cos((7pi)/5)

{ "query": { "display": "$$\\cos\\left(\\frac{7π}{5}\\right)$$", "symbolab_question": "TRIG_EVALUATE#\\cos(\\frac{7π}{5})" }, "solution": { "level": "PERFORMED", "subject": "Trigonometry", "topic": "Evaluate Functions", "subTopic": "Simplified", "default": "-\\frac{\\sqrt{2}\\sqrt{3-\\sqrt{5}}}{4}", "decimal": "-0.30901…", "meta": { "showVerify": true } }, "steps": { "type": "interim", "title": "$$\\cos\\left(\\frac{7π}{5}\\right)=-\\frac{\\sqrt{2}\\sqrt{3-\\sqrt{5}}}{4}$$", "input": "\\cos\\left(\\frac{7π}{5}\\right)", "steps": [ { "type": "interim", "title": "Rewrite using trig identities:$${\\quad}-\\sin\\left(\\frac{9π}{10}\\right)$$", "input": "\\cos\\left(\\frac{7π}{5}\\right)", "result": "=-\\sin\\left(\\frac{9π}{10}\\right)", "steps": [ { "type": "step", "primary": "Use the following identity: $$\\cos\\left(x\\right)=\\sin\\left(\\frac{π}{2}-x\\right)$$", "result": "=\\sin\\left(\\frac{π}{2}-\\frac{7π}{5}\\right)" }, { "type": "interim", "title": "Simplify:$${\\quad}\\frac{π}{2}-\\frac{7π}{5}=-\\frac{9π}{10}$$", "input": "\\frac{π}{2}-\\frac{7π}{5}", "steps": [ { "type": "interim", "title": "Least Common Multiplier of $$2,\\:5:{\\quad}10$$", "input": "2,\\:5", "steps": [ { "type": "definition", "title": "Least Common Multiplier (LCM)", "text": "The LCM of $$a,\\:b$$ is the smallest positive number that is divisible by both $$a$$ and $$b$$" }, { "type": "interim", "title": "Prime factorization of $$2:{\\quad}2$$", "input": "2", "steps": [ { "type": "step", "primary": "$$2$$ is a prime number, therefore no factorization is possible", "result": "=2" } ], "meta": { "solvingClass": "Composite Integer", "interimType": "Prime Fac 1Eq", "gptData": "pG0PljGlka7rWtIVHz2xyoU2cWyPLDgE1QLLHeauuc3PHQdChPJ2JhfqHT+ZU0OMrfn8NOj0LUzuzje6xTyxRl8ZboA8wPLg0yhI4RzfjFw/y9DKGIPglJ+qMi9xDu2KE1OovxZAaXg7BtrFPk4UcCzRnGgMN6CYRfod7Mq0dp1+G9v2aKasChgV65VW8cTW" } }, { "type": "interim", "title": "Prime factorization of $$5:{\\quad}5$$", "input": "5", "steps": [ { "type": "step", "primary": "$$5$$ is a prime number, therefore no factorization is possible", "result": "=5" } ], "meta": { "solvingClass": "Composite Integer", "interimType": "Prime Fac 1Eq", "gptData": "pG0PljGlka7rWtIVHz2xyoU2cWyPLDgE1QLLHeauuc3PHQdChPJ2JhfqHT+ZU0OMrfn8NOj0LUzuzje6xTyxRjl/dE9e0owjU0NK6lxSAv4/y9DKGIPglJ+qMi9xDu2KE1OovxZAaXg7BtrFPk4UcCzRnGgMN6CYRfod7Mq0dp3mWpvzkJh0pk9SzVPr3Sj8" } }, { "type": "step", "primary": "Multiply each factor the greatest number of times it occurs in either $$2$$ or $$5$$", "result": "=2\\cdot\\:5" }, { "type": "step", "primary": "Multiply the numbers: $$2\\cdot\\:5=10$$", "result": "=10" } ], "meta": { "solvingClass": "LCM", "interimType": "LCM Top 1Eq" } }, { "type": "interim", "title": "Adjust Fractions based on the LCM", "steps": [ { "type": "step", "primary": "Multiply each numerator by the same amount needed to multiply its<br/>corresponding denominator to turn it into the LCM $$10$$" }, { "type": "step", "primary": "For $$\\frac{π}{2}:\\:$$multiply the denominator and numerator by $$5$$", "result": "\\frac{π}{2}=\\frac{π5}{2\\cdot\\:5}=\\frac{π5}{10}" }, { "type": "step", "primary": "For $$\\frac{7π}{5}:\\:$$multiply the denominator and numerator by $$2$$", "result": "\\frac{7π}{5}=\\frac{7π2}{5\\cdot\\:2}=\\frac{14π}{10}" } ], "meta": { "interimType": "LCD Adjust Fractions 1Eq" } }, { "type": "step", "result": "=\\frac{π5}{10}-\\frac{14π}{10}" }, { "type": "step", "primary": "Since the denominators are equal, combine the fractions: $$\\frac{a}{c}\\pm\\frac{b}{c}=\\frac{a\\pm\\:b}{c}$$", "result": "=\\frac{π5-14π}{10}" }, { "type": "step", "primary": "Add similar elements: $$5π-14π=-9π$$", "result": "=\\frac{-9π}{10}" }, { "type": "step", "primary": "Apply the fraction rule: $$\\frac{-a}{b}=-\\frac{a}{b}$$", "result": "=-\\frac{9π}{10}" } ], "meta": { "solvingClass": "Solver", "interimType": "Generic Simplify Title 0Eq" } }, { "type": "step", "result": "=\\sin\\left(-\\frac{9π}{10}\\right)" }, { "type": "step", "primary": "Use the following property: $$\\sin\\left(-x\\right)=-\\sin\\left(x\\right)$$", "secondary": [ "$$\\sin\\left(-\\frac{9π}{10}\\right)=-\\sin\\left(\\frac{9π}{10}\\right)$$" ], "result": "=-\\sin\\left(\\frac{9π}{10}\\right)" } ], "meta": { "interimType": "Trig Rewrite Using Trig identities Title 0Eq" } }, { "type": "interim", "title": "Rewrite using trig identities:$${\\quad}\\sin\\left(\\frac{9π}{10}\\right)=\\frac{\\sqrt{2}\\sqrt{3-\\sqrt{5}}}{4}$$", "input": "\\sin\\left(\\frac{9π}{10}\\right)", "steps": [ { "type": "interim", "title": "Rewrite using trig identities:$${\\quad}\\sin\\left(\\frac{π}{10}\\right)$$", "input": "\\sin\\left(\\frac{9π}{10}\\right)", "result": "=\\sin\\left(\\frac{π}{10}\\right)", "steps": [ { "type": "step", "primary": "Use the basic trigonometric identity: $$\\sin\\left(x\\right)=\\sin\\left(π-x\\right)$$", "result": "=\\sin\\left(π-\\frac{9π}{10}\\right)" }, { "type": "interim", "title": "Simplify:$${\\quad}π-\\frac{9π}{10}=\\frac{π}{10}$$", "input": "π-\\frac{9π}{10}", "steps": [ { "type": "step", "primary": "Convert element to fraction: $$π=\\frac{π10}{10}$$", "result": "=\\frac{π10}{10}-\\frac{9π}{10}" }, { "type": "step", "primary": "Since the denominators are equal, combine the fractions: $$\\frac{a}{c}\\pm\\frac{b}{c}=\\frac{a\\pm\\:b}{c}$$", "result": "=\\frac{π10-9π}{10}" }, { "type": "step", "primary": "Add similar elements: $$10π-9π=π$$", "result": "=\\frac{π}{10}" } ], "meta": { "solvingClass": "Solver", "interimType": "Generic Simplify Title 0Eq" } }, { "type": "step", "result": "=\\sin\\left(\\frac{π}{10}\\right)" } ], "meta": { "interimType": "Trig Rewrite Using Trig identities Title 0Eq" } }, { "type": "interim", "title": "Rewrite using trig identities:$${\\quad}\\sqrt{\\frac{1-\\cos\\left(\\frac{π}{5}\\right)}{2}}$$", "input": "\\sin\\left(\\frac{π}{10}\\right)", "result": "=\\sqrt{\\frac{1-\\cos\\left(\\frac{π}{5}\\right)}{2}}", "steps": [ { "type": "step", "primary": "Write $$\\sin\\left(\\frac{π}{10}\\right)\\:$$as $$\\sin\\left(\\frac{\\frac{π}{5}}{2}\\right)$$", "result": "=\\sin\\left(\\frac{\\frac{π}{5}}{2}\\right)" }, { "type": "interim", "title": "Use