Paul's Online Notes
Home / Algebra Trig Review / Trigonometry / Trig Formulas
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 2.3 : Trig Formulas

This is not a complete list of trig formulas. This is just a list of formulas that I’ve found to be the most useful in a Calculus class. For a complete listing of trig formulas you can download my Trig Cheat Sheet.

Complete the following formulas. Show All Solutions Hide All Solutions

1. $${\sin ^2}\left( \theta \right) + {\cos ^2}\left( \theta \right) =$$
Show Solution
${\sin ^2}\left( \theta \right) + {\cos ^2}\left( \theta \right) = 1$

Note that this is true for ANY argument as long as it is the same in both the sine and the cosine. So, for example :

${\sin ^2}\left( {3{x^4} - 5{x^2} + 87} \right) + {\cos ^2}\left( {3{x^4} - 5{x^2} + 87} \right) = 1$
2. $${\tan ^2}\left( \theta \right) + 1 =$$
Show Solution
${\tan ^2}\left( \theta \right) + 1 = {\sec ^2}\left( \theta \right)$

If you know the formula from Problem 1 in this section you can get this one for free.

\begin{align*}{\sin ^2}\left( \theta \right) + {\cos ^2}\left( \theta \right) & = 1\\ \frac{{{{\sin }^2}\left( \theta \right)}}{{{{\cos }^2}\left( \theta \right)}} + \frac{{{{\cos }^2}\left( \theta \right)}}{{{{\cos }^2}\left( \theta \right)}} & = \frac{1}{{{{\cos }^2}\left( \theta \right)}}\\ {\tan ^2}\left( \theta \right) + 1 & = {\sec ^2}\left( \theta \right)\end{align*}

Can you come up with a similar formula relating $${\cot ^2}\left( \theta \right)$$ and $${\csc ^2}\left( \theta \right)$$?

3. $$\sin \left( {2t} \right) =$$
Show Solution
$\sin \left( {2t} \right) = 2\sin \left( t \right)\cos \left( t \right)$

This formula is often used in reverse so that a product of a sine and cosine (with the same argument of course) can be written as a single sine. For example,

\begin{align*}{\sin ^3}\left( {3{x^2}} \right){\cos ^3}\left( {3{x^2}} \right) & = {\left( {\sin \left( {3{x^2}} \right)\cos \left( {3{x^2}} \right)} \right)^3}\\ & = {\left( {\frac{1}{2}\sin \left( {2\left( {3{x^2}} \right)} \right)} \right)^3}\\ & = \frac{1}{8}{\sin ^3}\left( {6{x^2}} \right)\end{align*}

You will find that using this formula in reverse can significantly reduce the complexity of some of the problems that you’ll face in a Calculus class.

4. $$\cos \left( {2x} \right) =$$       (Three possible formulas)
Show Solution

As noted there are three possible formulas to use here.

\begin{align*}\cos \left( {2x} \right) & = {\cos ^2}\left( x \right) - {\sin ^2}\left( x \right)\\ \cos \left( {2x} \right) & = 2{\cos ^2}\left( x \right) - 1\\ \cos \left( {2x} \right) & = 1 - 2{\sin ^2}\left( x \right)\end{align*}

You can get the second formula by substituting $${\sin ^2}\left( x \right) = 1 - {\cos ^2}\left( x \right)$$ (see Problem 1 from this section) into the first. Likewise, you can substitute $${\cos ^2}\left( x \right) = 1 - {\sin ^2}\left( x \right)$$ into the first formula to get the third formula.

5. $${\cos ^2}(x) =$$       (In terms of cosine to the first power)
Show Solution
${\cos ^2}(x) = \frac{1}{2}\left( {1 + \cos \left( {2x} \right)} \right)$

This is really the second formula from Problem 4 in this section rearranged and is VERY useful for eliminating even powers of cosines. For example,

\begin{align*}5{\cos ^2}\left( {3x} \right) & = 5\left( {\frac{1}{2}\left( {1 + \cos \left( {2\left( {3x} \right)} \right)} \right)} \right)\\ & = \frac{5}{2}\left( {1 + \cos \left( {6x} \right)} \right)\end{align*}

Note that you probably saw this formula written as

$\cos \left( {\frac{x}{2}} \right) = \pm \sqrt {\frac{1}{2}\left( {1 + \cos \left( x \right)} \right)}$

in a trig class and called a half-angle formula.

6. $${\sin ^2}(x) =$$       (In terms of cosine to the first power)
Show Solution
${\sin ^2}(x) = \frac{1}{2}\left( {1 - \cos \left( {2x} \right)} \right)$

As with the previous problem this is really the third formula from Problem 4 in this section rearranged and is very useful for eliminating even powers of sine. For example,

\begin{align*}4{\sin ^4}\left( {2t} \right) & = 4{\left( {{{\sin }^2}\left( {2t} \right)} \right)^2}\\ & = 4{\left( {\frac{1}{2}\left( {1 - \cos \left( {4t} \right)} \right)} \right)^2}\\ & = 4\left( {\frac{1}{4}} \right)\left( {1 - 2\cos \left( {4t} \right) + {{\cos }^2}\left( {4t} \right)} \right)\\ & = 1 - 2\cos \left( {4t} \right) + \frac{1}{2}\left( {1 + \cos \left( {8t} \right)} \right)\\ & = \frac{3}{2} - 2\cos \left( {4t} \right) + \frac{1}{2}\cos \left( {8t} \right)\end{align*}

As shown in this example you may have to use both formulas and more than once if the power is larger than 2 and the answer will often have multiple cosines with different arguments.

Again, in a trig class, this was probably called a half-angle formula and written as,

$\sin \left( {\frac{x}{2}} \right) = \pm \sqrt {\frac{1}{2}\left( {1 - \cos \left( x \right)} \right)}$
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}