Skip to main content
K12 LibreTexts

3.2.2: Simplifying Trigonometric Expressions

  • Page ID
    4226
  • Convert to sine/cosine and use basic trig identities to simplify.

    How could you write the trigonometric function \(\cos \theta +\cos \theta (\tan^2 \theta )\) more simply?

    Simplifying Trigonometric Expressions

    Now that you are more familiar with trig identities, we can use them to simplify expressions. Remember, that you can use any of the following identities.

    Reciprocal Identities: \(\csc \theta =\dfrac{1}{\sin \theta}\), \(\sec \theta =\dfrac{1}{\cos \theta}\), and \(\cot \theta =\dfrac{1}{\tan \theta}\)

    Tangent and Cotangent Identities: \(\tan \theta =\dfrac{\sin \theta}{\cos \theta}\) and \(\cot \theta =\dfrac{\cos \theta}{\sin \theta}\)

    Pythagorean Identities: \(\sin^2 \theta +\cos ^2 \theta =1\), \(1+\tan^2 \theta =\sec^2 \theta\), and \(1+\cot^2 \theta =\csc^2 \theta\)

    Cofunction Identities: \(\sin(\dfrac{\pi}{2}− \theta )=\cos \theta\), \(\cos(\dfrac{\pi}{2}− \theta )=\sin \theta\), and \(\tan(\dfrac{\pi}{2}− \theta )=\cot \theta\)

    Negative Angle Identities: \(\sin(− \theta )=−sin \theta\), \(\cos(− \theta )=\cos \theta\), and \(\tan(− \theta )=−\tan \theta\)

    Let's simplify the following expressions.

    1. \(\dfrac{\sec x}{\sec x−1}\)

    When simplifying trigonometric expressions, one approach is to change everything into sine or cosine. First, we can change secant to cosine using the Reciprocal Identity.

    \(\dfrac{\sec x}{\sec x−1} \rightarrow \dfrac{\dfrac{1}{\cos x}}{\dfrac{1}{\cos x}−1}\)

    Now, combine the denominator into one fraction by multiplying 1 by \(\dfrac{\cos x}{\cos x}\).

    \(\dfrac{\dfrac{1}{\cos x}}{\dfrac{1}{\cos x}-1} \rightarrow \dfrac{\dfrac{1}{\cos x}}{\dfrac{1}{\cos x}-\dfrac{\cos x}{\cos x}} \rightarrow \dfrac{\dfrac{1}{\cos x}}{\dfrac{1-\cos x}{\cos x}}\)

    Change this problem into a division problem and simplify.

    \(\begin{aligned}
    \dfrac{\dfrac{1}{\cos x}}{\dfrac{1-\cos x}{\cos x}} & \rightarrow \dfrac{1}{\cos x} \div \dfrac{1-\cos x}{\cos x} \\
    & \dfrac{1}{\cancel{\cos x}} \cdot \dfrac{\cancel{\cos x}}{1-\cos x} \\
    & \dfrac{1}{1-\cos x}
    \end{aligned}\)

    1. \(\dfrac{\sin^4 x−\cos^4 x}{\sin^2 x−\cos ^2 x}\)

    With this problem, we need to factor the numerator and denominator and see if anything cancels. The rules of factoring a quadratic and the special quadratic formulas can be used in this scenario.

    \(\dfrac{\sin^4 x−\cos^4 x}{\sin^2 x−\cos ^2 x} \rightarrow \dfrac{\cancel{(\sin^2 x−\cos ^2 x)} (\sin^2 x+\cos ^2 x)}{\cancel{(\sin^2 x−\cos ^2 x)}}\rightarrow \sin^2 x+\cos ^2 x\rightarrow 1\)

    In the last step, we simplified to the left hand side of the Pythagorean Identity. Therefore, this expression simplifies to 1.

    1. \(\sec \theta \tan^2 \theta +\sec \theta\)

    First, pull out the GCF.

    \(\sec \theta \tan^2 \theta + \sec \theta \rightarrow \sec \theta (\tan^2 \theta +1)\)

    Now, \(\tan^2 \theta +1=\sec^2 \theta\) from the Pythagorean Identities, so simplify further.

    \(\sec \theta (\tan^2 \theta +1)\rightarrow \sec \theta \cdot \sec^2 \theta \rightarrow \sec^3 \theta\)

    Example \(\PageIndex{1}\)

    Earlier, you were asked to simplify the trigonometric function \(\cos \theta +\cos \theta (\tan^2 \theta )\).

