5.3 Sum and Difference Identities for Cosine

The sum and difference identities are identities that express sine, cosine, and tangent of the sum or difference of two angles in terms of the sine, cosine, and tangent of the individual angles.

Let’s denote the two angles as alpha (α) and beta (β).

\text{cos}(\alpha+\beta)=\text{cos}\alpha\text{cos}\beta-\text{sin}\alpha\text{sin}\beta

cos(\alpha-\beta) = \text{cos}\alpha\text{cos}\beta+\text{sin}\alpha\text{sin}\beta

Be sure to note that:

\text{cos}(\alpha+\beta) \neq \text{cos}\alpha+\text{cos}\beta

\text{cos}(\alpha-\beta)\neq\text{cos}\alpha-\text{cos}\beta

Cofunction Identities

\text{cos}(\frac{\pi}{2}-\theta) = \text{sin}\theta, \text{sin}(\frac{\pi}{2}-\theta)=\text{cos}\theta

\text{sec}(\frac{\pi}{2}-\theta)=\text{csc}\theta, \text{csc}(\frac{\pi}{2}-\theta)=\text{sec}\theta

\text{tan}(\frac{\pi}{2}-\theta)=\text{cot}\theta, \text{cot}(\frac{\pi}{2}-\theta)=\text{tan}\theta

Examples

1.

\text{cos}(\frac{\pi}{2}-\theta) = \text{cos}\frac{\pi} {2}\text{cos}\theta+\text{sin}\frac{\pi}{2}\text{sin}\theta \\ =0(\text{cos}\theta)+(1)\text{sin}\theta \\ = \text{sin}\theta

Recall where pi over 2 falls on our unit circle:

The two x and points refer to cosine and sine, respectively. Plug these in for our identity and you get your answer of simply sinθ.

2.

\text{sin}(\frac{\pi}{2}-x) = \text{cos}x \ \ \ \text{(cofunction identity)}

\text{cos}(x-\frac{\pi}{2}) = \text{cos}x\text{cos}\frac{\pi}{2}+\text{sin}x\text{sin}\frac{\pi}{2} \\ =\text{cos}x(0)+\text{sin}x(1) \\ =\text{sin}x

\text{sin}(x-\frac{\pi}{2}) = \text{cos}(\frac{\pi}{2}-(x-\frac{\pi}{2}) \ *\text{cofunction identity with different formula}

Distribute the negative sign after pi over two to the x and to the negative pi over two.

=\text{cos}(\frac{\pi}{2}-x+\frac{\pi}{2}) \\ = \text{cos}(\pi+x) \\ = \text{cos}\pi\text{cos}x+\text{sin}\pi\text{sin}x \\ =(-1)\text{cos}x+0\text{sin}x \\ =\boxed{-\text{cos}x}

3.

\text{cos}15\degree = \text{cos}(60\degree-45\degree) = \text{cos}60\degree\text{cos}45\degree+\text{sin}60\degree\text{sin}45\degree \\ =(\frac{1}{2})(\frac{\sqrt{2}}{2}) +(\frac{\sqrt{3}}{2})(\frac{\sqrt{3}}{2}) \\=\frac{\sqrt{2}}{4} +\frac{\sqrt{6}}{4} = \boxed{\frac{\sqrt{2}+\sqrt{6}}{4}}

4.

\text{cos}(-105\degree)=\text{cos}(105\degree)=\text{cos}(60\degree+45\degree) \\ = \text{cos}60\degree\text{cos}45\degree-\text{sin}60\degree\text{sin}45\degree \\ = (\frac{1}{2})(\frac{\sqrt{2}}{2})-(\frac{\sqrt{3}}{2})(\frac{\sqrt{3}}{2}) \\ =\frac{\sqrt{2}}{4}-\frac{\sqrt{6}}{4} = \boxed{\frac{\sqrt{2}-\sqrt{6}}{4}}

5.

\text{cos}(-\frac{5\pi}{12})=\text{cos}(\frac{5\pi}{12}) = \text{cos}(75\degree) \\ =\text{cos}(45\degree+30\degree) \\ = \text{cos}45\degree\text{cos}30\degree-\text{sin}45\degree\text{sin}30\degree \\ = (\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2})-(\frac{\sqrt{2}}{2})(\frac{1}{2}) \\ =\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4} = \boxed{\frac{\sqrt{6}-\sqrt{2}}{4}}

If it’s not an angle you know, you have to use the sum and difference formula.

6.

Simplify:

(\text{cos}40\degree)(\text{cos}50\degree)-(\text{sin}40\degree)(\text{sin}50\degree) \\ = \text{cos}(40\degree+50\degree) \\ = \text{cos}(90\degree) = \boxed{0}

7.