the Half Angle identity:$${\\quad}\\sin\\left(\\frac{θ}{2}\\right)=\\sqrt{\\frac{1-\\cos\\left(θ\\right)}{2}}$$", "steps": [ { "type": "step", "primary": "Use the Double Angle identity", "result": "\\cos\\left(2θ\\right)=1-2\\sin^{2}\\left(θ\\right)" }, { "type": "step", "primary": "Substitute $$θ$$ with $$\\frac{θ}{2}$$", "result": "\\cos\\left(θ\\right)=1-2\\sin^{2}\\left(\\frac{θ}{2}\\right)" }, { "type": "step", "primary": "Switch sides", "result": "2\\sin^{2}\\left(\\frac{θ}{2}\\right)=1-\\cos\\left(θ\\right)" }, { "type": "step", "primary": "Divide both sides by $$2$$", "result": "\\sin^{2}\\left(\\frac{θ}{2}\\right)=\\frac{\\left(1-\\cos\\left(θ\\right)\\right)}{2}" }, { "type": "step", "primary": "Square root both sides", "secondary": [ "Choose the root sign according to the quadrant of $$\\frac{\\theta}{2}$$:<br/>$$\\begin{array}{|c|c|c|c|}\\hline \\mathrm{range}&\\mathrm{quadrant}&\\sin&\\cos\\\\\\hline [0, \\frac{π}{2}]&I&\\mathrm{positive}&\\mathrm{positive}\\\\\\hline [\\frac{π}{2}, π]&II&\\mathrm{positive}&\\mathrm{negative}\\\\\\hline [π, \\frac{3π}{2}]&III&\\mathrm{negative}&\\mathrm{negative}\\\\\\hline [\\frac{3π}{2}, 2π]&IV&\\mathrm{negative}&\\mathrm{positive}\\\\\\hline \\end{array}$$" ], "result": "\\sin\\left(\\frac{θ}{2}\\right)=\\sqrt{\\frac{\\left(1-\\cos\\left(θ\\right)\\right)}{2}}" } ], "meta": { "interimType": "Trig Half Angle Identity Title 0Eq" } }, { "type": "step", "result": "=\\sqrt{\\frac{1-\\cos\\left(\\frac{π}{5}\\right)}{2}}" } ], "meta": { "interimType": "Trig Rewrite Using Trig identities Title 0Eq" } }, { "type": "interim", "title": "Rewrite using trig identities:$${\\quad}\\cos\\left(\\frac{π}{5}\\right)=\\frac{\\sqrt{5}+1}{4}$$", "input": "\\cos\\left(\\frac{π}{5}\\right)", "steps": [ { "type": "interim", "title": "Show that: $$\\cos\\left(\\frac{π}{5}\\right)-\\sin\\left(\\frac{π}{10}\\right)=\\frac{1}{2}$$", "steps": [ { "type": "step", "primary": "Use the following product to sum identity: $$2\\sin\\left(x\\right)\\cos\\left(y\\right)=\\sin\\left(x+y\\right)-\\sin\\left(x-y\\right)$$", "result": "2\\cos\\left(\\frac{π}{5}\\right)\\sin\\left(\\frac{π}{10}\\right)=\\sin\\left(\\frac{3π}{10}\\right)-\\sin\\left(\\frac{π}{10}\\right)" }, { "type": "interim", "title": "Show that: $$2\\cos\\left(\\frac{π}{5}\\right)\\sin\\left(\\frac{π}{10}\\right)=\\frac{1}{2}$$", "steps": [ { "type": "step", "primary": "Use the Double Angle identity: $$\\sin\\left(2x\\right)=2\\sin\\left(x\\right)\\cos\\left(x\\right)$$", "secondary": [ "$$\\sin\\left(\\frac{2π}{5}\\right)=2\\sin\\left(\\frac{π}{5}\\right)\\cos\\left(\\frac{π}{5}\\right)$$", "$$\\sin\\left(\\frac{π}{5}\\right)=2\\sin\\left(\\frac{π}{10}\\right)\\cos\\left(\\frac{π}{10}\\right)$$", "Multiply the two equations" ], "result": "\\sin\\left(\\frac{2π}{5}\\right)\\sin\\left(\\frac{π}{5}\\right)=4\\sin\\left(\\frac{π}{5}\\right)\\sin\\left(\\frac{π}{10}\\right)\\cos\\left(\\frac{π}{5}\\right)\\cos\\left(\\frac{π}{10}\\right)" }, { "type": "step", "primary": "Divide both sides by $$\\sin\\left(\\frac{π}{5}\\right)$$", "result": "\\sin\\left(\\frac{2π}{5}\\right)=4\\sin\\left(\\frac{π}{10}\\right)\\cos\\left(\\frac{π}{5}\\right)\\cos\\left(\\frac{π}{10}\\right)" }, { "type": "step", "primary": "Use the following identity: $$\\sin\\left(x\\right)=\\cos\\left(\\frac{π}{2}-x\\right)$$", "secondary": [ "$$\\sin\\left(\\frac{2π}{5}\\right)=\\cos\\left(\\frac{π}{2}-\\frac{2π}{5}\\right)$$" ], "result": "\\cos\\left(\\frac{π}{2}-\\frac{2π}{5}\\right)=4\\sin\\left(\\frac{π}{10}\\right)\\cos\\left(\\frac{π}{5}\\right)\\cos\\left(\\frac{π}{10}\\right)" }, { "type": "step", "result": "\\cos\\left(\\frac{π}{10}\\right)=4\\sin\\left(\\frac{π}{10}\\right)\\cos\\left(\\frac{π}{5}\\right)\\cos\\left(\\frac{π}{10}\\right)" }, { "type": "step", "primary": "Divide both sides by $$\\cos\\left(\\frac{π}{10}\\right)$$", "result": "1=4\\sin\\left(\\frac{π}{10}\\right)\\cos\\left(\\frac{π}{5}\\right)" }, { "type": "step", "primary": "Divide both sides by $$2$$", "result": "\\frac{1}{2}=2\\sin\\left(\\frac{π}{10}\\right)\\cos\\left(\\frac{π}{5}\\right)" } ], "meta": { "interimType": "Show That 1Eq" } }, { "type": "step", "primary": "Substitute $$\\frac{1}{2}=2\\sin\\left(\\frac{π}{10}\\right)\\cos\\left(\\frac{π}{5}\\right)$$", "result": "\\frac{1}{2}=\\sin\\left(\\frac{3π}{10}\\right)-\\sin\\left(\\frac{π}{10}\\right)" }, { "type": "step", "primary": "$$\\sin\\left(\\frac{3π}{10}\\right)=\\cos\\left(\\frac{π}{2}-\\frac{3π}{10}\\right)$$", "result": "\\frac{1}{2}=\\cos\\left(\\frac{π}{2}-\\frac{3π}{10}\\right)-\\sin\\left(\\frac{π}{10}\\right)" }, { "type": "step", "result": "\\frac{1}{2}=\\cos\\left(\\frac{π}{5}\\right)-\\sin\\left(\\frac{π}{10}\\right)" } ], "meta": { "interimType": "Show That 1Eq" } }, { "type": "interim", "title": "Show that: $$\\cos\\left(\\frac{π}{5}\\right)+\\sin\\left(\\frac{π}{10}\\right)=\\sqrt{\\frac{5}{4}}$$", "steps": [ { "type": "step", "primary": "Use the factorization rule: $$a^2-b^2=\\:\\left(a+b\\right)\\left(a-b\\right)$$", "secondary": [ "$$a=\\cos\\left(\\frac{π}{5}\\right)+\\sin\\left(\\frac{π}{10}\\right)$$", "$$b=\\cos\\left(\\frac{π}{5}\\right)-\\sin\\left(\\frac{π}{10}\\right)$$" ], "result": "\\left(\\cos\\left(\\frac{π}{5}\\right)+\\sin\\left(\\frac{π}{10}\\right)\\right)^{2}-\\left(\\cos\\left(\\frac{π}{5}\\right)-\\sin\\left(\\frac{π}{10}\\right)\\right)^{2}=\\left(\\left(\\cos\\left(\\frac{π}{5}\\right)+\\sin\\left(\\frac{π}{10}\\right)\\right)+\\left(\\cos\\left(\\frac{π}{5}\\right)-\\sin\\left(\\frac{π}{10}\\right)\\right)\\right)\\left(\\left(\\cos\\left(\\frac{π}{5}\\right)+\\sin\\left(\\frac{π}{10}\\right)\\right)-\\left(\\cos\\left(\\frac{π}{5}\\right)-\\sin\\left(\\frac{π}{10}\\right)\\right)\\right)" }, { "type": "step", "primary": "Refine", "result": "\\left(\\cos\\left(\\frac{π}{5}\\right)+\\sin\\left(\\frac{π}{10}\\right)\\right)^{2}-\\left(\\cos\\left(\\frac{π}{5}\\right)-\\sin\\left(\\frac{π}{10}\\right)\\right)^{2}=2\\left(2\\cos\\left(\\frac{π}{5}\\right)\\sin\\left(\\frac{π}{10}\\right)\\right)" }, { "type": "interim", "title": "Show that: $$2\\cos\\left(\\frac{π}{5}\\right)\\sin\\left(\\frac{π}{10}\\right)=\\frac{1}{2}$$", "steps": [ { "type": "step", "primary": "Use the Double Angle identity: $$\\sin\\left(2x\\right)=2\\sin\\left(x\\right)\\cos\\left(x\\right)$$", "secondary": [ "$$\\sin\\left(\\frac{2π}{5}\\right)=2\\sin\\left(\\frac{π}{5}\\right)\\cos\\left(\\frac{π}{5}\\right)$$", "$$\\sin\\left(\\frac{π}{5}\\right)=2\\sin\\left(\\frac{π}{10}\\right)\\cos\\left(\\frac{π}{10}\\right)$$", "Multiply the two equations" ], "result": "\\sin\\left(\\frac{2π}{5}\\right)\\sin\\left(\\frac{π}{5}\\right)=4\\sin\\left(\\frac{π}{5}\\right)\\sin\\left(\\frac{π}{10}\\right)\\cos\\left(\\frac{π}{5}\\right)\\cos\\left(\\frac{π}{10}\\right)" }, { "type": "step", "primary": "Divide both sides by $$\\sin\\left(\\frac{π}{5}\\right)$$", "result": "\\sin\\left(\\frac{2π}{5}\\right)=4\\sin\\left(\\frac{π}{10}\\right)\\cos\\left(\\frac{π}{5}\\right)\\cos\\left(\\frac{π}{10}\\right)" }, { "type": "step", "primary": "Use the following identity: $$\\sin\\left(x\\right)=\\cos\\left(\\frac{π}{2}-x\\right)$$", "secondary": [ "$$\\sin\\left(\\frac{2π}{5}\\right)=\\cos\\left(\\frac{π}{2}-\\frac{2π}{5}\\right)$$" ], "result": "\\cos\\left(\\frac{π}{2}-\\frac{2π}{5}\\right)=4\\sin\\left(\\frac{π}{10}\\right)\\cos\\left(\\frac{π}{5}\\right)\\cos\\left(\\frac{π}{10}\\right)" }, { "type": "step", "result": "\\cos\\left(\\frac{π}{10}\\right)=4\\sin\\left(\\frac{π}{10}\\right)\\cos\\left(\\frac{π}{5}\\right)\\cos\\left(\\frac{π}{10}\\right)" }, { "type": "step", "primary": "Divide both sides by $$\\cos\\left(\\frac{π}{10}\\right)$$", "result": "1=4\\sin\\left(\\frac{π}{10}\\right)\\cos\\left(\\frac{π}{5}\\right)" }, { "type": "step", "primary": "Divide both sides by $$2$$", "result": "\\frac{1}{2}=2\\sin\\left(\\frac{π}{10}\\right)\\cos\\left(\\frac{π}{5}\\right)" } ], "meta": { "interimType": "Show That 1Eq" } }, { "type": "step", "primary": "Substitute $$2\\cos\\left(\\frac{π}{5}\\right)\\sin\\left(\\frac{π}{10}\\right)=\\frac{1}{2}$$", "result": "\\left(\\cos\\left(\\frac{π}{5}\\right)+\\sin\\left(\\frac{π}{10}\\right)\\right)^{2}-\\left(\\cos\\left(\\frac{π}{5}\\right)-\\sin\\left(\\frac{π}{10}\\right)\\right)^{2}=1" }, { "type": "step", "primary": "Substitute $$\\cos\\left(\\frac{π}{5}\\right)-\\sin\\left(\\frac{π}{10}\\right)=\\frac{1}{2}$$", "result": "\\left(\\cos\\left(\\frac{π}{5}\\right)+\\sin\\left(\\frac{π}{10}\\right)\\right)^{2}-\\left(\\frac{1}{2}\\right)^{2}=1" }, { "type": "step", "primary": "Refine", "result": "\\left(\\cos\\left(\\frac{π}{5}\\right)+\\sin\\left(\\frac{π}{10}\\right)\\right)^{2}-\\frac{1}{4}=1" }, { "type": "step", "primary": "Add $$\\frac{1}{4}$$ to both sides", "result": "\\left(\\cos\\left(\\frac{π}{5}\\right)+\\sin\\left(\\frac{π}{10}\\right)\\right)^{2}-\\frac{1}{4}+\\frac{1}{4}=1+\\frac{1}{4}" }, { "type": "step", "primary": "Refine", "result": "\\left(\\cos\\left(\\frac{π}{5}\\right)+\\sin\\left(\\frac{π}{10}\\right)\\right)^{2}=\\frac{5}{4}" }, { "type": "step", "primary": "Take the square root of both