2+(8cos((pi*3)/3))-(2cos((2pi*3)/3))

{ "query": { "display": "$$2+\\left(8\\cos\\left(\\frac{π\\cdot\\:3}{3}\\right)\\right)-\\left(2\\cos\\left(\\frac{2π\\cdot\\:3}{3}\\right)\\right)$$", "symbolab_question": "TRIG_EVALUATE#2+(8\\cos(\\frac{π\\cdot 3}{3}))-(2\\cos(\\frac{2π\\cdot 3}{3}))" }, "solution": { "level": "PERFORMED", "subject": "Trigonometry", "topic": "Evaluate Functions", "subTopic": "Simplified", "default": "-8", "meta": { "showVerify": true } }, "steps": { "type": "interim", "title": "$$2+\\left(8\\cos\\left(\\frac{π3}{3}\\right)\\right)-\\left(2\\cos\\left(\\frac{2π3}{3}\\right)\\right)=-8$$", "input": "2+\\left(8\\cos\\left(\\frac{π3}{3}\\right)\\right)-\\left(2\\cos\\left(\\frac{2π3}{3}\\right)\\right)", "steps": [ { "type": "step", "result": "=2+8\\cos\\left(\\frac{π3}{3}\\right)-2\\cos\\left(\\frac{2π3}{3}\\right)" }, { "type": "step", "primary": "Simplify", "result": "=2+8\\cos\\left(π\\right)-2\\cos\\left(2π\\right)" }, { "type": "interim", "title": "$$\\cos\\left(2π\\right)=\\cos\\left(0\\right)$$", "input": "\\cos\\left(2π\\right)", "result": "=2+8\\cos\\left(π\\right)-2\\cos\\left(0\\right)", "steps": [ { "type": "step", "primary": "Rewrite $$2π$$ as $$2π+0$$", "result": "=\\cos\\left(2π+0\\right)" }, { "type": "step", "primary": "Apply the periodicity of $$\\cos$$: $$\\cos\\left(x+2π\\right)=\\cos\\left(x\\right)$$", "secondary": [ "$$\\cos\\left(2π+0\\right)=\\cos\\left(0\\right)$$" ], "result": "=\\cos\\left(0\\right)" } ], "meta": { "interimType": "N/A" } }, { "type": "interim", "title": "Use the following trivial identity:$${\\quad}\\cos\\left(π\\right)=\\left(-1\\right)$$", "input": "\\cos\\left(π\\right)", "steps": [ { "type": "step", "primary": "$$\\cos\\left(x\\right)$$ periodicity table with $$2πn$$ cycle:<br/>$$\\begin{array}{|c|c|c|c|}\\hline x&\\cos(x)&x&\\cos(x)\\\\\\hline 0&1&π&-1\\\\\\hline \\frac{π}{6}&\\frac{\\sqrt{3}}{2}&\\frac{7π}{6}&-\\frac{\\sqrt{3}}{2}\\\\\\hline \\frac{π}{4}&\\frac{\\sqrt{2}}{2}&\\frac{5π}{4}&-\\frac{\\sqrt{2}}{2}\\\\\\hline \\frac{π}{3}&\\frac{1}{2}&\\frac{4π}{3}&-\\frac{1}{2}\\\\\\hline \\frac{π}{2}&0&\\frac{3π}{2}&0\\\\\\hline \\frac{2π}{3}&-\\frac{1}{2}&\\frac{5π}{3}&\\frac{1}{2}\\\\\\hline \\frac{3π}{4}&-\\frac{\\sqrt{2}}{2}&\\frac{7π}{4}&\\frac{\\sqrt{2}}{2}\\\\\\hline \\frac{5π}{6}&-\\frac{\\sqrt{3}}{2}&\\frac{11π}{6}&\\frac{\\sqrt{3}}{2}\\\\\\hline \\end{array}$$" }, { "type": "step", "result": "=\\left(-1\\right)" } ], "meta": { "interimType": "Trig Trivial Angle Value Title 0Eq" } }, { "type": "interim", "title": "Use the following trivial identity:$${\\quad}\\cos\\left(0\\right)=1$$", "input": "\\cos\\left(0\\right)", "steps": [ { "type": "step", "primary": "$$\\cos\\left(x\\right)$$ periodicity table with $$2πn$$ cycle:<br/>$$\\begin{array}{|c|c|c|c|}\\hline x&\\cos(x)&x&\\cos(x)\\\\\\hline 0&1&π&-1\\\\\\hline \\frac{π}{6}&\\frac{\\sqrt{3}}{2}&\\frac{7π}{6}&-\\frac{\\sqrt{3}}{2}\\\\\\hline \\frac{π}{4}&\\frac{\\sqrt{2}}{2}&\\frac{5π}{4}&-\\frac{\\sqrt{2}}{2}\\\\\\hline \\frac{π}{3}&\\frac{1}{2}&\\frac{4π}{3}&-\\frac{1}{2}\\\\\\hline \\frac{π}{2}&0&\\frac{3π}{2}&0\\\\\\hline \\frac{2π}{3}&-\\frac{1}{2}&\\frac{5π}{3}&\\frac{1}{2}\\\\\\hline \\frac{3π}{4}&-\\frac{\\sqrt{2}}{2}&\\frac{7π}{4}&\\frac{\\sqrt{2}}{2}\\\\\\hline \\frac{5π}{6}&-\\frac{\\sqrt{3}}{2}&\\frac{11π}{6}&\\frac{\\sqrt{3}}{2}\\\\\\hline \\end{array}$$" }, { "type": "step", "result": "=1" } ], "meta": { "interimType": "Trig Trivial Angle Value Title 0Eq" } }, { "type": "step", "result": "=2+8\\left(-1\\right)-2\\cdot\\:1" }, { "type": "step", "primary": "Simplify", "result": "=-8" } ], "meta": { "solvingClass": "Trig Evaluate", "practiceLink": "/practice/trigonometry-practice#area=main&subtopic=Evaluate%20Functions", "practiceTopic": "Evaluate Functions" } }, "meta": { "showVerify": true } }