Write in terms of cofunction

\text{csc}\frac{5\pi}{24}

=\text{sec}(\frac{\pi}{2}-\frac{5\pi}{24}) \\ =\text{sec}(\frac{\pi}{2}\cdot\frac{12}{12}-\frac{5\pi}{24}) \\ =\text{sec}(\frac{12\pi}{24} - \frac{5\pi}{24}) = \boxed{\text{sec}\frac{7\pi}{24}}

8.

\text{csc}\frac{\pi}{3} = \underline{\space\space\space\space\space\space\space\space}\frac{\pi}{6} \\ \text{csc}60\degree = \text{sec}30\degree

\text{csc}\frac{\pi}{3} = \text{sec}(\frac{\pi}{2}-\frac{\pi}{3})

Multiply pi over two by 3 over 3 to get 3 pi over 6, and multiply pi over 3 by 2 over 2 to get 2 pi over 6.

=\text{sec}(\frac{3\pi}{6}-\frac{2\pi}{6}) \\ = \text{sec}\frac{\pi}{6}

9.

\text{csc}\theta = \text{sec}(4\theta+65)

Find theta

\theta+4\theta+65=90\degree \\5\theta+65=90\degree \\5\theta=25 \\\theta=5\degree

\text{sec}(90-\theta) = \text{sec}(4\theta+65) \\90-\theta=4\theta+65 \\90=5\theta+65 \\25=5\theta \\\theta+5

10.

Find cos(s+t) and cos(s-t)

\text{sin s } = \frac{4}{5}, \text{ sin t } = -\frac{12}{13}, \text{ s is in Q1, t is in Q3}

Method 1 – Draw a picture

Because of SOH CAH TOA, we know that sin is opposite over hypotenuse, so we can go ahead and make and label our triangles.

Sin s

sin t

\text{cos}(s+t)= \text{cos}s\text{cos}t-\text{sin}s\text{sin}t \\ =(\frac{3}{5})(-\frac{5}{13})-(\frac{4}{5})(-\frac{12}{13}) \\ =-\frac{15}{65}--\frac{48}{65}=-\frac{15}{65}+\frac{48}{65}=\boxed{\frac{33}{65}}

\text{cos}(s-t)=\text{cos}s\text{cos}t+\text{sin}s\text{sin}t\\ =(\frac{3}{5})(-\frac{5}{13})+(\frac{4}{5})(-\frac{12}{13}) \\ =-\frac{15}{65}+-\frac{48}{65}=\boxed{-\frac{63}{65}}

11.

\text{sin}s=\frac{3}{5}, \text{cos}t=-\frac{12}{13}, \text{ s and t are in quadrant 2}

Find cos(s+t).

Method 2

=\text{cos}s\text{cos}t-\text{sin}s\text{sin}t

Use the Pythagorean Identity:

\text{sin}^2\theta+\text{cos}^2\theta=1 \\(\frac{3}{5})^2+\text{cos}^2s=1 \\\frac{9}{25}+\text{cos}^2s=\frac{25}{25}-\frac{9}{25} \\ \text{cos}^2s=\frac{16}{25} \\\text{cos}^2s=\pm\sqrt{\frac{16}{25}} \\ \text{cos}s=-\frac{4}{5}

\text{sin}^2t+\text{cos}^2t=1

\text{sin}^2t+(-\frac{12}{13}) = 1 \\ \text{sin}^2t+\frac{144}{169}=1 \\ \text{sin}^2t=\frac{169}{169}-\frac{144}{169} \\ \text{sin}^2t=\frac{25}{169} \\ \text{sin}t=\pm\sqrt{\frac{25}{169}}=\frac{5}{13}

\text{cos}(s+t)=(-\frac{4}{5})(-\frac{12}{13}-(\frac{3}{5})(\frac{5}{13})=\frac{48}{65}-\frac{15}{65} = \boxed{\frac{33}{65}}

12.

Verify:

\text{Hint: cos2}x=\text{cos}(x+x)

\text{cos}(2x)=\text{cos}^2x-\text{sin}^2x

\text{cos}(2x) \\=\text{cos}(x+x) \\=\text{cos}x\text{cos}x-\text{sin}x\text{sin}x \\=\text{cos}^2x-\text{sin}^2x\checkmark

13.

\text{Note the common household electric current called ac... voltage V}\\\text{ in a typical 115v outlet is expressed by the function}\\ \text{V(t)=163sin}\omega\text{t, where }\omega \text{ is angular speed (in radians/sec) of rotating} \\ \text{ generator at electric plant and t is time in seconds. It is essential} \\ \text{ to rotate at precisely 60 x/sec}. \\ \text{How many times does it rotate in .5 seconds?}

\text{a) } \frac{60 \text{cycles}}{1 \text{sec}}\cdot\frac{0.50\text{sec}}{1}=30\text{times}

Is it always equal to 115?

V(t)=163\text{sin}\omega\text{t}

Amp = 163, Range: [-163, 163]

Max = 163V, Min = -163V

No, it is not always equal to 115.