3. Trigonometry

d. Trig Identities

The Sum and Difference Identities give the values of the trig functions for the sum and difference of two angles.

4. Sum and Difference Identities

Sum Identities

Sum Identities: Sine and Cosine
sin⁡(A+B)=sin⁡(A)cos⁡(B)+cos⁡(A)sin⁡(B)cos⁡(A+B)=cos⁡(A)cos⁡(B)−sin⁡(A)sin⁡(B)\begin{aligned} \sin(A+B)&=\sin(A)\cos(B)+\cos(A)\sin(B) \\ \cos(A+B)&=\cos(A)\cos(B)-\sin(A)\sin(B) \end{aligned} ⇐⇐\Leftarrow\Leftarrow Read it! It depends on the distance formula in the plane.
⇐⇐\Leftarrow\Leftarrow Read it! It's an easy consequence.

Proof

In the diagram, look at the triangles: △PORand△QOS \triangle POR \qquad \text{and} \qquad \triangle QOS For both triangles, the central angle is A+BA+B. So the length of the opposite sides PRPR and QSQS are equal.
The coordinates are: P=(1,0)Q=(cos⁡(B),sin⁡(B))R=(cos⁡(A+B),sin⁡(A+B))S=(cos⁡(A),−sin⁡(A))\begin{aligned} P&=(1,0) \\ Q&=(\cos(B),\sin(B)) \\ R&=(\cos(A+B),\sin(A+B)) \\ S&=(\cos(A),-\sin(A)) \end{aligned}

pf_cos_sum

So the distances are: d(P,R)=(cos⁡(A+B)−1)2+(sin⁡(A+B)−0)2=cos⁡2(A+B)−2cos⁡(A+B)+1+sin⁡2(A+B)=2−2cos⁡(A+B)d(Q,S)=(cos⁡(A)−cos⁡(B))2+(−sin⁡(A)−sin⁡(B))2=cos⁡2(A)−2cos⁡(A)cos⁡(B)+cos⁡2(B)+sin⁡2(A)+2sin⁡(A)sin⁡(B)+sin⁡2(B)=2−2cos⁡(A)cos⁡(B)+2sin⁡(A)sin⁡(B)\begin{aligned} d(P,R)&=\sqrt{(\cos(A+B)-1)^2+(\sin(A+B)-0)^2} \\ &=\sqrt{\cos^2(A+B)-2\cos(A+B)+1+\sin^2(A+B)} \\ &=\sqrt{2-2\cos(A+B)} \\ d(Q,S)&=\sqrt{(\cos(A)-\cos(B))^2+(-\sin(A)-\sin(B))^2} \\ &=\sqrt{\begin{array}{ll} \cos^2(A)-2\cos(A)\cos(B)+\cos^2(B) \\ \quad+\sin^2(A)+2\sin(A)\sin(B)+\sin^2(B) \end{array}} \\ &=\sqrt{2-2\cos(A)\cos(B)+2\sin(A)\sin(B)} \end{aligned} Equating these and squaring both sides gives: 2−2cos⁡(A+B)=2−2cos⁡(A)cos⁡(B)+2sin⁡(A)sin⁡(B) 2-2\cos(A+B)=2-2\cos(A)\cos(B)+2\sin(A)\sin(B) Subtracting 22 from both sides and then dividing both sides by −2-2 leads to the result: cos⁡(A+B)=cos⁡(A)cos⁡(B)−sin⁡(A)sin⁡(B) \cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)

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Proof

We start by using the Complementary Angle Identity: sin⁡(A+B)=cos⁡[π2−(A+B)]=cos⁡[(π2−A)−B] \sin(A+B)=\cos\left[\dfrac{\pi}{2}-(A+B)\right] =\cos\left[\left(\dfrac{\pi}{2}-A\right)-B\right] Now we apply the cos⁡(A+B)\cos(A+B) formula with AA replaced by π2−A\dfrac{\pi}{2}-A and BB replaced by −B-B: sin⁡(A+B)=cos⁡(π2−A)cos⁡(−B)−sin⁡(π2−A)sin⁡(−B) \sin(A+B)=\cos\left(\dfrac{\pi}{2}-A\right)\cos(-B) -\sin\left(\dfrac{\pi}{2}-A\right)\sin(-B) Finally, we use the Complementary Angle Identities in each of the first factors and Parity Identities in the second factors: sin⁡(A+B)=sin⁡(A)cos⁡(B)+cos⁡(A)sin⁡(B) \sin(A+B)=\sin(A)\cos(B)+\cos(A)\sin(B)

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The Trig Rap

Rhythmically Repeat: Sin    Cos    Cos   Sin Cos   Cos    Sin    Sin

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As a consequence:

Sum Identities: Tangent
tan⁡(A+B)=tan⁡(A)+tan⁡(B)1−tan⁡(A)tan⁡(B) \tan(A+B)=\dfrac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)} ⇐⇐\Leftarrow\Leftarrow Read it! It's just algebra.

