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Popular Trigonometry >

sin(c)= 2/(pi(-cos(c)+1))

  • Pre Algebra
  • Algebra
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Solution

sin(c)=π(−cos(c)+1)2​

Solution

c=1.23822…+2πn,c=2.80812…+2πn
+1
Degrees
c=70.94503…∘+360∘n,c=160.89345…∘+360∘n
Solution steps
sin(c)=π(−cos(c)+1)2​
Square both sidessin2(c)=(π(−cos(c)+1)2​)2
Subtract (π(−cos(c)+1)2​)2 from both sidessin2(c)−π2(−cos(c)+1)24​=0
Simplify sin2(c)−π2(−cos(c)+1)24​:π2(−cos(c)+1)2π2sin2(c)(−cos(c)+1)2−4​
sin2(c)−π2(−cos(c)+1)24​
Convert element to fraction: sin2(c)=π2(−cos(c)+1)2sin2(c)π2(−cos(c)+1)2​=π2(−cos(c)+1)2sin2(c)π2(−cos(c)+1)2​−π2(−cos(c)+1)24​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=π2(−cos(c)+1)2sin2(c)π2(−cos(c)+1)2−4​
π2(−cos(c)+1)2π2sin2(c)(−cos(c)+1)2−4​=0
g(x)f(x)​=0⇒f(x)=0π2sin2(c)(−cos(c)+1)2−4=0
Rewrite using trig identities
−4+(1−cos(c))2sin2(c)π2
Use the Pythagorean identity: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=−4+(1−cos(c))2(1−cos2(c))π2
−4+(1−cos(c))2(1−cos2(c))π2=0
Solve by substitution
−4+(1−cos(c))2(1−cos2(c))π2=0
Let: cos(c)=u−4+(1−u)2(1−u2)π2=0
−4+(1−u)2(1−u2)π2=0:u≈0.32647…,u≈−0.94491…
−4+(1−u)2(1−u2)π2=0
Expand −4+(1−u)2(1−u2)π2:−4+2π2u3−π2u4−2π2u+π2
−4+(1−u)2(1−u2)π2
(1−u)2=1−2u+u2
(1−u)2
Apply Perfect Square Formula: (a−b)2=a2−2ab+b2a=1,b=u
=12−2⋅1⋅u+u2
Simplify 12−2⋅1⋅u+u2:1−2u+u2
12−2⋅1⋅u+u2
Apply rule 1a=112=1=1−2⋅1⋅u+u2
Multiply the numbers: 2⋅1=2=1−2u+u2
=1−2u+u2
=−4+π2(u2−2u+1)(−u2+1)
=−4+π2(1−2u+u2)(1−u2)
Expand (1−2u+u2)(1−u2)π2:2π2u3−π2u4−2π2u+π2
Expand (1−2u+u2)(1−u2):2u3−u4−2u+1
(1−2u+u2)(1−u2)
Distribute parentheses=1⋅1+1⋅(−u2)+(−2u)⋅1+(−2u)(−u2)+u2⋅1+u2(−u2)
Apply minus-plus rules+(−a)=−a,(−a)(−b)=ab=1⋅1−1⋅u2−2⋅1⋅u+2u2u+1⋅u2−u2u2
Simplify 1⋅1−1⋅u2−2⋅1⋅u+2u2u+1⋅u2−u2u2:2u3−u4−2u+1
1⋅1−1⋅u2−2⋅1⋅u+2u2u+1⋅u2−u2u2
Group like terms=−1⋅u2+2u2u+1⋅u2−u2u2−2⋅1⋅u+1⋅1
Add similar elements: −1⋅u2+1⋅u2=0=2u2u−u2u2−2⋅1⋅u+1⋅1
2u2u=2u3
2u2u
Apply exponent rule: ab⋅ac=ab+cu2u=u2+1=2u2+1
Add the numbers: 2+1=3=2u3
u2u2=u4
u2u2
Apply exponent rule: ab⋅ac=ab+cu2u2=u2+2=u2+2
Add the numbers: 2+2=4=u4
2⋅1⋅u=2u
2⋅1⋅u
Multiply the numbers: 2⋅1=2=2u
1⋅1=1
1⋅1
Multiply the numbers: 1⋅1=1=1
=2u3−u4−2u+1
=2u3−u4−2u+1
=π2(2u3−u4−2u+1)
Expand π2(2u3−u4−2u+1):2π2u3−π2u4−2π2u+π2
π2(2u3−u4−2u+1)
