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    Home / Algebra Trig Review / Trigonometry / Trig Formulas
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    I apologize for the outage on the site yesterday and today. Lamar University is in Beaumont Texas and Hurricane Laura came through here and caused a brief power outage at Lamar. Things should be up and running at this point and (hopefully) will stay that way, at least until the next hurricane comes through here which seems to happen about once every 10-15 years. Note that I wouldn't be too suprised if there are brief outages over the next couple of days as they work to get everything back up and running properly. I apologize for the inconvienence.

    August 27, 2020

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    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,

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