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GRE Subject Test: Math : Trigonometric Functions

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Example Questions

Example Question #1 : Trigonometric Functions

Evaluate:

$\int\sin(75x)\cos(52x)dx$

Possible Answers:

$\frac{1}{2}\cos(23x)+\frac{1}{2}\cos(127x)+C$

$-\frac{1}{46}\cos(23x)-\frac{1}{254}\cos(127x)+C$

$-\frac{1}{46}\sin(23x)-\frac{1}{254}\sin(127x)+C$

$\frac{1}{46}\cos(75x)+\frac{1}{254}\cos(52x)+C$

$\frac{1}{52}\sin^2(75x)\cos(52x)+C$

Correct answer:

$-\frac{1}{46}\cos(23x)-\frac{1}{254}\cos(127x)+C$

Explanation:

Evaluating this integral requires use of the "Product to Sum Formulas of Trigonometry":

For:

$\sin(\alpha x)\cos(\beta x)=\frac{1}{2}[\sin((\alpha - \beta)x)+\sin((\alpha + \beta)x)]$

So for our given integral, we can rewrite like so:

$\int \sin(75x)\cos(52x)dx\rightarrow \frac{1}{2}\int[\sin(23x)+\sin(127x)]dx$

This can be rewritten as two separate integrals and solved using a simple substitution.

$\frac{1}{2}\int[\sin(23x)+\sin(127x)]dx\rightarrow \frac{1}{2}\int \sin(23x)dx+\frac{1}{2}\int\sin(127x)dx$

Solving each integral individually, we have:

$\frac{1}{2}\int\sin(23x)dx\rightarrow u=23x, du=23dx$

$\frac{1}{23}du=dx$

Substituting this into the integral results in:

$\frac{1}{2}\int\frac{1}{23}\sin(u)du=-\frac{1}{46}\cos(u)+C\rightarrow -\frac{1}{46}\cos(23x)+C$

The other integral is solved the same way:

$\frac{1}{2}\int\sin(127x)dx\rightarrow u=127x, du=127dx$

$\frac{1}{127}du=dx$

Substituting this into the integral results in:

$\frac{1}{2}\int\frac{1}{127}\sin(u)du=-\frac{1}{254}\cos(u)+C\rightarrow-\frac{1}{254}\cos(127x)+C$

Now combining these two statements together results in one of the answer choices:

$-\frac{1}{46}\cos(23x)-\frac{1}{254}\cos(127x)+C$

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Example Question #2 : Trigonometric Functions

Evaluate:

$\int\sin^4(2x)\cos^5(2x)dx$

Possible Answers:

$\frac{1}{18}\cos^9(2x)-\frac{1}{7}\cos^7(2x)+\frac{1}{10}\cos^5(2x)+C$

$\frac{1}{10}\cos^9(2x)-\frac{1}{7}\cos^7(2x)+\frac{1}{18}\cos^5(2x)+C$

$\frac{1}{10}\sin^9(2x)-\frac{1}{7}\sin^7(2x)+\frac{1}{18}\sin^5(2x)+C$

$\frac{1}{18}\sin^9(2x)-\frac{1}{7}\sin^7(2x)+\frac{1}{10}\sin^5(2x)+C$

$\frac{1}{5}\sin^9(2x)-\frac{2}{7}\sin^7(2x)+\frac{1}{9}\sin^5(2x)+C$

Correct answer:

$\frac{1}{18}\sin^9(2x)-\frac{1}{7}\sin^7(2x)+\frac{1}{10}\sin^5(2x)+C$

Explanation:

This integral can be easily evaluated by following the rules outlined for integrating powers of sine and cosine.