Distribute parentheses=π2⋅2u3+π2(−u4)+π2(−2u)+π2⋅1
Apply minus-plus rules+(−a)=−a=2π2u3−π2u4−2π2u+1⋅π2
Multiply: 1⋅π2=π2=2π2u3−π2u4−2π2u+π2
=2π2u3−π2u4−2π2u+π2
=−4+2π2u3−π2u4−2π2u+π2
−4+2π2u3−π2u4−2π2u+π2=0
Write in the standard form an​xn+…+a1​x+c=0−π2u4+2π2u3−2π2u−4+π2=0
Find one solution for −9.86960…u4+19.73920…u3−19.73920…u+5.86960…=0 using Newton-Raphson:u≈0.32647…
−9.86960…u4+19.73920…u3−19.73920…u+5.86960…=0
Newton-Raphson Approximation Definition
f(u)=−9.86960…u4+19.73920…u3−19.73920…u+5.86960…
Find f′(u):−39.47841…u3+59.21762…u2−19.73920…
dud​(−9.86960…u4+19.73920…u3−19.73920…u+5.86960…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=−dud​(9.86960…u4)+dud​(19.73920…u3)−dud​(19.73920…u)+dud​(5.86960…)
dud​(9.86960…u4)=39.47841…u3
dud​(9.86960…u4)
Take the constant out: (a⋅f)′=a⋅f′=9.86960…dud​(u4)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=9.86960…⋅4u4−1
Simplify=39.47841…u3
dud​(19.73920…u3)=59.21762…u2
dud​(19.73920…u3)
Take the constant out: (a⋅f)′=a⋅f′=19.73920…dud​(u3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=19.73920…⋅3u3−1
Simplify=59.21762…u2
dud​(19.73920…u)=19.73920…
dud​(19.73920…u)
Take the constant out: (a⋅f)′=a⋅f′=19.73920…dudu​
Apply the common derivative: dudu​=1=19.73920…⋅1
Simplify=19.73920…
dud​(5.86960…)=0
dud​(5.86960…)
Derivative of a constant: dxd​(a)=0=0
=−39.47841…u3+59.21762…u2−19.73920…+0
Simplify=−39.47841…u3+59.21762…u2−19.73920…
Let u0​=0Compute un+1​ until Δun+1​<0.000001
u1​=0.29735…:Δu1​=0.29735…
f(u0​)=−9.86960…⋅04+19.73920…⋅03−19.73920…⋅0+5.86960…=5.86960…f′(u0​)=−39.47841…⋅03+59.21762…⋅02−19.73920…=−19.73920…u1​=0.29735…
Δu1​=∣0.29735…−0∣=0.29735…Δu1​=0.29735…
u2​=0.32578…:Δu2​=0.02843…
f(u1​)=−9.86960…⋅0.29735…4+19.73920…⋅0.29735…3−19.73920…⋅0.29735…+5.86960…=0.44183…f′(u1​)=−39.47841…⋅0.29735…3+59.21762…⋅0.29735…2−19.73920…=−15.54109…u2​=0.32578…
Δu2​=∣0.32578…−0.29735…∣=0.02843…Δu2​=0.02843…
u3​=0.32647…:Δu3​=0.00068…
f(u2​)=−9.86960…⋅0.32578…4+19.73920…⋅0.32578…3−19.73920…⋅0.32578…+5.86960…=0.01017…f′(u2​)=−39.47841…⋅0.32578…3+59.21762…⋅0.32578…2−19.73920…=−14.81908…u3​=0.32647…
Δu3​=∣0.32647…−0.32578…∣=0.00068…Δu3​=0.00068…
u4​=0.32647…:Δu4​=4.14683E−7
f(u3​)=−9.86960…⋅0.32647…4+19.73920…⋅0.32647…3−19.73920…⋅0.32647…+5.86960…=6.13781E−6f′(u3​)=−39.47841…⋅0.32647…3+59.21762…⋅0.32647…2−19.73920…=−14.80121…u4​=0.32647…
