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3.2.3: Proofs of Trigonometric Identities

  • Page ID
    4227
  • Convert to sine/cosine, use basic identities, and simplify sides of the equation.

    Verify that \(\dfrac{\sin^2 x}{\tan^2 x}=1−\sin^2 x\).

    Verifying Trigonometric Identities

    Now that you are comfortable simplifying expressions, we will extend the idea to verifying entire identities. Here are a few helpful hints to verify an identity:

    • Change everything into terms of sine and cosine.
    • Use the identities when you can.
    • Start with simplifying the left-hand side of the equation, then, once you get stuck, simplify the right-hand side. As long as the two sides end up with the same final expression, the identity is true.

    Let's verify the following identities.

    1. \(\dfrac{\cot^2 x}{\csc x}=\csc x−\sin x\)

    Rather than have an equal sign between the two sides of the equation, we will draw a vertical line so that it is easier to see what we do to each side of the equation. Start with changing everything into sine and cosine.

    \(\begin{array}{l|l}
    \dfrac{\cot ^{2} x}{\csc x} & \csc x-\sin x \\
    \dfrac{\dfrac{\cos ^{2} x}{\sin ^{2} x}}{\dfrac{1}{\sin x}} & \dfrac{1}{\sin x} \\
    \dfrac{\cos ^{2} x}{\sin x} &
    \end{array}\)

    Now, it looks like we are at an impasse with the left-hand side. Let’s combine the right-hand side by giving them same denominator.

    \(\begin{array}{|c}
    \dfrac{1}{\sin x}-\dfrac{\sin ^{2} x}{\sin x} \\
    \dfrac{1-\sin ^{2} x}{\sin x} \\
    \dfrac{\cos ^{2} x}{\sin x}
    \end{array}\)

    The two sides reduce to the same expression, so we can conclude this is a valid identity. In the last step, we used the Pythagorean Identity, \(\sin^2 \theta +\cos^2 \theta =1\), and isolated the \(\cos^2 x=1−\sin^2x\).

    There are usually more than one way to verify a trig identity. When proving this identity in the first step, rather than changing the cotangent to\(\dfrac{\cos^2 x}{\sin^2 x}\), we could have also substituted the identity \(\cot^2 x=\csc^2 x−1\).

    1. \(\dfrac{\sin x}{1−\cos x}=\dfrac{1+\cos x}{\sin x}\)

    Multiply the left-hand side of the equation by \(\dfrac{1+\cos x}{1+\cos x}\).

    \(\begin{aligned}
    \dfrac{\sin x}{1-\cos x} &=\dfrac{1+\cos x}{\sin x} \\
    \dfrac{1+\cos x}{1+\cos x} \cdot \dfrac{\sin x}{1-\cos x} &=\\
    \dfrac{\sin (1+\cos x)}{1-\cos ^{2} x} &=\\
    \dfrac{\sin (1+\cos x)}{\sin ^{2} x} &=\\
    \dfrac{1+\cos x}{\sin x} &=
    \end{aligned}\)

    The two sides are the same, so we are done.

    1. \(\sec(−x)=\sec x\)

    Change secant to cosine.

    \(\sec(−x)=\dfrac{1}{cos(−x)}\)

    From the Negative Angle Identities, we know that \(\cos(−x)=\cos x\).

    \(\begin{aligned} &=\dfrac{1}{\cos x} \\&=\sec x\end{aligned}\)

    Example \(\PageIndex{1}\)

    Earlier, you were asked to verify that \(\sin^2 x \tan^2 x=1−\sin^2 x\). 

    Solution

    Start by simplifying the left-hand side of the equation.

    \(\sin ^2 x \tan ^2 x=\dfrac{\sin ^2 x}{\dfrac{\sin ^2 x}{\cos ^2 x}}=\cos ^2 x\)

    Now simplify the right-hand side of the equation. By manipulating the Trigonometric Identity,

    \(\sin ^2 x+\cos ^2 x=1\), we get \(\cos ^2 x=1−\sin ^2 x\).

    \(\cos ^2 x=\cos ^2 x\) and the equation is verified.

    Verify the following identities.

    Example \(\PageIndex{2}\)

    \(\cos x \sec x=1\)

    Solution

    Change secant to cosine.

    \(\begin{aligned} \cos x \sec x&=\cos x \cdot \dfrac{1}{\cos x} \\&=1\end{aligned}\)

    Example \(\PageIndex{3}\)

    \(2−sec^2 x=1−\tan ^2 x\)

    Solution

    Use the identity \(1+\tan^2 \theta =\sec^2 \theta\).

    \(\begin{aligned} 2−\sec^2 x &=2−(1+\tan ^2 x) \\&=2−1−\tan ^2 x \\&=1−\tan ^2 x\end{aligned}\)

    Example \(\PageIndex{4}\)

    \(\dfrac{\cos(−x)}{1+\sin(−x)}=\sec x+\tan x\)

    Solution

    Here, start with the Negative Angle Identities and multiply the top and bottom by \(\dfrac{1+\sin x}{1+\sin x}\) to make the denominator a monomial.

    \(\begin{aligned}
    \dfrac{\cos (-x)}{1+\sin (-x)} &=\dfrac{\cos x}{1-\sin x} \cdot \dfrac{1+\sin x}{1+\sin x} \\
    &=\dfrac{\cos x(1+\sin x)}{1-\sin ^{2} x} \\
    &=\dfrac{\cos x(1+\sin x)}{\cos ^{2} x} \\
    &=\dfrac{1+\sin x}{\cos x} \\
    &=\dfrac{1}{\cos x}+\dfrac{\sin x}{\cos x} \\
    &=\sec x+\tan x
    \end{aligned}\)

    Review

    Verify the following identities.

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

    Answers for Review Problems

    To see the Review answers, open this PDF file and look for section 14.9. 

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