But first a substitution needs to be made:

$\int\sin^4(2x)\cos^5(2x)dx$

$u=2x;du=2dx\rightarrow\frac{1}{2}du=dx$

$\frac{1}{2}\int\sin^4(u)cos^5(u)du$

Now that we've made this substitution, we will use the rules outlined for integrating powers of sine and cosine:

In General:

$\int\sin^m(x)\cos^n(x)dx$

1. If "m" is odd, then we make the substitution $u=\sin(x)$ , and we use the identity $\cos^2(x)=1-\sin^2(x)$ .

2. If "n" is odd, then we make the substitution $u=\cos(x)$ , and we use the identity $\sin^2(x)=1-\cos^2(x)$ .

For our given problem statement we will use the first rule, and alter the integral like so:

$\frac{1}{2}\int\sin^4(u)\cos^5(u)du\rightarrow\frac{1}{2}\int\sin^4(u)\cos^4(u)\cos(u)du$

$\frac{1}{2}\int\sin^4(u)[\cos^2(u)]^2\cos(u)du\rightarrow\frac{1}{2}\int\sin^4(u)[1-\sin^2(u)]^2\cos(u)du$

$v=\sin(u);dv=\cos(u)du$

$\frac{1}{2}\int v^4[1-v^2]^2dv\rightarrow\frac{1}{2}\int v^4[1-2v^2+v^4]dv\rightarrow\frac{1}{2}\int (v^4-2v^6+v^8)dv$

$\frac{1}{2}[\frac{1}{5}v^5-\frac{2}{7}v^7+\frac{1}{9}v^9]+C\rightarrow\frac{1}{10}v^5-\frac{1}{7}v^7+\frac{1}{18}v^9+C$

Now we need to substitute back into v:

$\frac{1}{10}\sin^5(u)-\frac{1}{7}\sin^7(u)+\frac{1}{18}\sin^9(u)+C$

Now we need to substitute back into u, and rearrange to make it look like one of the answer choices:

$\frac{1}{18}\sin^9(2x)-\frac{1}{7}\sin^7(2x)+\frac{1}{10}\sin^5(2x)+C$

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Example Question #1 : Trigonometric Functions

Find :

$sin(x^{\circ})=\frac{\sqrt2}{2}$

Possible Answers:

Correct answer:

Explanation:

Step 1: Draw a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle..

The short sides have a length of and the hypotenuse has a length of $\sqrt 2$ .

Step 2: Find Sin (Angle A):

$sin(A)=sin(45^{\circ})=\frac {opp}{hyp}=\frac {1}{\sqrt 2}$

Step 3: Rationalize the root at the bottom:

$sin(45^{\circ})=\frac {1}{\sqrt 2}\cdot\frac {\sqrt 2}{\sqrt 2}=\frac {\sqrt 2}{2}$

x=45

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Example Question #1 : Trigonometric Functions

Evaluate the derivative $\frac{\mathrm{d} }{\mathrm{d} x} sin^{-1}(x)$

Possible Answers:

$\frac{1}{\sqrt{1+x^2}}$

$\frac{1}{\sqrt{1+x}}$

$\frac{1}{\sqrt{1-x}}$

$\frac{1}{\sqrt{1-x^2}}$

Correct answer:

$\frac{1}{\sqrt{1-x^2}}$

Explanation:

The inverse trig functions should be memorized.

$\frac{\mathrm{d} }{\mathrm{d} x} sin^{-1}(x)=\frac{1}{\sqrt{1-x^2}}$

The other common inverse trig functions are

$\frac{\mathrm{d} }{\mathrm{d} x} cos^{-1}(x)=-\frac{1}{\sqrt{1-x^2}}$

$\frac{\mathrm{d} }{\mathrm{d} x} tan^{-1}(x)=\frac{1}{1+x^2}$

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GRE Subject Test: Math : Trigonometric Functions

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #1 : Trigonometric Functions

Evaluate:

Example Question #2 : Trigonometric Functions

Evaluate:

Example Question #1 : Trigonometric Functions

Example Question #1 : Trigonometric Functions

All GRE Subject Test: Math Resources

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