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,

   \[\sin \left( {\frac{x}{2}} \right) = \pm \sqrt {\frac{1}{2}\left( {1 - \cos \left( x \right)} \right)} \]