Homework 6

Directions:

  1. Show each step of your work and fully simplify each expression.
  2. Turn in your answers in class on a physical piece of paper.
  3. Staple multiple sheets together.
  4. Feel free to use Desmos for graphing.

Helpful hint:

Because \[\sin(2x) = 2\sin x\cos x\] you can replace all the $x$ with $2x$ to get \[\sin(4x) = \sin(2\cdot 2x) = 2\sin 2x \cos 2x\] Can do this for even coefficients of $x$, for example \[\sin(4x) \qquad \sin(6x) \qquad \sin(8x)\]

Answer the following:

  1. Prove the following identities.
    1. $\dfrac{(\sin x + \cos x)^2}{\sin^2x - \cos^2x} = \dfrac{\sin^2 x - \cos^2 x}{(\sin x - \cos x)^2}$
    2. $\dfrac{1}{1 - \sin^2 y} = 1 + \tan^2 y$
    3. $2 \csc x = \dfrac{1 - \cos x}{\sin x} + \dfrac{\sin x}{1 - \cos x}$
    4. $\dfrac{\cos x}{1 - \sin x} = \dfrac{\sin x - \csc x}{\cos x - \cot x}$
    5. $\cos^2 x - \sin^2 x = 2 \cos^2 x - 1$
  2. Assume $0 < \theta < \pi/2$. Make the indicated trigonometric substitution and simplify.
    1. $\dfrac{x}{\sqrt{1 - x^2}}, \quad x = \sin \theta$
    2. $\sqrt{9 - x^2}, \quad x = 3\sin \theta$
  3. Use addition/subtraction formulas to find the following values:
    1. $\sin 75^\circ$
    2. $\cos \dfrac{11\pi}{12}$
    3. $\sin18^\circ \cos27^\circ + \cos18^\circ\sin27^\circ$
    4. $\tan 105^\circ$
    5. $\csc\left(\dfrac{5\pi}{12}\right)$
    6. $\cos\left(\sin^{-1}\dfrac{\sqrt{3}}{2} + \cos^{-1} \dfrac{1}{2} + \dfrac{\pi}{12}\right)$
  4. Prove these identities. Remember: do not assume the statement you are trying to prove. Start from one side!
    1. $\cos(x + y)\cos(x - y) = \cos^2 x - \sin^2 y$
    2. $\sin(x + y) - \sin(x - y) = 2\cos x \sin y$
    3. $\tan(x - \pi) = \tan x$
    4. $1 - \tan x \tan y = \dfrac{\cos(x + y)}{\cos x \cos y}$
    5. $\cos\left(x + \dfrac{\pi}{3}\right) + \sin\left(x - \dfrac{\pi}{6}\right) = 0$
  5. Suppose $f(x) = \sin(x)$. Evaluate and simplify the following expressing using the addition identity: \[\dfrac{f(x + h) - f(x)}{h}\] You should end up with the expression \[\left(\dfrac{\sin h}{h}\right)\cos x - \sin x \left(\dfrac{1 - \cos h}{h}\right)\]
  6. On next week's quiz, what five types of trig identity formulas will you need to be familiar with?
  7. Evaluate the following using a double-angle or half-angle formula.
    1. $\sin 15^\circ$
    2. $2\sin 22.5^\circ \cos 22.5^\circ$
    3. $\tan \dfrac{\pi}{8}$
    4. $\cos \dfrac{3\pi}{8}$
    5. $\cos^215^\circ - \sin^2 15^\circ$
    6. $\dfrac{2\tan15^\circ}{1 - \tan^2 15^\circ}$
    7. $\cos 112.5^\circ$
  8. Convert global terms to global factors.
    1. $\cos 9x + \cos 2x$
    2. $\sin 5x + \sin 3x$
    3. $\sin x - \sin 4x$
  9. How many global factors are there in the expression \[\cos 3x \sin 4x \tan 5x\]
  10. Now rewrite the previous expression with parentheses so the factors are visually more separated.
  11. Convert global factors to global terms.
    1. $\sin x \sin 5x$
    2. $3\cos 4x \cos 7x$
    3. $\cos 5x\cos3x$
  12. Prove the following identities.
    1. $\cos^2 5x - \sin^2 5x = \cos 10x$
    2. $\dfrac{\sin 3x + \cos 3x}{\cos x - \sin x} = 1 + 4\sin x \cos x$
    3. $\cos^4 x - \sin^4x = \cos 2x$
    4. $\dfrac{\sin 10 x}{\sin 9x + \sin x} = \dfrac{\cos 5x}{\cos 4x}$
    5. $\dfrac{\sin 4x}{\sin x} = 4\cos x \cos 2x$
    6. $\dfrac{\sin 2x}{1 + \cos 2x} = \tan x$
    7. $\dfrac{2\tan x}{1 + \tan^2x} = \sin 2x$