Δu4​=∣0.32647…−0.32647…∣=4.14683E−7Δu4​=4.14683E−7
u≈0.32647…
Apply long division:u−0.32647…−π2u4+2π2u3−2π2u−4+π2​=−9.86960…u3+16.51702…u2+5.39239…u−17.97872…
−9.86960…u3+16.51702…u2+5.39239…u−17.97872…≈0
Find one solution for −9.86960…u3+16.51702…u2+5.39239…u−17.97872…=0 using Newton-Raphson:u≈−0.94491…
−9.86960…u3+16.51702…u2+5.39239…u−17.97872…=0
Newton-Raphson Approximation Definition
f(u)=−9.86960…u3+16.51702…u2+5.39239…u−17.97872…
Find f′(u):−29.60881…u2+33.03405…u+5.39239…
dud​(−9.86960…u3+16.51702…u2+5.39239…u−17.97872…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=−dud​(9.86960…u3)+dud​(16.51702…u2)+dud​(5.39239…u)−dud​(17.97872…)
dud​(9.86960…u3)=29.60881…u2
dud​(9.86960…u3)
Take the constant out: (a⋅f)′=a⋅f′=9.86960…dud​(u3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=9.86960…⋅3u3−1
Simplify=29.60881…u2
dud​(16.51702…u2)=33.03405…u
dud​(16.51702…u2)
Take the constant out: (a⋅f)′=a⋅f′=16.51702…dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=16.51702…⋅2u2−1
Simplify=33.03405…u
dud​(5.39239…u)=5.39239…
dud​(5.39239…u)
Take the constant out: (a⋅f)′=a⋅f′=5.39239…dudu​
Apply the common derivative: dudu​=1=5.39239…⋅1
Simplify=5.39239…
dud​(17.97872…)=0
dud​(17.97872…)
Derivative of a constant: dxd​(a)=0=0
=−29.60881…u2+33.03405…u+5.39239…−0
Simplify=−29.60881…u2+33.03405…u+5.39239…
Let u0​=−1Compute un+1​ until Δun+1​<0.000001
u1​=−0.94732…:Δu1​=0.05267…
f(u0​)=−9.86960…(−1)3+16.51702…(−1)2+5.39239…(−1)−17.97872…=3.01551…f′(u0​)=−29.60881…(−1)2+33.03405…(−1)+5.39239…=−57.25047…u1​=−0.94732…
Δu1​=∣−0.94732…−(−1)∣=0.05267…Δu1​=0.05267…
u2​=−0.94491…:Δu2​=0.00241…
f(u1​)=−9.86960…(−0.94732…)3+16.51702…(−0.94732…)2+5.39239…(−0.94732…)−17.97872…=0.12652…f′(u1​)=−29.60881…(−0.94732…)2+33.03405…(−0.94732…)+5.39239…=−52.47351…u2​=−0.94491…
Δu2​=∣−0.94491…−(−0.94732…)∣=0.00241…Δu2​=0.00241…
u3​=−0.94491…:Δu3​=4.95571E−6
f(u2​)=−9.86960…(−0.94491…)3+16.51702…(−0.94491…)2+5.39239…(−0.94491…)−17.97872…=0.00025…f′(u2​)=−29.60881…(−0.94491…)2+33.03405…(−0.94491…)+5.39239…=−52.25876…u3​=−0.94491…
Δu3​=∣−0.94491…−(−0.94491…)∣=4.95571E−6Δu3​=4.95571E−6
u4​=−0.94491…:Δu4​=2.09107E−11
f(u3​)=−9.86960…(−0.94491…)3+16.51702…(−0.94491…)2+5.39239…(−0.94491…)−17.97872…=1.09276E−9f′(u3​)=−29.60881…(−0.94491…)2+33.03405…(−0.94491…)+5.39239…=−52.25832…u4​=−0.94491…
