\frac{1-\cos^{2}2A}{4} = \frac{1-\cos^{4}A -2\cos^{2}A\sin^{2}A+\sin^{4}A}{4}$

The next step we take is to remove the double angle, since there is no double angle in the alleged solution:

$\frac{1 - \left(\cos^{2}A -\sin^{2}A\right )^{2}}{4}=\frac{1-\cos^{4}A -2\cos^{2}A\sin^{2}A+\sin^{4}A}{4}$

Finally we multiply the binomials in the numerator on the left hand side to determine if the equivalence holds:

$\frac{1-\cos^{4}A +2\cos^{2}A\sin^{2}A-\sin^{4}A}{4} = \frac{1-\cos^{4}A -2\cos^{2}A\sin^{2}A+\sin^{4}A}{4}$

We see that the equivalence does not hold.

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Example Question #1 : Trigonometric Identities

Simplify the expression:

$\csc\theta(\sin\theta+\cos\theta \cot\theta)$

Possible Answers:

$\csc^{2}\theta$

$-\sec^{2}\theta$

$\cos\theta$

$\frac{1}{\csc^{2}\theta }$

Correct answer:

$\csc^{2}\theta$

Explanation:

The first step in solving this equation is to distribute :

$\csc\theta(\sin\theta +\cos\theta \cot\theta ) = \csc \theta\sin\theta+ \csc\theta\cos\theta\cot\theta$

At this point, simplify using known Pythagorean identities. The left quantity simplifies such:

$\csc\theta\sin\theta = \frac{\sin\theta }{\sin\theta } = 1$

and the right quantity simplifies such:

$\csc\theta\cos\theta\cot\theta = (\frac{1}{\sin\theta })(\cos\theta )(\frac{\cos\theta }{\sin\theta })=\frac{\cos^{2}\theta }{\sin^{2}\theta }=\cot^{2}\theta$

Thus, we end up with:

$1+\cot^2\theta$ ,

which our Pythagorean identity tells us is equivalent to $\csc^{2}\theta$ .

Thus,

$\csc\theta(\sin\theta+\cos\theta\cot\theta) = \csc^{2}\theta$

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Example Question #1 : Identities Of Squared Trigonometric Functions

Simplify using identities:

$\frac{-\cos x+\sec x}{\tan x}$

Possible Answers:

$\csc x$

$\sin x$

$\cos x$

$\sec x$

$\tan x$

Correct answer:

$\sin x$

Explanation:

First we expand the inverse identities into fractional form:

$\frac{-\cos x+\sec x}{\tan x}= \frac{-\cos x+\frac{1}{\cos x}}{\frac{\sin x}{\cos x}}$

Invert the bottom fraction and distribute into the top, keeping track of the negative:

$\frac{-cos x+\frac{1}{\cos x}}{\frac{\sin x}{\cos x}} = \bigg(\frac{1-\cos^{2}x}{\cos x}\bigg)\bigg(\frac{\cos x}{\sin x}\bigg)$

Using the Pythagorean identity $1-\cos^{2}x = \sin^{2}x$ , our equation becomes:

$\bigg(\frac{\sin^{2}x}{\cos x}\bigg)\bigg(\frac{\cos x}{\sin x}\bigg)$

At this point, cross-cancel to obtain $\sin x$ .

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Example Question #2 : Trigonometric Identities

Determine the value of: $(1+cot^2(\theta))(1-cos^2(\theta))$

Possible Answers:

$1-cos(\theta)$

$1-csc(\theta)$

$sin(\theta)-cos(\theta)$

Correct answer:

Explanation:

In order to solve $(1+cot^2(\theta))(1-cos^2(\theta))$ , first identify all the recognizable identities.

The identities that can be used for this scenario are:

$sin^2(\theta)+cos^2(\theta)=1$

$1+cot^2(\theta)=csc^2(\theta)$

The first identity can also be manipulated.

$1-cos^2(\theta)=sin^2(\theta)$

Replace $(1+cot^2(\theta))(1-cos^2(\theta))$ with the correct identities and simplify.

$csc^2(\theta)\times sin^2(\theta)= \frac{1}{sin^2(\theta)}\times sin^2(\theta) =1$

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All SAT II Math I Resources

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SAT II Math I : SAT Subject Test in Math I

Study concepts, example questions & explanations for SAT II Math I

All SAT II Math I Resources

Example Questions

Example Question #4 : Right Triangles And Similar Triangles

Example Question #1 : Trigonometric Identities

Example Question #1 : Trigonometric Identities

Example Question #2 : Trigonometric Identities

Example Question #51 : Trigonometric Identities

Example Question #52 : Trigonometric Identities

Example Question #1 : Identities Of Squared Trigonometric Functions

Example Question #1 : Trigonometric Identities

Example Question #1 : Identities Of Squared Trigonometric Functions

Example Question #2 : Trigonometric Identities

All SAT II Math I Resources

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