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The cosine and the sine of a sum and of a difference

Trigonometry, grade 11 (5 units) Lesson 3 of 18Learning

The cosine and the sine of a sum and of a difference

cos⁡(α−β)\displaystyle \cos(\alpha-\beta) is not cos⁡α−cos⁡β\displaystyle \cos\alpha-\cos\beta

The cosine of a difference is not a difference of cosines. For example, cos⁡(π2−0)=0\displaystyle {\cos\left(\frac{\pi}{2}-0\right)}={0}, but cos⁡π2−cos⁡0=−1\displaystyle {\cos\frac{\pi}{2}-\cos0}={-1}.

The right formula brings in the sines as well:

cos⁡(α−β)\displaystyle \cos(\alpha-\beta) =\displaystyle = cos⁡α⋅cos⁡β+sin⁡α⋅sin⁡β\displaystyle \cos\alpha\cdot\cos\beta+\sin\alpha\cdot\sin\beta
A proof of the formula

The proof below works out the distance between two points of the unit circle in two ways – by the distance formula and by the law of cosines – and compares them.
A=(cos⁡α, sin⁡α)\displaystyle {A}={(\cos\alpha,\ \sin\alpha)} and B=(cos⁡β, sin⁡β)\displaystyle {B}={(\cos\beta,\ \sin\beta)} are the points of the angles α\displaystyle \alpha and β\displaystyle \beta, and O\displaystyle O is the origin.
The angle AOB\displaystyle AOB is α−β\displaystyle \alpha-\beta (when 0<α−β<π\displaystyle 0<\alpha-\beta<\pi; the other cases go the same way).

ClaimReason
1AB2=(cos⁡α−cos⁡β)2+(sin⁡α−sin⁡β)2\displaystyle {AB^2}={(\cos\alpha-\cos\beta)^2+(\sin\alpha-\sin\beta)^2}The distance between two points
2AB2=cos⁡2α+sin⁡2α+cos⁡2β+sin⁡2β−2(cos⁡αcos⁡β+sin⁡αsin⁡β)\displaystyle {AB^2}={\cos^2\alpha+\sin^2\alpha+\cos^2\beta+\sin^2\beta-2(\cos\alpha\cos\beta+\sin\alpha\sin\beta)}Expanding the brackets
3AB2=2−2(cos⁡αcos⁡β+sin⁡αsin⁡β)\displaystyle {AB^2}={2-2(\cos\alpha\cos\beta+\sin\alpha\sin\beta)}sin⁡2+cos⁡2=1\displaystyle {\sin^2+\cos^2}={1}, twice
4AB2=12+12−2⋅1⋅1⋅cos⁡(α−β)=2−2cos⁡(α−β)\displaystyle {AB^2}={1^2+1^2-2\cdot1\cdot1\cdot\cos(\alpha-\beta)}={2-2\cos(\alpha-\beta)}The law of cosines in triangle AOB\displaystyle AOB: OA=OB=1\displaystyle {OA}={OB}={1}
5cos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β\displaystyle {\cos(\alpha-\beta)}={\cos\alpha\cos\beta+\sin\alpha\sin\beta}Comparing (3) and (4)
The other three formulas

Three identities you already know give the other formulas:

  • Put −β\displaystyle -\beta in place of β\displaystyle \beta, and use cos⁡(−β)=cos⁡β\displaystyle {\cos(-\beta)}={\cos\beta} and sin⁡(−β)=−sin⁡β\displaystyle {\sin(-\beta)}={-\sin\beta}:
cos⁡(α+β)\displaystyle \cos(\alpha+\beta) =\displaystyle = cos⁡α⋅cos⁡β−sin⁡α⋅sin⁡β\displaystyle \cos\alpha\cdot\cos\beta-\sin\alpha\cdot\sin\beta
  • sin⁡(α+β)=cos⁡(π2−α−β)=cos⁡((π2−α)−β)\displaystyle {\sin(\alpha+\beta)}={\cos\left(\frac{\pi}{2}-\alpha-\beta\right)}={\cos\left(\left(\frac{\pi}{2}-\alpha\right)-\beta\right)}, and by the first formula:
sin⁡(α+β)\displaystyle \sin(\alpha+\beta) =\displaystyle = sin⁡α⋅cos⁡β+cos⁡α⋅sin⁡β\displaystyle \sin\alpha\cdot\cos\beta+\cos\alpha\cdot\sin\beta
  • Put −β\displaystyle -\beta in place of β\displaystyle \beta:
sin⁡(α−β)\displaystyle \sin(\alpha-\beta) =\displaystyle = sin⁡α⋅cos⁡β−cos⁡α⋅sin⁡β\displaystyle \sin\alpha\cdot\cos\beta-\cos\alpha\cdot\sin\beta

Mind the signs: the cosine of a sum has a minus, and the sine of a sum a plus.

The exact value of cos⁡15∘\displaystyle \cos15^\circ

An angle that is a sum or a difference of angles from the table gets an exact value:

cos⁡15∘=cos⁡(45∘−30∘)=cos⁡45∘cos⁡30∘+sin⁡45∘sin⁡30∘=22⋅32+22⋅12=6+24\begin{aligned}\cos15^\circ&=\cos(45^\circ-30^\circ)\\&=\cos45^\circ\cos30^\circ+\sin45^\circ\sin30^\circ\\&=\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\cdot\frac{1}{2}\\&=\frac{\sqrt{6}+\sqrt{2}}{4}\end{aligned}

In radians this is cos⁡π12\displaystyle \cos\frac{\pi}{12}, since π12=π4−π6\displaystyle {\frac{\pi}{12}}={\frac{\pi}{4}-\frac{\pi}{6}}.

Exercise 1

Find the exact value. A hint: 75∘=45∘+30∘\displaystyle {75^\circ}={45^\circ+30^\circ}.

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    Exercise 2

    Find the exact value. A hint: 7π12=π3+π4\displaystyle {\frac{7\pi}{12}}={\frac{\pi}{3}+\frac{\pi}{4}}.

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      Exercise 3

      Find the exact value.

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        Exercise 4

        Work it out without a calculator. Which formula is written here from right to left?

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          Exercise 5

          Simplify until you reach cos⁡x\displaystyle \cos x.

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            Exercise 6

            Simplify until you reach cos⁡x\displaystyle \cos x.

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