Proof

tan⁡(A+B)=sin⁡(A+B)cos⁡(A+B)=sin⁡(A)cos⁡(B)+cos⁡(A)sin⁡(B)cos⁡(A)cos⁡(B)−sin⁡(A)sin⁡(B) \tan(A+B)=\dfrac{\sin(A+B)}{\cos(A+B)} =\dfrac{\sin(A)\cos(B)+\cos(A)\sin(B)}{\cos(A)\cos(B)-\sin(A)\sin(B)} Now we divide top and bottom by cos⁡(A)cos⁡(B)\cos(A)\cos(B): tan⁡(A+B)=tan⁡(A)+tan⁡(B)1−tan⁡(A)tan⁡(B) \tan(A+B)=\dfrac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)}

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Difference Identities

Difference Identities: Sine and Cosine
sin⁡(A−B)=sin⁡(A)cos⁡(B)−cos⁡(A)sin⁡(B)cos⁡(A−B)=cos⁡(A)cos⁡(B)+sin⁡(A)sin⁡(B)\begin{aligned} \sin(A-B)&=\sin(A)\cos(B)-\cos(A)\sin(B) \\ \cos(A-B)&=\cos(A)\cos(B)+\sin(A)\sin(B) \end{aligned} ⇐⇐\Leftarrow\Leftarrow Read it! It's straightforward.

Proof

We use the Sum Formulas with BB replaced by −B-B followed by the Parity Identities: sin⁡(A−B)=sin⁡(A)cos⁡(−B)+cos⁡(A)sin⁡(−B)=sin⁡(A)cos⁡(B)−cos⁡(A)sin⁡(B)\begin{aligned} \sin(A-B) &=\sin(A)\cos(-B)+\cos(A)\sin(-B) \\ &=\sin(A)\cos(B)-\cos(A)\sin(B) \end{aligned} cos⁡(A−B)=cos⁡(A)cos⁡(−B)−sin⁡(A)sin⁡(−B)=cos⁡(A)cos⁡(B)+sin⁡(A)sin⁡(B)\begin{aligned} \cos(A-B) &=\cos(A)\cos(-B)-\sin(A)\sin(-B) \\ &=\cos(A)\cos(B)+\sin(A)\sin(B) \end{aligned}

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As a consequence:

Difference Identities: Tangent
tan⁡(A−B)=tan⁡(A)−tan⁡(B)1+tan⁡(A)tan⁡(B) \tan(A-B)=\dfrac{\tan(A)-\tan(B)}{1+\tan(A)\tan(B)} ⇐⇐\Leftarrow\Leftarrow Read it! It's just algebra.

Proof

tan⁡(A−B)=sin⁡(A−B)cos⁡(A−B)=sin⁡(A)cos⁡(B)−cos⁡(A)sin⁡(B)cos⁡(A)cos⁡(B)+sin⁡(A)sin⁡(B) \tan(A-B)=\dfrac{\sin(A-B)}{\cos(A-B)} =\dfrac{\sin(A)\cos(B)-\cos(A)\sin(B)}{\cos(A)\cos(B)+\sin(A)\sin(B)} Now we divide top and bottom by cos⁡(A)cos⁡(B)\cos(A)\cos(B): tan⁡(A−B)=tan⁡(A)−tan⁡(B)1+tan⁡(A)tan⁡(B) \tan(A-B)=\dfrac{\tan(A)-\tan(B)}{1+\tan(A)\tan(B)}

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Find sin⁡75∘\sin75^\circ and cos⁡75∘\cos75^\circ.