Δu4​=∣−0.94491…−(−0.94491…)∣=2.09107E−11Δu4​=2.09107E−11
u≈−0.94491…
Apply long division:u+0.94491…−9.86960…u3+16.51702…u2+5.39239…u−17.97872…​=−9.86960…u2+25.84293…u−19.02688…
−9.86960…u2+25.84293…u−19.02688…≈0
Find one solution for −9.86960…u2+25.84293…u−19.02688…=0 using Newton-Raphson:No Solution for u∈R
−9.86960…u2+25.84293…u−19.02688…=0
Newton-Raphson Approximation Definition
f(u)=−9.86960…u2+25.84293…u−19.02688…
Find f′(u):−19.73920…u+25.84293…
dud​(−9.86960…u2+25.84293…u−19.02688…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=−dud​(9.86960…u2)+dud​(25.84293…u)−dud​(19.02688…)
dud​(9.86960…u2)=19.73920…u
dud​(9.86960…u2)
Take the constant out: (a⋅f)′=a⋅f′=9.86960…dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=9.86960…⋅2u2−1
Simplify=19.73920…u
dud​(25.84293…u)=25.84293…
dud​(25.84293…u)
Take the constant out: (a⋅f)′=a⋅f′=25.84293…dudu​
Apply the common derivative: dudu​=1=25.84293…⋅1
Simplify=25.84293…
dud​(19.02688…)=0
dud​(19.02688…)
Derivative of a constant: dxd​(a)=0=0
=−19.73920…u+25.84293…−0
Simplify=−19.73920…u+25.84293…
Let u0​=1Compute un+1​ until Δun+1​<0.000001
u1​=1.50027…:Δu1​=0.50027…
f(u0​)=−9.86960…⋅12+25.84293…⋅1−19.02688…=−3.05355…f′(u0​)=−19.73920…⋅1+25.84293…=6.10372…u1​=1.50027…
Δu1​=∣1.50027…−1∣=0.50027…Δu1​=0.50027…
u2​=0.84530…:Δu2​=0.65497…
f(u1​)=−9.86960…⋅1.50027…2+25.84293…⋅1.50027…−19.02688…=−2.47014…f′(u1​)=−19.73920…⋅1.50027…+25.84293…=−3.77137…u2​=0.84530…
Δu2​=∣0.84530…−1.50027…∣=0.65497…Δu2​=0.65497…
u3​=1.30766…:Δu3​=0.46235…
f(u2​)=−9.86960…⋅0.84530…2+25.84293…⋅0.84530…−19.02688…=−4.23396…f′(u2​)=−19.73920…⋅0.84530…+25.84293…=9.15728…u3​=1.30766…
Δu3​=∣1.30766…−0.84530…∣=0.46235…Δu3​=0.46235…
u4​=70.11579…:Δu4​=68.80812…
f(u3​)=−9.86960…⋅1.30766…2+25.84293…⋅1.30766…−19.02688…=−2.10989…f′(u3​)=−19.73920…⋅1.30766…+25.84293…=0.03066…u4​=70.11579…
Δu4​=∣70.11579…−1.30766…∣=68.80812…Δu4​=68.80812…
u5​=35.71095…:Δu5​=34.40483…
f(u4​)=−9.86960…⋅70.11579…2+25.84293…⋅70.11579…−19.02688…=−46728.21617…f′(u4​)=−19.73920…⋅70.11579…+25.84293…=−1358.18729…u5​=35.71095…
Δu5​=∣35.71095…−70.11579…∣=34.40483…Δu5​=34.40483…
u6​=18.50697…:Δu6​=17.20397…
f(u5​)=−9.86960…⋅35.71095…2+25.84293…⋅35.71095…−19.02688…=−11682.58153…f′(u5​)=−19.73920…⋅35.71095…+25.84293…=−679.06298…u6​=18.50697…