sides", "result": "\\cos\\left(\\frac{π}{5}\\right)+\\sin\\left(\\frac{π}{10}\\right)=\\pm\\:\\sqrt{\\frac{5}{4}}" }, { "type": "step", "primary": "$$\\cos\\left(\\frac{π}{5}\\right)\\:$$cannot be negative", "secondary": [ "$$\\sin\\left(\\frac{π}{10}\\right)\\:$$cannot be negative" ], "result": "\\cos\\left(\\frac{π}{5}\\right)+\\sin\\left(\\frac{π}{10}\\right)=\\sqrt{\\frac{5}{4}}" } ], "meta": { "interimType": "Show That 1Eq" } }, { "type": "step", "primary": "Add the following equations", "secondary": [ "$$\\cos\\left(\\frac{π}{5}\\right)+\\sin\\left(\\frac{π}{10}\\right)=\\frac{\\sqrt{5}}{2}$$", "$$\\cos\\left(\\frac{π}{5}\\right)-\\sin\\left(\\frac{π}{10}\\right)=\\frac{1}{2}$$" ], "result": "\\left(\\left(\\cos\\left(\\frac{π}{5}\\right)+\\sin\\left(\\frac{π}{10}\\right)\\right)+\\left(\\cos\\left(\\frac{π}{5}\\right)-\\sin\\left(\\frac{π}{10}\\right)\\right)\\right)=\\left(\\frac{\\sqrt{5}}{2}+\\frac{1}{2}\\right)" }, { "type": "step", "primary": "Refine", "result": "\\cos\\left(\\frac{π}{5}\\right)=\\frac{\\sqrt{5}+1}{4}" }, { "type": "step", "result": "=\\frac{\\sqrt{5}+1}{4}" } ], "meta": { "interimType": "Trig Rewrite Using Trig identities Title 0Eq" } }, { "type": "step", "result": "=\\sqrt{\\frac{1-\\frac{\\sqrt{5}+1}{4}}{2}}" }, { "type": "interim", "title": "Simplify $$\\sqrt{\\frac{1-\\frac{\\sqrt{5}+1}{4}}{2}}:{\\quad}\\frac{\\sqrt{2}\\sqrt{3-\\sqrt{5}}}{4}$$", "input": "\\sqrt{\\frac{1-\\frac{\\sqrt{5}+1}{4}}{2}}", "result": "=\\frac{\\sqrt{2}\\sqrt{3-\\sqrt{5}}}{4}", "steps": [ { "type": "interim", "title": "$$\\frac{1-\\frac{\\sqrt{5}+1}{4}}{2}=\\frac{3-\\sqrt{5}}{8}$$", "input": "\\frac{1-\\frac{\\sqrt{5}+1}{4}}{2}", "steps": [ { "type": "interim", "title": "Join $$1-\\frac{\\sqrt{5}+1}{4}:{\\quad}\\frac{3-\\sqrt{5}}{4}$$", "input": "1-\\frac{\\sqrt{5}+1}{4}", "result": "=\\frac{\\frac{3-\\sqrt{5}}{4}}{2}", "steps": [ { "type": "step", "primary": "Convert element to fraction: $$1=\\frac{1\\cdot\\:4}{4}$$", "result": "=\\frac{1\\cdot\\:4}{4}-\\frac{\\sqrt{5}+1}{4}" }, { "type": "step", "primary": "Since the denominators are equal, combine the fractions: $$\\frac{a}{c}\\pm\\frac{b}{c}=\\frac{a\\pm\\:b}{c}$$", "result": "=\\frac{1\\cdot\\:4-\\left(\\sqrt{5}+1\\right)}{4}" }, { "type": "step", "primary": "Multiply the numbers: $$1\\cdot\\:4=4$$", "result": "=\\frac{4-\\left(1+\\sqrt{5}\\right)}{4}" }, { "type": "interim", "title": "Expand $$4-\\left(\\sqrt{5}+1\\right):{\\quad}3-\\sqrt{5}$$", "input": "4-\\left(\\sqrt{5}+1\\right)", "result": "=\\frac{3-\\sqrt{5}}{4}", "steps": [ { "type": "interim", "title": "$$-\\left(\\sqrt{5}+1\\right):{\\quad}-\\sqrt{5}-1$$", "input": "-\\left(\\sqrt{5}+1\\right)", "result": "=4-\\sqrt{5}-1", "steps": [ { "type": "step", "primary": "Distribute parentheses", "result": "=-\\left(\\sqrt{5}\\right)-\\left(1\\right)" }, { "type": "step", "primary": "Apply minus-plus rules", "secondary": [ "$$+\\left(-a\\right)=-a$$" ], "result": "=-\\sqrt{5}-1" } ], "meta": { "interimType": "N/A" } }, { "type": "step", "primary": "Subtract the numbers: $$4-1=3$$", "result": "=3-\\sqrt{5}" } ], "meta": { "interimType": "Algebraic