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Sum and Difference Identities

Sine, cosine, or tangent of two angles that are added or subtracted.

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Pythagorean Identities

The Pythagorean Theorem works on right triangles.  If you consider the \begin{align*}x\end{align*}x coordinate of a point along the unit circle to be the cosine and the \begin{align*}y\end{align*}y coordinate of the point to be the sine and the distance to the origin to be 1 then the Pythagorean Theorem immediately yields the identity:

\begin{align*}& y^2+x^2=1\\ & \sin^2 x+\cos^2 x=1\end{align*}y2+x2=1sin2x+cos2x=1

An observant student may guess that other Pythagorean identities exist with the rest of the trigonometric functions.  Is \begin{align*}\tan^2 x+\cot^2 x=1\end{align*}tan2x+cot2x=1 a legitimate identity? 

Watch This

http://www.youtube.com/watch?v=OmJ5fxyXrfg James Sousa: Fundamental Identities: Reciprocal, Quotient, Pythagorean

Guidance

The proof of the Pythagorean identity for sine and cosine is essentially just drawing a right triangle in a unit circle, identifying the cosine as the \begin{align*}x\end{align*}x coordinate, the sine as the \begin{align*}y\end{align*}y coordinate and 1 as the hypotenuse.

\begin{align*}\cos^2 x+\sin^2 x=1\end{align*}cos2x+sin2x=1

Most people rewrite the order of the sine and cosine so that the sine comes first.

\begin{align*}\sin^2 x+\cos^2 x=1\end{align*}sin2x+cos2x=1

The two other Pythagorean identities are:

  • \begin{align*}1+\cot^2 x=\csc^2 x\end{align*}1+cot2x=csc2x
  • \begin{align*}\tan^2 x+1=\sec^2 x\end{align*}tan2x+1=sec2x

To derive these two Pythagorean identities, divide the original Pythagorean identity by \begin{align*}\sin^2 x\end{align*}sin2x and \begin{align*}\cos^2 x\end{align*}cos2x respectively.

Example A

Derive the following Pythagorean identity: \begin{align*}1+\cot^2 x=\csc^2 x\end{align*}1+cot2x=csc2x.

Solution:  First start with the original Pythagorean identity and then divide through by \begin{align*}\sin^2 x\end{align*}sin2x and simplify.

\begin{align*}\frac{\sin^2 x}{\sin^2 x}+\frac{\cos^2 x}{\sin^2 x} &= \frac{1}{\sin^2 x}\\ 1+\cot^2 x &= \csc^2 x\end{align*}sin2xsin2x+cos2xsin2x1+cot2x=1sin2x=csc2x

Example B

Simplify the following expression: \begin{align*}\frac{\sin x (\csc x-\sin x)}{1-\sin x}\end{align*}sinx(cscxsinx)1sinx.

Solution:

\begin{align*}\frac{\sin x (\csc x-\sin x)}{1-\sin x} &= \frac{\sin x \cdot \csc x-\sin^2 x}{1-\sin x}\\ &= \frac{1-\sin^2 x}{1-\sin x}\\ &= \frac{(1-\sin x)(1+\sin x)}{1-\sin x}\\ &= 1+\sin x\end{align*}sinx(cscxsinx)1sinx=sinxcscxsin2x1sinx=1sin2x1sinx=(1sinx)(1+sinx)1sinx=1+sinx

Note that factoring the Pythagorean identity is one of the most powerful applications.  This is very common and is a technique that you should feel comfortable using. 

Example C

Prove the following trigonometric identity: \begin{align*}(\sec^2 x+\csc^2 x)-(\tan^2 x+\cot^2 x)=2\end{align*}(sec2x+csc2x)(tan2x+cot2x)=2.

Solution:  Group the terms and apply a different form of the second two Pythagorean identities which are \begin{align*}1+\cot^2 x=\csc^2 x\end{align*}1+cot2x=csc2x and \begin{align*}\tan^2 x+1=\sec^2 x\end{align*}tan2x+1=sec2x.

\begin{align*}(\sec^2 x+\csc^2 x)-(\tan^2 x+\cot^2 x) &= \sec^2 x-\tan^2 x+\csc^2 x-\cot^2 x\\ &= 1+1\\ &= 2\end{align*}(sec2x+csc2x)(tan2x+cot2x)=sec2xtan2x+csc2xcot2x=1+1=2

Concept Problem Revisited

Cofunctions are not always connected directly through a Pythagorean identity.

\begin{align*}\tan^2 x+\cot^2 x \neq 1\end{align*}tan2x+cot2x1

Visually, the right triangle connecting tangent and secant can also be observed in the unit circle.  Most people do not know that tangent is named “tangent” because it refers to the distance of the line tangent from the point on the unit circle to the \begin{align*}x\end{align*}x axis.  Look at the picture below and think about why it makes sense that \begin{align*}\tan x\end{align*} and \begin{align*}\sec x\end{align*} are as marked. \begin{align*}\tan x=\frac{opp}{adj}\end{align*}.  Since the adjacent side is equal to 1 (the radius of the circle), \begin{align*}\tan x\end{align*} simply equals the opposite side.  Similar logic can explain the placement of \begin{align*}\sec x\end{align*}.

Vocabulary

The Pythagorean Theorem states that the sum of the squares of the two legs in a right triangle will always be the square of the hypotenuse. 

The Pythagorean Identity states that since sine and cosine are equal to two legs in a right triangle with a hypotenuse of 1, then their relationship is that of the Pythagorean Theorem. 

Guided Practice

1. Derive the following Pythagorean identity:

\begin{align*}\tan^2 x+1=\sec^2 x\end{align*}

2. Simplify the following expression.

\begin{align*}(\sec^2 x)(1-\sin^2 x)-\left(\frac{\sin x}{\csc x}+\frac{\cos x}{\sec x}\right)\end{align*}

3. Simplify the following expression.

\begin{align*}(\cos t-\sin t)^2+(\cos t+\sin t)^2\end{align*}

Answers:

1. 

\begin{align*}\frac{\sin^2 x}{\cos^2 x}+\frac{\cos^2 x}{\cos^2 x}&= \frac{1}{\cos^2 x}\\ \tan^2 x+1 &= \sec^2 x\end{align*}

2.

\begin{align*}&(\sec^2 x)(1-\sin^2 x)-\left(\frac{\sin x}{\csc x}+\frac{\cos x}{\sec x}\right)\\ &=\sec^2 x \cdot \cos^2 x-(\sin^2 x+\cos^2 x)\\ &= 1-1\\ &= 0\end{align*}

3. Note that initially, the expression is not the same as the Pythagorean identity.

\begin{align*}&(\cos t-\sin t)^2+(\cos t+\sin t)^2 \\ &= \cos^2 t-2 \cos t \sin t+\sin^2 t +\cos^2 t+2 \cos t \sin t+\sin^2 t\\ &= 1-2 \cos t \sin t+1+2 \cos t \sin t\\ &= 2\end{align*}

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