Compound Angle flash cards

1. 1.State the general solution of $\sin\theta = \sin\alpha$.

   $\theta = n\pi + (-1)^n\alpha$, $n \in \mathbb{Z}$

   Hint: Sine repeats with alternating sign pattern over $\pi$

2. 2.Write the formulas for $\sin(A+B)$ and $\cos(A+B)$.

   $\sin(A+B)=\sin A\cos B+\cos A\sin B$; $\cos(A+B)=\cos A\cos B-\sin A\sin B$

   Hint: Compound angle expansion

3. 3.Express $\tan(A+B)$ in terms of $\tan A$ and $\tan B$.

   $\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$

   Hint: Divide sine sum formula by cosine sum formula

4. 4.What are the sum-to-product formulas for $\sin C + \sin D$ and $\cos C + \cos D$?

   $\sin C+\sin D=2\sin\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)$; $\cos C+\cos D=2\cos\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)$

   Hint: Used to convert sums into products for simplification

5. 5.State the maximum and minimum values of $a\sin\theta+b\cos\theta$.

   Maximum $=\sqrt{a^2+b^2}$, Minimum $=-\sqrt{a^2+b^2}$

   Hint: Write as $R\sin(\theta+\phi)$ where $R=\sqrt{a^2+b^2}$

6. 6.Evaluate $\cos20°\cos40°\cos60°\cos80°$.

   Using $\cos\theta\cos(60°-\theta)\cos(60°+\theta)=\dfrac14\cos3\theta$ with $\theta=20°$: $\cos20°\cos40°\cos80°=\dfrac14\cos60°=\dfrac18$. Multiplying by $\cos60°=\dfrac12$ gives the full product $=\dfrac1{16}$.

   Hint: Group $\cos20°,\cos40°,\cos80°$ using the $60°\pm\theta$ product identity.

7. 7.Prove $\sin\theta\sin(60°-\theta)\sin(60°+\theta)=\dfrac14\sin3\theta$, and hence find $\sin20°\sin40°\sin80°$.

   Since $\sin(60°-\theta)\sin(60°+\theta)=\sin^260°-\sin^2\theta=\dfrac34-\sin^2\theta$, multiplying by $\sin\theta$ gives $\dfrac34\sin\theta-\sin^3\theta$. Using $\sin^3\theta=\dfrac{3\sin\theta-\sin3\theta}4$, this simplifies exactly to $\dfrac14\sin3\theta$. With $\theta=20°$: $\sin20°\sin40°\sin80°=\dfrac14\sin60°=\dfrac{\sqrt3}8$.

   Hint: Express $\sin(60°-\theta)\sin(60°+\theta)$ as $\sin^260°-\sin^2\theta$ first.

8. 8.Find the value of $\sin10°\sin30°\sin50°\sin70°$.

   By the identity $\sin\theta\sin(60°-\theta)\sin(60°+\theta)=\dfrac14\sin3\theta$ with $\theta=10°$: $\sin10°\sin50°\sin70°=\dfrac14\sin30°=\dfrac18$. Multiplying by $\sin30°=\dfrac12$ gives $\dfrac1{16}$.

   Hint: Group $\sin10°,\sin50°,\sin70°$ via the $60°\pm\theta$ product identity, then bring in $\sin30°$.

9. 9.Simplify $\dfrac{\sin5x+\sin3x}{\cos5x+\cos3x}$ to a single trig ratio.

   By sum-to-product, $\sin5x+\sin3x=2\sin4x\cos x$ and $\cos5x+\cos3x=2\cos4x\cos x$. Dividing, the $2\cos x$ factors cancel (for $\cos x\ne0$), leaving $\tan4x$.

   Hint: Apply sum-to-product formulas to numerator and denominator separately.

10. 10.If $\cos\alpha+\cos\beta=a$ and $\sin\alpha+\sin\beta=b$, find $\cos(\alpha-\beta)$ in terms of $a,b$.

    Squaring and adding: $a^2+b^2=2+2\cos\alpha\cos\beta+2\sin\alpha\sin\beta=2+2\cos(\alpha-\beta)$. Solving, $\cos(\alpha-\beta)=\dfrac{a^2+b^2-2}2$.

    Hint: Square both given equations and add them.

Open the interactive deck for the other 75 cards, with self-grading so the ones you keep missing come back.