Prove that sin 4x = 4sin x cos³x − 4cos x sin³x Prove that \[ \sin4x=4\sin x\cos^3x-4\cos x\sin^3x \] Proof: Using the double angle identity: \[ \sin4x=2\sin2x\cos2x \] Also, \[ \sin2x=2\sin x\cos x \] and \[ \cos2x=\cos^2x-\sin^2x \] Substituting these values: \[ \sin4x = 2(2\sin x\cos x)(\cos^2x-\sin^2x) \] \[ = 4\sin x\cos x(\cos^2x-\sin^2x) \] Multiplying: […]

Prove that cos 4x = 1 − 8cos²x + 8cos⁴x Prove that \[ \cos4x=1-8\cos^2x+8\cos^4x \] Proof: Using the double angle identity: \[ \cos4x=1-2\sin^22x \] Also, \[ \sin2x=2\sin x\cos x \] Squaring both sides: \[ \sin^22x=4\sin^2x\cos^2x \] Substituting: \[ \cos4x = 1-2(4\sin^2x\cos^2x) \] \[ = 1-8\sin^2x\cos^2x \] Using \[ \sin^2x=1-\cos^2x \] we get \[ \cos4x =

Prove that cos²(π/4 − x) − sin²(π/4 − x) = sin 2x Prove that \[ \cos^2\left(\frac{\pi}{4}-x\right) – \sin^2\left(\frac{\pi}{4}-x\right) = \sin2x \] Proof: Using the identity \[ \cos^2A-\sin^2A=\cos2A \] let \[ A=\frac{\pi}{4}-x \] Then, \[ LHS = \cos\left[ 2\left( \frac{\pi}{4}-x \right) \right] \] \[ = \cos\left( \frac{\pi}{2}-2x \right) \] Using the identity \[ \cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta \] we

Prove that (sin 3x + sin x)sin x + (cos 3x − cos x)cos x = 0 Prove that \[ (\sin3x+\sin x)\sin x+(\cos3x-\cos x)\cos x=0 \] Proof: \[ LHS=(\sin3x+\sin x)\sin x+(\cos3x-\cos x)\cos x \] Expanding: \[ =\sin3x\sin x+\sin^2x+\cos3x\cos x-\cos^2x \] Grouping terms: \[ =(\sin3x\sin x+\cos3x\cos x) +(\sin^2x-\cos^2x) \] Using the identity \[ \cos(A-B)=\cos A\cos B+\sin

Prove that cos³2x + 3cos2x = 4(cos⁶x − sin⁶x) Prove that \[ \cos^32x+3\cos2x=4(\cos^6x-\sin^6x) \] Proof: Start with the right-hand side: \[ RHS=4(\cos^6x-\sin^6x) \] Using the identity \[ a^3-b^3=(a-b)(a^2+ab+b^2) \] we get \[ \cos^6x-\sin^6x = (\cos^2x-\sin^2x) (\cos^4x+\cos^2x\sin^2x+\sin^4x) \] Using \[ \cos2x=\cos^2x-\sin^2x \] therefore, \[ RHS = 4\cos2x (\cos^4x+\cos^2x\sin^2x+\sin^4x) \] Now, \[ \cos^4x+\sin^4x = (\cos^2x+\sin^2x)^2 -2\sin^2x\cos^2x \]

Prove that 1 + cos²2x = 2(cos⁴x + sin⁴x) Prove that \[ 1+\cos^22x=2(\cos^4x+\sin^4x) \] Proof: Start with the right-hand side: \[ RHS=2(\cos^4x+\sin^4x) \] Using the identity \[ a^2+b^2=(a+b)^2-2ab \] we get \[ \cos^4x+\sin^4x = (\cos^2x+\sin^2x)^2 -2\sin^2x\cos^2x \] Using \[ \sin^2x+\cos^2x=1 \] therefore, \[ \cos^4x+\sin^4x = 1-2\sin^2x\cos^2x \] Hence, \[ RHS = 2\left(1-2\sin^2x\cos^2x\right) \] \[ =

Prove that sin²(π/8 + x/2) − sin²(π/8 − x/2) = (1/√2) sin x Prove that \[ \sin^2\left(\frac{\pi}{8}+\frac{x}{2}\right) – \sin^2\left(\frac{\pi}{8}-\frac{x}{2}\right) = \frac{1}{\sqrt2}\sin x \] Proof: Using the identity \[ \sin^2A-\sin^2B=(\sin A-\sin B)(\sin A+\sin B) \] let \[ A=\frac{\pi}{8}+\frac{x}{2} \] and \[ B=\frac{\pi}{8}-\frac{x}{2} \] Then, \[ LHS= (\sin A-\sin B)(\sin A+\sin B) \] Using the identities \[

Prove that (cos α + cos β)² + (sin α + sin β)² = 4cos²((α − β)/2) Prove that \[ (\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2 =4\cos^2\left(\frac{\alpha-\beta}{2}\right) \] Proof: \[ LHS=(\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2 \] Expanding both squares: \[ =\cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta \] \[ +\sin^2\alpha+\sin^2\beta+2\sin\alpha\sin\beta \] Grouping terms: \[ =(\cos^2\alpha+\sin^2\alpha) +(\cos^2\beta+\sin^2\beta) \] \[ +2(\cos\alpha\cos\beta+\sin\alpha\sin\beta) \] Using \[ \cos^2\theta+\sin^2\theta=1 \] and \[ \cos(\alpha-\beta) =\cos\alpha\cos\beta+\sin\alpha\sin\beta \] we

Prove that sin²(π/8) + sin²(3π/8) + sin²(5π/8) + sin²(7π/8) = 2 Prove that \[ \sin^2\frac{\pi}{8} +\sin^2\frac{3\pi}{8} +\sin^2\frac{5\pi}{8} +\sin^2\frac{7\pi}{8} =2 \] Proof: Using the identity \[ \sin(\pi-\theta)=\sin\theta \] therefore, \[ \sin^2\frac{5\pi}{8}=\sin^2\frac{3\pi}{8} \] and \[ \sin^2\frac{7\pi}{8}=\sin^2\frac{\pi}{8} \] Hence, \[ LHS =2\sin^2\frac{\pi}{8} +2\sin^2\frac{3\pi}{8} \] \[ =2\left( \sin^2\frac{\pi}{8} +\sin^2\frac{3\pi}{8} \right) \] Using \[ \sin^2A=\frac{1-\cos2A}{2} \] we get \[ LHS

Prove that cos²(π/8) + cos²(3π/8) + cos²(5π/8) + cos²(7π/8) = 2 Prove that \[ \cos^2\frac{\pi}{8}+\cos^2\frac{3\pi}{8}+\cos^2\frac{5\pi}{8}+\cos^2\frac{7\pi}{8}=2 \] Proof: Using the identity \[ \cos(\pi-\theta)=-\cos\theta \] therefore, \[ \cos^2\frac{5\pi}{8}=\cos^2\frac{3\pi}{8} \] and \[ \cos^2\frac{7\pi}{8}=\cos^2\frac{\pi}{8} \] Hence, \[ LHS=2\cos^2\frac{\pi}{8}+2\cos^2\frac{3\pi}{8} \] \[ =2\left(\cos^2\frac{\pi}{8}+\cos^2\frac{3\pi}{8}\right) \] Using \[ \cos^2A=\frac{1+\cos2A}{2} \] we get \[ LHS=2\left( \frac{1+\cos\frac{\pi}{4}}{2} + \frac{1+\cos\frac{3\pi}{4}}{2} \right) \] \[ =2\left( \frac{1+\frac{\sqrt2}{2}}{2} +