3. Trigonometry

Starting with the Sum Identities: \[\begin{aligned} \sin(A+B)&=\sin(A)\cos(B)+\cos(A)\sin(B) \\ \cos(A+B)&=\cos(A)\cos(B)-\sin(A)\sin(B) \end{aligned}\] Replace \(B\) by \(A\) to get: \[\begin{aligned} \sin(2A)&=2\sin(A)\cos(A) \\ \cos(2A)&=\cos^2(A)-\sin^2(A) \end{aligned}\] There are two other forms of the cosine identity obtained by using the Pythagorean Identity \(\sin^2(A)+\cos^2(A)=1\). In the formula for \(\cos(2A)\), replace \(\sin^2(A)\) by \(1-\cos^2(A)\): \[ \cos(2A)=\cos^2(A)-(1-\cos^2(A))=2\cos^2(A)-1 \] Again in the formula for \(\cos(2A)\), this time replace \(\cos^2(A)\) by \(1-\sin^2(A)\): \[ \cos(2A)=1-\sin^2(A)-\sin^2(A)=1-2\sin^2(A) \]

For tangent, we have: \[ \tan(2A)=\dfrac{\sin(2A)}{\cos(2A)} =\dfrac{2\sin(A)\cos(A)}{\cos^2(A)-\sin^2(A)} \] Now divide top and bottom by \(\cos^2(A)\): \[ \tan(2A)=\dfrac{2\tan(A)}{1-\tan^2(A)} \] For cotangent, we have: \[ \cot(2A)=\dfrac{\cos(2A)}{\sin(2A)} =\dfrac{\cos^2(A)-\sin^2(A)}{2\sin(A)\cos(A)} \] Now divide top and bottom by \(\sin^2(A)\): \[ \cot(2A)=\dfrac{\cos(2A)}{\sin(2A)} =\dfrac{\cot^2(A)-1}{2\cot(A)} \] For secant, we have: \[ \sec(2A)=\dfrac{1}{\cos(2A)} =\dfrac{1}{\cos^2(A)-\sin^2(A)} \] Now divide top and bottom by \(\cos^2(A)\): \[ \sec(2A)=\dfrac{\sec^2(A)}{1-\tan^2(A)} \] For cosecant, we have: \[ \csc(2A)=\dfrac{1}{\sin(2A)} =\dfrac{1}{2\sin(A)\cos(A)} \] Now divide top and bottom by \(\sin^2(A)\): \[ \csc(2A)=\dfrac{\csc^2(A)}{2\cot(A)} \]

Start with the identity: \[ \cos(2A)=1-2\sin^2(A) \] Add \(2\sin^2(A)\) to both sides and subtract \(\cos(2A)\) from both sides: \[ 2\sin^2(A)=1-\cos(2A) \] Now divide by \(2\): \[ \sin^2(A)=\dfrac{1-\cos(2A)}{2} \] Now start with the identity: \[ \cos(2A)=2\cos^2(A)-1 \] Add \(1\) to both sides: \[ 1+\cos(2A)=2\cos^2(A) \] Now divide by \(2\): \[ \dfrac{1+\cos(2A)}{2}=\cos^2(A) \]

In the Square Identities \[ \sin^2(A)=\dfrac{1-\cos(2A)}{2} \qquad \cos^2(A)=\dfrac{1+\cos(2A)}{2} \] replace \(A\) by \(\dfrac{A}{2}\): \[ \sin^2\left(\dfrac{A}{2}\right)=\dfrac{1-\cos(A)}{2} \qquad \cos^2\left(\dfrac{A}{2}\right)=\dfrac{1+\cos(A)}{2} \] Finally take the square root of both sides and remember to insert a \(\pm\). The sign is determined by the quadrant of the angle \(\dfrac{A}{2}\).

We add and subtract the identities \[\begin{aligned} \cos(A+B)&=\cos A\cos B-\sin A\sin B \\ \cos(A-B)&=\cos A\cos B+\sin A\sin B \end{aligned}\] to get \[\begin{aligned} \cos(A+B)+\cos(A-B)&=2\cos A\cos B \\ \cos(A-B)-\cos(A+B)&=2\sin A\sin B \end{aligned}\] Dividing by \(2\) gives: \[\begin{aligned} \sin A\sin B&=\dfrac{1}{2}[\cos(A-B)-\cos(A+B)] \\ \cos A\cos B&=\dfrac{1}{2}[\cos(A+B)+\cos(A-B)] \end{aligned}\] Now we add and subtract the identities \[\begin{aligned} \sin(A+B)&=\sin A\cos B+\cos A\sin B \\ \sin(A-B)&=\sin A\cos B-\cos A\sin B \end{aligned}\] to get \[\begin{aligned} \sin(A+B)+\sin(A-B)&=2\sin A\cos B \\ \sin(A+B)-\sin(A-B)&=2\cos A\sin B \end{aligned}\] Dividing by \(2\) gives: \[\begin{aligned} \sin A\cos B&=\dfrac{1}{2}[\sin(A+B)+\sin(A-B)] \\ \cos A\sin B&=\dfrac{1}{2}[\sin(A+B)-\sin(A-B)] \end{aligned}\] These last two formulas are the same but with the roles of \(A\) and \(B\) reversed.