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

\[\begin{aligned} \sin(A+B)&=\sin(A)\cos(B)+\cos(A)\sin(B) \\ \cos(A+B)&=\cos(A)\cos(B)-\sin(A)\sin(B) \end{aligned}\] \(\Leftarrow\Leftarrow\) Read it! It depends on the distance formula in the plane.
\(\Leftarrow\Leftarrow\) Read it! It's an easy consequence.

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

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

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

The Trig Rap

Rhythmically Repeat: Sin Cos Cos Sin Cos Cos Sin Sin

As a consequence:

\[ \tan(A+B)=\dfrac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)} \] \(\Leftarrow\Leftarrow\) Read it! It's just algebra.

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

Difference Identities

\[\begin{aligned} \sin(A-B)&=\sin(A)\cos(B)-\cos(A)\sin(B) \\ \cos(A-B)&=\cos(A)\cos(B)+\sin(A)\sin(B) \end{aligned}\] \(\Leftarrow\Leftarrow\) Read it! It's straightforward.

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

As a consequence:

\[ \tan(A-B)=\dfrac{\tan(A)-\tan(B)}{1+\tan(A)\tan(B)} \] \(\Leftarrow\Leftarrow\) Read it! It's just algebra.

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

Find \(\sin75^\circ\) and \(\cos75^\circ\).

Knowing that: \[\begin{aligned} \sin(A+B)&=\sin(A)\cos(B)+\cos(A)\sin(B) \\ \cos(A+B)&=\cos(A)\cos(B)-\sin(A)\sin(B) \end{aligned}\] We use \(A=45^\circ\) and \(B=30^\circ\) and see: \[\begin{aligned} \sin(75^\circ)&=\sin(45^\circ)\cos(30^\circ)+\cos(45^\circ)\sin(30^\circ) \\ &=\dfrac{1}{\sqrt{2}}\dfrac{\sqrt{3}}{2}+\dfrac{1}{\sqrt{2}}\dfrac{1}{2} =\dfrac{\sqrt{3}+1}{2\sqrt{2}} \approx0.966\\ \cos(75^\circ)&=\cos(45^\circ)\cos(30^\circ)-\sin(45^\circ)\sin(30^\circ) \\ &=\dfrac{1}{\sqrt{2}}\dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}}\dfrac{1}{2} =\dfrac{\sqrt{3}-1}{2\sqrt{2}} \approx0.259 \end{aligned}\]

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

Find \(\sin15^\circ\) and \(\cos15^\circ\).

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

\(\begin{aligned} \sin(15^\circ)=\dfrac{\sqrt{3}-1}{2\sqrt{2}}\approx0.259 \\ \cos(15^\circ)=\dfrac{\sqrt{3}+1}{2\sqrt{2}}\approx0.966 \end{aligned}\)

Knowing that: \[\begin{aligned} \sin(A-B)&=\sin(A)\cos(B)-\cos(A)\sin(B) \\ \cos(A-B)&=\cos(A)\cos(B)+\sin(A)\sin(B) \end{aligned}\] We use \(A=45^\circ\) and \(B=30^\circ\) and see: \[\begin{aligned} \sin(15^\circ)&=\sin(45^\circ)\cos(30^\circ)-\cos(45^\circ)\sin(30^\circ) \\ &=\dfrac{1}{\sqrt{2}}\dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}}\dfrac{1}{2} =\dfrac{\sqrt{3}-1}{2\sqrt{2}} \approx0.259 \\ \cos(15^\circ)&=\cos(45^\circ)\cos(30^\circ)+\sin(45^\circ)\sin(30^\circ) \\ &=\dfrac{1}{\sqrt{2}}\dfrac{\sqrt{3}}{2}+\dfrac{1}{\sqrt{2}}\dfrac{1}{2} =\dfrac{\sqrt{3}+1}{2\sqrt{2}} \approx0.966 \end{aligned}\]

We check that \[\begin{aligned} \sin^2(15^\circ)+\cos^2(15^\circ) &=\left(\dfrac{\sqrt{3}-1}{2\sqrt{2}}\right)^2 +\left(\dfrac{\sqrt{3}+1}{2\sqrt{2}}\right)^2 \\ &=\dfrac{3-2\sqrt{3}+1}{8} +\dfrac{3+2\sqrt{3}+1}{8} \\ &=\dfrac{4+4}{8}=1 \end{aligned}\]