Δu6​=∣18.50697…−35.71095…∣=17.20397…Δu6​=17.20397…
u7​=9.90188…:Δu7​=8.60509…
f(u6​)=−9.86960…⋅18.50697…2+25.84293…⋅18.50697…−19.02688…=−2921.17294…f′(u6​)=−19.73920…⋅18.50697…+25.84293…=−339.47016…u7​=9.90188…
Δu7​=∣9.90188…−18.50697…∣=8.60509…Δu7​=8.60509…
u8​=5.59311…:Δu8​=4.30877…
f(u7​)=−9.86960…⋅9.90188…2+25.84293…⋅9.90188…−19.02688…=−730.82108…f′(u7​)=−19.73920…⋅9.90188…+25.84293…=−169.61239…u8​=5.59311…
Δu8​=∣5.59311…−9.90188…∣=4.30877…Δu8​=4.30877…
u9​=3.42621…:Δu9​=2.16689…
f(u8​)=−9.86960…⋅5.59311…2+25.84293…⋅5.59311…−19.02688…=−183.23426…f′(u8​)=−19.73920…⋅5.59311…+25.84293…=−84.56065…u9​=3.42621…
Δu9​=∣3.42621…−5.59311…∣=2.16689…Δu9​=2.16689…
u10​=2.31722…:Δu10​=1.10898…
f(u9​)=−9.86960…⋅3.42621…2+25.84293…⋅3.42621…−19.02688…=−46.34217…f′(u9​)=−19.73920…⋅3.42621…+25.84293…=−41.78781…u10​=2.31722…
Δu10​=∣2.31722…−3.42621…∣=1.10898…Δu10​=1.10898…
u11​=1.70718…:Δu11​=0.61004…
f(u10​)=−9.86960…⋅2.31722…2+25.84293…⋅2.31722…−19.02688…=−12.13817…f′(u10​)=−19.73920…⋅2.31722…+25.84293…=−19.89727…u11​=1.70718…
Δu11​=∣1.70718…−2.31722…∣=0.61004…Δu11​=0.61004…
u12​=1.23961…:Δu12​=0.46756…
f(u11​)=−9.86960…⋅1.70718…2+25.84293…⋅1.70718…−19.02688…=−3.67298…f′(u11​)=−19.73920…⋅1.70718…+25.84293…=−7.85553…u12​=1.23961…
Δu12​=∣1.23961…−1.70718…∣=0.46756…Δu12​=0.46756…
u13​=2.81013…:Δu13​=1.57051…
f(u12​)=−9.86960…⋅1.23961…2+25.84293…⋅1.23961…−19.02688…=−2.15767…f′(u12​)=−19.73920…⋅1.23961…+25.84293…=1.37386…u13​=2.81013…
Δu13​=∣2.81013…−1.23961…∣=1.57051…Δu13​=1.57051…
u14​=1.98846…:Δu14​=0.82167…
f(u13​)=−9.86960…⋅2.81013…2+25.84293…⋅2.81013…−19.02688…=−24.34357…f′(u13​)=−19.73920…⋅2.81013…+25.84293…=−29.62687…u14​=1.98846…
Δu14​=∣1.98846…−2.81013…∣=0.82167…Δu14​=0.82167…
u15​=1.49147…:Δu15​=0.49698…
f(u14​)=−9.86960…⋅1.98846…2+25.84293…⋅1.98846…−19.02688…=−6.66341…f′(u14​)=−19.73920…⋅1.98846…+25.84293…=−13.40771…u15​=1.49147…
Δu15​=∣1.49147…−1.98846…∣=0.49698…Δu15​=0.49698…
u16​=0.81389…:Δu16​=0.67758…
f(u15​)=−9.86960…⋅1.49147…2+25.84293…⋅1.49147…−19.02688…=−2.43772…f′(u15​)=−19.73920…⋅1.49147…+25.84293…=−3.59765…u16​=0.81389…
Δu16​=∣0.81389…−1.49147…∣=0.67758…Δu16​=0.67758…
Cannot find solution
The solutions areu≈0.32647…,u≈−0.94491…
Substitute back u=cos(c)cos(c)≈0.32647…,cos(c)≈−0.94491…
cos(c)≈0.32647…,cos(c)≈−0.94491…
cos(c)=0.32647…:c=arccos(0.32647…)+2πn,c=2π−arccos(0.32647…)+2πn
