\( \begin{array}{ll}5 & \text { Use compound and double angles to prove that: } \\ 5.1 & \sin 3 x=3 \sin x-4 \sin ^{3} x \\ 5.2 & \cos 3 x=\cos x\left(1-4 \sin ^{2} x\right) \\ 5.3 & \cos 4 x=8 \cos ^{4} x-8 \cos ^{2} x+1 \\ 5.4 & 1-\cos 4 x=2 \sin ^{2} 2 x \\ 5.5 & \sin 4 x=\cos x\left(4 \sin x-8 \sin ^{3}\right) \\ 5.6 & \sin 4 x=4 \sin x \cos ^{3} x-4 \cos x \sin ^{3} x\end{array} \)

**5.1 Proof that \( \sin 3x = 3\sin x - 4\sin^3 x \):** 1. Write \( 3x \) as \( 2x + x \): \[ \sin 3x = \sin(2x + x) = \sin 2x \cos x + \cos 2x \sin x \] 2. Use the double‐angle identities: \[ \sin 2x = 2\sin x \cos x,\quad \cos 2x = 1 - 2\sin^2 x \] Substitute these: \[ \sin 3x = (2\sin x \cos x)\cos x + (1-2\sin^2 x)\sin x \] 3. Simplify: \[ \sin 3x = 2\sin x \cos^2 x + \sin x - 2\sin^3 x \] 4. Express \( \cos^2 x \) in terms of \( \sin^2 x \): \[ \cos^2 x = 1-\sin^2 x \] Then, \[ \sin 3x = 2\sin x(1-\sin^2 x) + \sin x - 2\sin^3 x = 2\sin x - 2\sin^3 x + \sin x - 2\sin^3 x \] 5. Combine like terms: \[ \sin 3x = 3\sin x - 4\sin^3 x \] --- **5.2 Proof that \( \cos 3x = \cos x\left(1 - 4\sin^2 x\right) \):** 1. Express \( 3x \) as \( 2x + x \): \[ \cos 3x = \cos(2x + x) = \cos 2x\cos x - \sin 2x\sin x \] 2. Substitute the double‐angle formulas: \[ \cos 2x = 1-2\sin^2 x,\quad \sin 2x = 2\sin x\cos x \] Then, \[ \cos 3x = (1-2\sin^2 x)\cos x - (2\sin x\cos x)\sin x \] 3. Simplify: \[ \cos 3x = \cos x-2\sin^2 x\cos x-2\sin^2 x\cos x = \cos x - 4\sin^2 x\cos x \] 4. Factor out \( \cos x \): \[ \cos 3x = \cos x\left(1-4\sin^2 x\right) \] --- **5.3 Proof that \( \cos 4x = 8\cos^4 x - 8\cos^2 x + 1 \):** 1. Write \( 4x \) as \( 2(2x) \): \[ \cos 4x = 2\cos^2 2x - 1 \] 2. Express \( \cos 2x \) in terms of \( \cos x \): \[ \cos 2x = 2\cos^2 x - 1 \] Thus, \[ \cos^2 2x = (2\cos^2 x - 1)^2 = 4\cos^4 x - 4\cos^2 x + 1 \] 3. Substitute back into the formula for \( \cos4x \): \[ \cos 4x = 2(4\cos^4 x - 4\cos^2 x + 1) - 1 = 8\cos^4 x - 8\cos^2 x + 2 - 1 \] 4. Simplify to obtain: \[ \cos 4x = 8\cos^4 x - 8\cos^2 x + 1 \] --- **5.4 Proof that \( 1-\cos 4x = 2\sin^2 2x \):** 1. Using the result from 5.3: \[ \cos 4x = 8\cos^4 x - 8\cos^2 x + 1 \] So, \[ 1 - \cos 4x = 1 - (8\cos^4 x - 8\cos^2 x + 1) = -8\cos^4 x + 8\cos^2 x \] 2. Factor out \( 8\cos^2 x \): \[ 1 - \cos 4x = 8\cos^2 x(1-\cos^2 x) \] 3. Recognize that \( 1-\cos^2 x = \sin^2 x \): \[ 1 - \cos 4x = 8\cos^2 x\sin^2 x \] 4. Now, use the double-angle identity for sine: \[ \sin 2x = 2\sin x\cos x \quad \Longrightarrow \quad \sin^2 2x = 4\sin^2 x\cos^2 x \] 5. Substitute to match the expression: \[ 8\cos^2 x\sin^2 x = 2\bigl(4\sin^2 x\cos^2 x\bigr) = 2\sin^2 2x \] Thus, \[ 1-\cos 4x = 2\sin^2 2x \] --- **5.5 Proof that \( \sin 4x = \cos x\left(4\sin x - 8\sin^3 x\right) \):** 1. Start with the double-angle expansion for \( \sin 4x \): \[ \sin 4x = 2\sin 2x \cos 2x \] 2. Write \( \sin 2x \) and \( \cos 2x \) in terms of \( \sin x \) and \( \cos x \): \[ \sin 2x = 2\sin x\cos x,\quad \cos 2x = 1-2\sin^2 x \] 3. Substitute these into the formula: \[ \sin 4x = 2\,(2\sin x\cos x)\,(1-2\sin^2 x)= 4\sin x\cos x\,(1-2\sin^2 x) \] 4. Distribute \( 4\sin x\cos x \): \[ \sin 4x = 4\sin x\cos x - 8\sin^3 x\cos x \] 5. Factor out \( \cos x \): \[ \sin 4x = \cos x\left(4\sin x-8\sin^3 x\right) \] --- **5.6 Proof that \( \sin 4x = 4\sin x\cos^3 x - 4\cos x\sin^3 x \):** 1. Begin with the expression from 5.5: \[ \sin 4x = 4\sin x\cos x - 8\sin^3 x\cos x \] 2. Alternatively, express \( \sin 4x \) using the sine addition formula for \( 2x+2x \): \[ \sin 4x = \sin(2x+2x) = \sin 2x\cos 2x + \cos 2x\sin 2x = 2\sin 2x \cos 2x \] 3. Substitute the double-angle identities: \[ \sin 2x = 2\sin x\cos x,\quad \cos 2x = 2\cos^2 x - 1 \] Then, \[ \sin 4x = 2\,(2\sin x\cos x)(2\cos^2 x-1) = 4\sin x\cos x(2\cos^2 x-1) \] 4. Expand the product: \[ \sin 4x = 8\sin x\cos^3 x - 4\sin x\cos x \] 5. Factor \( 4\sin x\cos x \) from the original forms: \[ \sin 4x = 4\sin x\cos x\left(2\cos^2 x - 1\right) \] 6. Express \( 2\cos^2 x-1 \) in a form that highlights a difference: \[ 2\cos^2 x - 1 = (\cos^2 x - \sin^2 x) \] (using the identity \( \cos^2 x - \sin^2 x = 2\cos^2 x-1 \)) 7. Thus, we have: \[ \sin 4x = 4\sin x\cos x(\cos^2 x - \sin^2 x) \] 8. Now, distribute to write the expression as: \[ \sin 4x = 4\sin x\cos^3 x - 4\sin x\cos x\sin^2 x \] 9. Recognize that \( \sin x\cos x\sin^2 x = \cos x\sin^3 x \), so: \[ \sin 4x = 4\sin x\cos^3 x - 4\cos x\sin^3 x \] Each of the identities is now proven using compound and double angle formulas.