Manipulation Expand Title 1Eq", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7/J3dGgB8pPTlGri5jViu89MvHyY50dhXPFfrjcmooUjHzrGrE/UqRiLXjZy0OtF4ca4U9Rxa0hK2LyJrWc715dbA+zX4bD3u3gx65o2NJhMjp5+1swD9TK8f03ykF6JXJvmp1p4a0lbu5DKRk2UXug==" } } ], "meta": { "interimType": "Algebraic Manipulation Join Concise Title 1Eq" } }, { "type": "step", "primary": "Apply the fraction rule: $$\\frac{\\frac{b}{c}}{a}=\\frac{b}{c\\:\\cdot\\:a}$$", "result": "=\\frac{3-\\sqrt{5}}{4\\cdot\\:2}" }, { "type": "step", "primary": "Multiply the numbers: $$4\\cdot\\:2=8$$", "result": "=\\frac{3-\\sqrt{5}}{8}" } ], "meta": { "solvingClass": "Solver", "interimType": "Solver", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7PndYYYIot5Qf6JPEzqJ6WOs7JGaKQiXU4o6aU/wy/k+YcM2VFKV06omE0l+DfOFfcJChiVhDxT5N/LHSTLMjyOtHxBP20+SYdC9Nfr3IMAsKnFYhPDBPYThI6650cVDw+kFkqWazQhzcnJ0o+hKZMBNl8iW2U2p5BaF79hdzYqP+cSas0wnwx68MY28z4QsvwNxEcIN5cs2cS0cGeOuMH8FGyAnbu3zZ2Xnj6PZ0lBY=" } }, { "type": "step", "result": "=\\sqrt{\\frac{3-\\sqrt{5}}{8}}" }, { "type": "step", "primary": "Apply radical rule: $$\\sqrt[n]{\\frac{a}{b}}=\\frac{\\sqrt[n]{a}}{\\sqrt[n]{b}},\\:\\quad$$ assuming $$a\\ge0,\\:b\\ge0$$", "result": "=\\frac{\\sqrt{3-\\sqrt{5}}}{\\sqrt{8}}", "meta": { "practiceLink": "/practice/radicals-practice", "practiceTopic": "Radical Rules" } }, { "type": "interim", "title": "$$\\sqrt{8}=2\\sqrt{2}$$", "input": "\\sqrt{8}", "result": "=\\frac{\\sqrt{3-\\sqrt{5}}}{2\\sqrt{2}}", "steps": [ { "type": "interim", "title": "Prime factorization of $$8:{\\quad}2^{3}$$", "input": "8", "result": "=\\sqrt{2^{3}}", "steps": [ { "type": "step", "primary": "$$8\\:$$divides by $$2\\quad\\:8=4\\cdot\\:2$$", "result": "=2\\cdot\\:4" }, { "type": "step", "primary": "$$4\\:$$divides by $$2\\quad\\:4=2\\cdot\\:2$$", "result": "=2\\cdot\\:2\\cdot\\:2" }, { "type": "step", "primary": "$$2$$ is a prime number, therefore no further factorization is possible", "result": "=2\\cdot\\:2\\cdot\\:2" }, { "type": "step", "result": "=2^{3}" } ], "meta": { "solvingClass": "Composite Integer", "interimType": "Prime Fac 1Eq", "gptData": "pG0PljGlka7rWtIVHz2xyoU2cWyPLDgE1QLLHeauuc3PHQdChPJ2JhfqHT+ZU0OMrfn8NOj0LUzuzje6xTyxRtXnBik8vDPXVw+nKWp28DI/y9DKGIPglJ+qMi9xDu2KE1OovxZAaXg7BtrFPk4UcCzRnGgMN6CYRfod7Mq0dp1RcgsS082tQWmOBW6FvhEw" } }, { "type": "step", "primary": "Apply exponent rule: $$a^{b+c}=a^b\\cdot\\:a^c$$", "result": "=\\sqrt{2^{2}\\cdot\\:2}", "meta": { "practiceLink": "/practice/exponent-practice", "practiceTopic": "Expand FOIL" } }, { "type": "step", "primary": "Apply radical rule: $$\\sqrt[n]{ab}=\\sqrt[n]{a}\\sqrt[n]{b}$$", "result": "=\\sqrt{2}\\sqrt{2^{2}}", "meta": { "practiceLink": "/practice/radicals-practice", "practiceTopic": "Radical Rules" } }, { "type": "step", "primary": "Apply radical rule: $$\\sqrt[n]{a^n}=a$$", "secondary": [ "$$\\sqrt{2^{2}}=2$$" ], "result": "=2\\sqrt{2}", "meta": { "practiceLink": "/practice/radicals-practice", "practiceTopic": "Radical Rules" } } ], "meta": { "interimType": "N/A" } }, { "type": "interim", "title": "Rationalize $$\\frac{\\sqrt{3-\\sqrt{5}}}{2\\sqrt{2}}:{\\quad}\\frac{\\sqrt{2}\\sqrt{3-\\sqrt{5}}}{4}$$", "input": "\\frac{\\sqrt{3-\\sqrt{5}}}{2\\sqrt{2}}", "result": "=\\frac{\\sqrt{2}\\sqrt{3-\\sqrt{5}}}{4}", "steps": [ { "type": "step", "primary": "Multiply by the conjugate $$\\frac{\\sqrt{2}}{\\sqrt{2}}$$", "result": "=\\frac{\\sqrt{3-\\sqrt{5}}\\sqrt{2}}{2\\sqrt{2}\\sqrt{2}}", "meta": { "title": { "extension": "To rationalize the denominator, multiply numerator and denominator by the conjugate of the radical $$\\sqrt{2}$$" } } }, { "type": "interim", "title": "$$2\\sqrt{2}\\sqrt{2}=4$$", "input": "2\\sqrt{2}\\sqrt{2}", "result": "=\\frac{\\sqrt{2}\\sqrt{3-\\sqrt{5}}}{4}", "steps": [ { "type": "step", "primary": "Apply exponent rule: $$a^b\\cdot\\:a^c=a^{b+c}$$", "secondary": [ "$$2\\sqrt{2}\\sqrt{2}=\\:2\\cdot\\:2^{\\frac{1}{2}}\\cdot\\:2^{\\frac{1}{2}}=\\:2^{1+\\frac{1}{2}+\\frac{1}{2}}$$" ], "result": "=2^{1+\\frac{1}{2}+\\frac{1}{2}}", "meta": { "practiceLink": "/practice/exponent-practice", "practiceTopic": "Expand FOIL" } }, { "type": "step", "primary": "Add similar elements: $$\\frac{1}{2}+\\frac{1}{2}=2\\cdot\\:\\frac{1}{2}$$", "result": "=2^{1+2\\cdot\\:\\frac{1}{2}}" }, { "type": "interim", "title": "$$2\\cdot\\:\\frac{1}{2}=1$$", "input": "2\\cdot\\:\\frac{1}{2}", "steps": [ { "type": "step", "primary": "Multiply fractions: $$a\\cdot\\frac{b}{c}=\\frac{a\\:\\cdot\\:b}{c}$$", "result": "=\\frac{1\\cdot\\:2}{2}" }, { "type": "step", "primary": "Cancel the common factor: $$2$$", "result": "=1" } ], "meta": { "solvingClass": "Solver", "interimType": "Solver", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7/OsC643lXZbU+VEjF1qviunYaDZZLRLMVMbjt+IYzOarju+5Z51e/ZZSD3gRHwjBE9/03SOiEv+BIHutWLr6nTQhqRfFhpwnpyuz2TS2M3xMz+u325qtellRtF+nqNeo" } }, { "type": "step", "result": "=2^{1+1}" }, { "type": "step", "primary": "Add the numbers: $$1+1=2$$", "result": "=2^{2}" }, { "type": "step", "primary": "$$2^{2}=4$$", "result": "=4" } ], "meta": { "solvingClass": "Solver", "interimType": "Solver", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7mGTgoVo05Sts7XaxvZWn37c/1aV0UaN4psdFpdDCCJpwkKGJWEPFPk38sdJMsyPIpxDcd3ce1jrro8deQTJg9Dy5FZVhty+dm32gdTi9c/N3RwQhNxCnUEv+iF15JJUV" } } ], "meta": { "interimType": "Rationalize Title 1Eq" } } ], "meta": { "solvingClass": "Solver", "interimType": "Generic Simplify Specific 1Eq", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7FJIkkmi1CWuhEmDQGlA0N2tmJEyNicOx9Ua+nriVUfIEVsE6gfrgu/hkYFccglMJ3oZCq59Hq2va8/E5S/sf7wVWfb8eY836P59a0g0AIKavHyURZf1bQrjajsxTSOQ1rWesMl3UTGoiMqFoPOhtVv8//6/nV5O4fb8Xgwi7mapyhd7tjiG+GxQNxDvGkZUlso+LtoJRIwsulWqdmg47D9Lklu58dNkF5ng1mQ4pBf0Dc2cNMhVCYAhuSboCp8VavzIPeEtDfcHv/z8uls8Teg==" } } ], "meta": { "interimType": "Trig Rewrite Using Trig identities Title 0Eq" } }, { "type": "step", "result": "=-\\frac{\\sqrt{2}\\sqrt{3-\\sqrt{5}}}{4}" } ], "meta": { "solvingClass": "Trig Evaluate", "practiceLink": "/practice/trigonometry-practice#area=main&subtopic=Evaluate%20Functions", "practiceTopic": "Evaluate Functions" } }, "meta": { "showVerify": true } }