cos(c)=0.32647…
Apply trig inverse properties
cos(c)=0.32647…
General solutions for cos(c)=0.32647…cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnc=arccos(0.32647…)+2πn,c=2π−arccos(0.32647…)+2πn
c=arccos(0.32647…)+2πn,c=2π−arccos(0.32647…)+2πn
cos(c)=−0.94491…:c=arccos(−0.94491…)+2πn,c=−arccos(−0.94491…)+2πn
cos(c)=−0.94491…
Apply trig inverse properties
cos(c)=−0.94491…
General solutions for cos(c)=−0.94491…cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnc=arccos(−0.94491…)+2πn,c=−arccos(−0.94491…)+2πn
c=arccos(−0.94491…)+2πn,c=−arccos(−0.94491…)+2πn
Combine all the solutionsc=arccos(0.32647…)+2πn,c=2π−arccos(0.32647…)+2πn,c=arccos(−0.94491…)+2πn,c=−arccos(−0.94491…)+2πn
Verify solutions by plugging them into the original equation
Check the solutions by plugging them into sin(c)=π(−cos(c)+1)2​
Remove the ones that don't agree with the equation.
Check the solution arccos(0.32647…)+2πn:True
arccos(0.32647…)+2πn
Plug in n=1arccos(0.32647…)+2π1
For sin(c)=π(−cos(c)+1)2​plug inc=arccos(0.32647…)+2π1sin(arccos(0.32647…)+2π1)=π(−cos(arccos(0.32647…)+2π1)+1)2​
Refine0.94520…=0.94520…
⇒True
Check the solution 2π−arccos(0.32647…)+2πn:False
2π−arccos(0.32647…)+2πn
Plug in n=12π−arccos(0.32647…)+2π1
For sin(c)=π(−cos(c)+1)2​plug inc=2π−arccos(0.32647…)+2π1sin(2π−arccos(0.32647…)+2π1)=π(−cos(2π−arccos(0.32647…)+2π1)+1)2​
Refine−0.94520…=0.94520…
⇒False
Check the solution arccos(−0.94491…)+2πn:True
arccos(−0.94491…)+2πn
Plug in n=1arccos(−0.94491…)+2π1
For sin(c)=π(−cos(c)+1)2​plug inc=arccos(−0.94491…)+2π1sin(arccos(−0.94491…)+2π1)=π(−cos(arccos(−0.94491…)+2π1)+1)2​
Refine0.32732…=0.32732…
⇒True
Check the solution −arccos(−0.94491…)+2πn:False
−arccos(−0.94491…)+2πn
Plug in n=1−arccos(−0.94491…)+2π1
For sin(c)=π(−cos(c)+1)2​plug inc=−arccos(−0.94491…)+2π1sin(−arccos(−0.94491…)+2π1)=π(−cos(−arccos(−0.94491…)+2π1)+1)2​
Refine−0.32732…=0.32732…
⇒False
c=arccos(0.32647…)+2πn,c=arccos(−0.94491…)+2πn
Show solutions in decimal formc=1.23822…+2πn,c=2.80812…+2πn

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Frequently Asked Questions (FAQ)

  • What is the general solution for sin(c)= 2/(pi(-cos(c)+1)) ?

    The general solution for sin(c)= 2/(pi(-cos(c)+1)) is c=1.23822…+2pin,c=2.80812…+2pin
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