Knowing that: sin⁡(A+B)=sin⁡(A)cos⁡(B)+cos⁡(A)sin⁡(B)cos⁡(A+B)=cos⁡(A)cos⁡(B)−sin⁡(A)sin⁡(B)\begin{aligned} \sin(A+B)&=\sin(A)\cos(B)+\cos(A)\sin(B) \\ \cos(A+B)&=\cos(A)\cos(B)-\sin(A)\sin(B) \end{aligned} We use A=45∘A=45^\circ and B=30∘B=30^\circ and see: sin⁡(75∘)=sin⁡(45∘)cos⁡(30∘)+cos⁡(45∘)sin⁡(30∘)=1232+1212=3+122≈0.966cos⁡(75∘)=cos⁡(45∘)cos⁡(30∘)−sin⁡(45∘)sin⁡(30∘)=1232−1212=3−122≈0.259\begin{aligned} \sin(75^\circ)&=\sin(45^\circ)\cos(30^\circ)+\cos(45^\circ)\sin(30^\circ) \\ &=\dfrac{1}{\sqrt{2}}\dfrac{\sqrt{3}}{2}+\dfrac{1}{\sqrt{2}}\dfrac{1}{2} =\dfrac{\sqrt{3}+1}{2\sqrt{2}} \approx0.966\\ \cos(75^\circ)&=\cos(45^\circ)\cos(30^\circ)-\sin(45^\circ)\sin(30^\circ) \\ &=\dfrac{1}{\sqrt{2}}\dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}}\dfrac{1}{2} =\dfrac{\sqrt{3}-1}{2\sqrt{2}} \approx0.259 \end{aligned}

You should check that sin⁡2(75∘)+cos⁡2(75∘)=1 \sin^2(75^\circ)+\cos^2(75^\circ)=1

Find sin⁡15∘\sin15^\circ and cos⁡15∘\cos15^\circ.

Hint

Use the Difference Formulas with A=45∘A=45^\circ and B=30∘B=30^\circ.

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Answer

sin⁡(15∘)=3−122≈0.259cos⁡(15∘)=3+122≈0.966\begin{aligned} \sin(15^\circ)=\dfrac{\sqrt{3}-1}{2\sqrt{2}}\approx0.259 \\ \cos(15^\circ)=\dfrac{\sqrt{3}+1}{2\sqrt{2}}\approx0.966 \end{aligned}

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Solution

Knowing that: sin⁡(A−B)=sin⁡(A)cos⁡(B)−cos⁡(A)sin⁡(B)cos⁡(A−B)=cos⁡(A)cos⁡(B)+sin⁡(A)sin⁡(B)\begin{aligned} \sin(A-B)&=\sin(A)\cos(B)-\cos(A)\sin(B) \\ \cos(A-B)&=\cos(A)\cos(B)+\sin(A)\sin(B) \end{aligned} We use A=45∘A=45^\circ and B=30∘B=30^\circ and see: sin⁡(15∘)=sin⁡(45∘)cos⁡(30∘)−cos⁡(45∘)sin⁡(30∘)=1232−1212=3−122≈0.259cos⁡(15∘)=cos⁡(45∘)cos⁡(30∘)+sin⁡(45∘)sin⁡(30∘)=1232+1212=3+122≈0.966\begin{aligned} \sin(15^\circ)&=\sin(45^\circ)\cos(30^\circ)-\cos(45^\circ)\sin(30^\circ) \\ &=\dfrac{1}{\sqrt{2}}\dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}}\dfrac{1}{2} =\dfrac{\sqrt{3}-1}{2\sqrt{2}} \approx0.259 \\ \cos(15^\circ)&=\cos(45^\circ)\cos(30^\circ)+\sin(45^\circ)\sin(30^\circ) \\ &=\dfrac{1}{\sqrt{2}}\dfrac{\sqrt{3}}{2}+\dfrac{1}{\sqrt{2}}\dfrac{1}{2} =\dfrac{\sqrt{3}+1}{2\sqrt{2}} \approx0.966 \end{aligned}

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Check

We check that sin⁡2(15∘)+cos⁡2(15∘)=(3−122)2+(3+122)2=3−23+18+3+23+18=4+48=1\begin{aligned} \sin^2(15^\circ)+\cos^2(15^\circ) &=\left(\dfrac{\sqrt{3}-1}{2\sqrt{2}}\right)^2 +\left(\dfrac{\sqrt{3}+1}{2\sqrt{2}}\right)^2 \\ &=\dfrac{3-2\sqrt{3}+1}{8} +\dfrac{3+2\sqrt{3}+1}{8} \\ &=\dfrac{4+4}{8}=1 \end{aligned}

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Remark

Comparing this exercise with the previous example, we see: sin⁡(15∘)=cos⁡(75∘)=3−122cos⁡(15∘)=sin⁡(75∘)=3+122\begin{aligned} \sin(15^\circ)&=\cos(75^\circ)=\dfrac{\sqrt{3}-1}{2\sqrt{2}} \\ \cos(15^\circ)&=\sin(75^\circ)=\dfrac{\sqrt{3}+1}{2\sqrt{2}} \end{aligned} This is not surprising, since 15∘15^\circ and 75∘75^\circ are complimentary angles.

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