Precalculus Analytic Trigonometry Flashcards

24 sample cards from the Analytic Trigonometry deck

Sampled from the end of the deck, so these are different cards from the ones shown on the syllabus page.

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

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

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

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

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

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

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

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

5. State the double-angle formula for sine.

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

6. State the double-angle formula for tangent.

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

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

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

8. 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 $\tfrac{\theta}{2}$.

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

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

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

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

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

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

12. 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.

13. 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.

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

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

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

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

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

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

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

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

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

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

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

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

20. 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).

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

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

22. 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 $\cos x=\tfrac12$ ($x=\tfrac{\pi}{3},\tfrac{5\pi}{3}$).

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

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

24. 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.

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