Precalculus Analytic Trigonometry Syllabus

What is the difference between finding solutions 'on an interval' versus finding the 'general solution' of a trig equation?

Interval solutions list only angles within a bounded range like $[0,2\pi)$; the general solution adds the period multiple to cover every coterminal angle, e.g. $+2\pi n$ for sine/cosine or $+\pi n$ for tangent.

Write the general solution of $\sin x=\tfrac12$.

$$x=\frac{\pi}{6}+2\pi n\quad\text{or}\quad x=\frac{5\pi}{6}+2\pi n,\quad n\in\mathbb{Z}$$

Write the general solution of $\tan x=1$.

$$x=\frac{\pi}{4}+\pi n,\quad n\in\mathbb{Z}$$ (period $\pi$)

State the sum and difference formulas for cosine.

$$\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B$$

State the sum and difference formulas for sine.

$$\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B$$

State the sum and difference formulas for tangent.

$$\tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}$$

Use a difference formula to find the exact value of $\cos 15^{\circ}$.

$$\cos(45^{\circ}-30^{\circ})=\cos45^{\circ}\cos30^{\circ}+\sin45^{\circ}\sin30^{\circ}=\frac{\sqrt{6}+\sqrt{2}}{4}$$

Find the exact value of $\sin\dfrac{7\pi}{12}$ using a sum formula.

$$\sin\!\left(\frac{\pi}{4}+\frac{\pi}{3}\right)=\sin\frac{\pi}{4}\cos\frac{\pi}{3}+\cos\frac{\pi}{4}\sin\frac{\pi}{3}=\frac{\sqrt{6}+\sqrt{2}}{4}$$

Simplify $\cos40^{\circ}\cos10^{\circ}+\sin40^{\circ}\sin10^{\circ}$ to an exact value.

This is $\cos(40^{\circ}-10^{\circ})=\cos30^{\circ}=\dfrac{\sqrt{3}}{2}$.

State the three standard forms of the double-angle formula for cosine.

$$\cos2\theta=\cos^{2}\theta-\sin^{2}\theta=2\cos^{2}\theta-1=1-2\sin^{2}\theta$$

State the double-angle formula for sine.

$$\sin2\theta=2\sin\theta\cos\theta$$

State the double-angle formula for tangent.

$$\tan2\theta=\frac{2\tan\theta}{1-\tan^{2}\theta}$$

Given $\sin\theta=\tfrac35$ with $\theta$ in Quadrant I, find $\sin2\theta$.

$\cos\theta=\tfrac45$, so $$\sin2\theta=2\cdot\tfrac35\cdot\tfrac45=\frac{24}{25}$$

State the half-angle formulas for sine and cosine.

$$\sin\frac{\theta}{2}=\pm\sqrt{\frac{1-\cos\theta}{2}},\qquad \cos\frac{\theta}{2}=\pm\sqrt{\frac{1+\cos\theta}{2}}$$ The sign depends on the quadrant of $\tfrac{\theta}{2}$.

State the half-angle formulas for tangent (three equivalent forms).

$$\tan\frac{\theta}{2}=\pm\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}=\frac{1-\cos\theta}{\sin\theta}=\frac{\sin\theta}{1+\cos\theta}$$

Use a half-angle formula to find the exact value of $\cos15^{\circ}$.

$$\cos\frac{30^{\circ}}{2}=\sqrt{\frac{1+\cos30^{\circ}}{2}}=\sqrt{\frac{1+\frac{\sqrt3}{2}}{2}}=\frac{\sqrt{2+\sqrt3}}{2}=\frac{\sqrt6+\sqrt2}{4}$$

State the three power-reducing formulas for $\sin^{2}\theta$, $\cos^{2}\theta$, and $\tan^{2}\theta$.

$$\sin^{2}\theta=\frac{1-\cos2\theta}{2},\quad \cos^{2}\theta=\frac{1+\cos2\theta}{2},\quad \tan^{2}\theta=\frac{1-\cos2\theta}{1+\cos2\theta}$$

How are the power-reducing formulas derived from the double-angle formulas?

Solve the cosine double-angle forms $\cos2\theta=1-2\sin^{2}\theta$ and $\cos2\theta=2\cos^{2}\theta-1$ for $\sin^{2}\theta$ and $\cos^{2}\theta$ respectively.

Express $\cos^{4}\theta$ using power-reducing formulas (first step).

$$\cos^{4}\theta=(\cos^{2}\theta)^{2}=\left(\frac{1+\cos2\theta}{2}\right)^{2}=\frac{1+2\cos2\theta+\cos^{2}2\theta}{4}$$ then reduce $\cos^{2}2\theta$ again.

State the product-to-sum formula for $\sin A\cos B$.

$$\sin A\cos B=\tfrac12\big[\sin(A+B)+\sin(A-B)\big]$$

State the product-to-sum formula for $\cos A\cos B$.

$$\cos A\cos B=\tfrac12\big[\cos(A-B)+\cos(A+B)\big]$$

State the product-to-sum formula for $\sin A\sin B$.

$$\sin A\sin B=\tfrac12\big[\cos(A-B)-\cos(A+B)\big]$$

State the sum-to-product formula for $\sin A+\sin B$.

$$\sin A+\sin B=2\sin\!\left(\frac{A+B}{2}\right)\cos\!\left(\frac{A-B}{2}\right)$$

State the sum-to-product formula for $\cos A+\cos B$.

$$\cos A+\cos B=2\cos\!\left(\frac{A+B}{2}\right)\cos\!\left(\frac{A-B}{2}\right)$$

State the sum-to-product formula for $\cos A-\cos B$.

$$\cos A-\cos B=-2\sin\!\left(\frac{A+B}{2}\right)\sin\!\left(\frac{A-B}{2}\right)$$

When are product-to-sum versus sum-to-product formulas useful?

Product-to-sum converts a product of trig functions into a sum (helpful for integration or simplifying products); sum-to-product converts a sum/difference into a product (helpful for factoring and solving equations set equal to zero).

Find the exact value of $\tan\dfrac{\pi}{12}$ using a difference formula.

$$\tan\!\left(\frac{\pi}{3}-\frac{\pi}{4}\right)=\frac{\tan\frac{\pi}{3}-\tan\frac{\pi}{4}}{1+\tan\frac{\pi}{3}\tan\frac{\pi}{4}}=\frac{\sqrt3-1}{1+\sqrt3}=2-\sqrt3$$

Solve $\sin2x=\sin x$ on $[0,2\pi)$ using a double-angle identity.

$2\sin x\cos x-\sin x=0\Rightarrow \sin x(2\cos x-1)=0$. So $\sin x=0$ ($x=0,\pi$) or $\cos x=\tfrac12$ ($x=\tfrac{\pi}{3},\tfrac{5\pi}{3}$).

To rewrite $\sin5x\cos3x$ as a sum, which formula applies and what is the result?

Use $\sin A\cos B=\tfrac12[\sin(A+B)+\sin(A-B)]$: $$\sin5x\cos3x=\tfrac12(\sin8x+\sin2x)$$

Why must you check for extraneous solutions after squaring both sides or dividing by a trig function when solving equations?

Squaring can introduce roots that do not satisfy the original equation, and dividing by a trig factor can drop valid solutions where that factor equals zero; always verify candidates in the original equation and factor rather than divide.