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Popular Trigonometry >

prove cos(2θ)=2cos^2(θ)-1

  • Pre Algebra
  • Algebra
  • Pre Calculus
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Solution

prove

Solution

True
Solution steps
cos(2θ)=2cos2(θ)−1
Manipulating left sidecos(2θ)
Rewrite using trig identities
cos(2θ)
Rewrite as=cos(θ+θ)
Use the Angle Sum identity: cos(s+s)=cos(s)cos(s)−sin(s)sin(s)=cos(θ)cos(θ)−sin(θ)sin(θ)
Simplify cos(θ)cos(θ)−sin(θ)sin(θ):cos2(θ)−sin2(θ)
cos(θ)cos(θ)−sin(θ)sin(θ)
cos(θ)cos(θ)=cos2(θ)
cos(θ)cos(θ)
Apply exponent rule: ab⋅ac=ab+ccos(θ)cos(θ)=cos1+1(θ)=cos1+1(θ)
Add the numbers: 1+1=2=cos2(θ)
sin(θ)sin(θ)=sin2(θ)
sin(θ)sin(θ)
Apply exponent rule: ab⋅ac=ab+csin(θ)sin(θ)=sin1+1(θ)=sin1+1(θ)
Add the numbers: 1+1=2=sin2(θ)
=cos2(θ)−sin2(θ)
=cos2(θ)−sin2(θ)
=cos2(θ)−sin2(θ)
Use the Pythagorean identity: cos2(θ)+sin2(θ)=1sin2(θ)=1−cos2(θ)=cos2(θ)−(1−cos2(θ))
Simplify cos2(θ)−(1−cos2(θ)):2cos2(θ)−1
cos2(θ)−(1−cos2(θ))
−(1−cos2(θ)):−1+cos2(θ)
−(1−cos2(θ))
Distribute parentheses=−1−(−cos2(θ))
Apply minus-plus rules−(−a)=a,−(a)=−a=−1+cos2(θ)
=cos2(θ)−1+cos2(θ)
Simplify cos2(θ)−1+cos2(θ):2cos2(θ)−1
cos2(θ)−1+cos2(θ)
Group like terms=cos2(θ)+cos2(θ)−1
Add similar elements: cos2(θ)+cos2(θ)=2cos2(θ)=2cos2(θ)−1
=2cos2(θ)−1
=2cos2(θ)−1
We showed that the two sides could take the same form⇒True

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sin((7pi)/8)sin((4pi)/3)sin^2(x)=3cos^2(x)sin(-pi)cos(5)

Frequently Asked Questions (FAQ)

  • Is cos(2θ)=2cos^2(θ)-1 ?

    The answer to whether cos(2θ)=2cos^2(θ)-1 is True
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