    Solution

    Notice that the terms in the expression \(\cos \theta +\cos \theta (\tan^2 \theta )\) have a common factor of \(cos \theta\), so start by factoring this common term out.

    \(\cos \theta +\cos \theta (\tan^2 \theta ) \\ \cos \theta (1+\tan^2 \theta )\)

    Now, use the trigonometric identity \(1+\tan^2 \theta =\sec^2 \theta\), substitute, and simplify.

    \(\begin{aligned} \cos \theta (1+tan^2 \theta ) &=\cos \theta (\sec^2 \theta ) \\ &=\cos \theta \left(\dfrac{1}{\cos ^2 \theta}\right) \\&=\dfrac{1}{\cos \theta}\\ &=\sec \theta \end{aligned}\)

    Simplify the following trigonometric expressions.

    Example \(\PageIndex{2}\)

    \(\cos\left(\dfrac{\pi}{2}−x\right)\cot x\)

    Solution

    Use the Cotangent Identity and the Cofunction Identity \(\cos(\dfrac{\pi}{2}− \theta )=\sin \theta\).

    \(\cos(\dfrac{\pi}{2}−x)\cot x\rightarrow \cancel{\sin x} \cdot \dfrac{\cos x}{\cancel{\sin x}}\rightarrow \cos x\)

    Example \(\PageIndex{3}\)

    \(\dfrac{\sin(−x)\cos x}{\tan x}\)

    Solution

    Use the Negative Angle Identity and the Tangent Identity.

    \(\dfrac{\sin (-x) \cos x}{\tan x} \rightarrow \dfrac{-\sin x \cos x}{\dfrac{\sin x}{\cos x}} \rightarrow -\cancel{\sin x} \cos x \cdot \dfrac{\cos x}{\sin x} \rightarrow-\cos ^{2} x\)

     

    Example \(\PageIndex{4}\)

    \(\dfrac{\cot x \cos x}{\tan (-x) \sin \left(\dfrac{\pi}{2}-x\right)}\)

    Solution

    In this problem, you will use several identities.

    \(\dfrac{\cot x \cos x}{\tan (-x) \sin \left(\dfrac{\pi}{2}-x\right)} \rightarrow \dfrac{\dfrac{\cos x}{\sin x} \cdot \cos x}{-\dfrac{\sin x}{\cancel{\cos x}} \cdot \cancel{\cos x}} \rightarrow \dfrac{\dfrac{\cos ^{2} x}{\sin x}}{-\sin x} \rightarrow \dfrac{\cos ^{2} x}{\sin x} \cdot-\dfrac{1}{\sin x} \rightarrow-\dfrac{\cos ^{2} x}{\sin ^{2} x} \rightarrow −\cot ^2 x\)

     

    Review

    Simplify the following expressions.

    1. \(\cot x \sin x\)
    2. \(\cos ^2 x \tan(−x)\)
    3. \(\dfrac{cos(−x)}{\sin(−x)}\)
    4. \(\sec x \cos(−x)−\sin^2 x\)
    5. \(\sin x(1+cot^2 x)\)
    6. \(1−\sin^2\left(\dfrac{\pi}{2}−x\right)\)
    7. \(1−\cos^2\left(\dfrac{\pi}{2}−x\right)\)
    8. \(\dfrac{\tan\left(\dfrac{\pi}{2}−x\right) \sec x}{1−\csc^2 x}\)
    9. \(\dfrac{\cos ^2 x \tan^2 x−1 }{\cos ^2 x}\)
    10. \(\cot^2 x+\sin ^2x+\cos^2(−x)\)
    11. \(\dfrac{\sec x\sin x+\cos\left(\dfrac{\pi}{2}−x\right)}{1+\cos x}\)
    12. \(\dfrac{\cos(−x)}{1+\sin(−x)}\)
    13. \(\dfrac{\sin^2(−x)}{\tan^2 x}\)
    14. \(\tan\left(\dfrac{\pi}{2}−x\right)\cot x−\csc^2 x\)
    15. \(\dfrac{\csc x(1−\cos ^2 x)}{\sin x\cos x}\)

    Answers for Review Problems

    To see the Review answers, open this PDF file and look for section 14.8. 

    Additional Resources

    Practice: Simplifying Trigonometric Expressions

     

    • Was this article helpful?