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Popular Trigonometry >

prove (sin(a+2a)+4sin^3(a))/3 =sin(a)

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Solution

prove 3sin(a+2a)+4sin3(a)​=sin(a)

Solution

True
Solution steps
3sin(a+2a)+4sin3(a)​=sin(a)
Manipulating left side3sin(a+2a)+4sin3(a)​
Simplify=3sin(3a)+4sin3(a)​
Rewrite using trig identities
3sin(3a)+4sin3(a)​
Use the following identity:sin(3x)=3sin(x)−4sin3(x)
sin(3x)
Rewrite using trig identities
sin(3x)
Rewrite as=sin(2x+x)
Use the Angle Sum identity: sin(s+t)=sin(s)cos(t)+cos(s)sin(t)=sin(2x)cos(x)+cos(2x)sin(x)
Use the Double Angle identity: sin(2x)=2sin(x)cos(x)=cos(2x)sin(x)+cos(x)2sin(x)cos(x)
Simplify cos(2x)sin(x)+cos(x)⋅2sin(x)cos(x):sin(x)cos(2x)+2cos2(x)sin(x)
cos(2x)sin(x)+cos(x)2sin(x)cos(x)
cos(x)⋅2sin(x)cos(x)=2cos2(x)sin(x)
cos(x)2sin(x)cos(x)
Apply exponent rule: ab⋅ac=ab+ccos(x)cos(x)=cos1+1(x)=2sin(x)cos1+1(x)
Add the numbers: 1+1=2=2sin(x)cos2(x)
=sin(x)cos(2x)+2cos2(x)sin(x)
=sin(x)cos(2x)+2cos2(x)sin(x)
=sin(x)cos(2x)+2cos2(x)sin(x)
Use the Double Angle identity: cos(2x)=1−2sin2(x)=(1−2sin2(x))sin(x)+2cos2(x)sin(x)
Use the Pythagorean identity: cos2(x)+sin2(x)=1cos2(x)=1−sin2(x)=(1−2sin2(x))sin(x)+2(1−sin2(x))sin(x)
Expand (1−2sin2(x))sin(x)+2(1−sin2(x))sin(x):−4sin3(x)+3sin(x)
(1−2sin2(x))sin(x)+2(1−sin2(x))sin(x)
=sin(x)(1−2sin2(x))+2sin(x)(1−sin2(x))
Expand sin(x)(1−2sin2(x)):sin(x)−2sin3(x)
sin(x)(1−2sin2(x))
Apply the distributive law: a(b−c)=ab−aca=sin(x),b=1,c=2sin2(x)=sin(x)1−sin(x)2sin2(x)
=1sin(x)−2sin2(x)sin(x)
Simplify 1⋅sin(x)−2sin2(x)sin(x):sin(x)−2sin3(x)
1sin(x)−2sin2(x)sin(x)
1⋅sin(x)=sin(x)
1sin(x)
Multiply: 1⋅sin(x)=sin(x)=sin(x)
2sin2(x)sin(x)=2sin3(x)
2sin2(x)sin(x)
Apply exponent rule: ab⋅ac=ab+csin2(x)sin(x)=sin2+1(x)=2sin2+1(x)
Add the numbers: 2+1=3=2sin3(x)
=sin(x)−2sin3(x)
=sin(x)−2sin3(x)
=sin(x)−2sin3(x)+2(1−sin2(x))sin(x)
Expand 2sin(x)(1−sin2(x)):2sin(x)−2sin3(x)
2sin(x)(1−sin2(x))
Apply the distributive law: a(b−c)=ab−aca=2sin(x),b=1,c=sin2(x)=2sin(x)1−2sin(x)sin2(x)
=2⋅1sin(x)−2sin2(x)sin(x)
Simplify 2⋅1⋅sin(x)−2sin2(x)sin(x):2sin(x)−2sin3(x)
2⋅1sin(x)−2sin2(x)sin(x)
2⋅1⋅sin(x)=2sin(x)
2⋅1sin(x)
Multiply the numbers: 2⋅1=2=2sin(x)
2sin2(x)sin(x)=2sin3(x)
2sin2(x)sin(x)
Apply exponent rule: ab⋅ac=ab+csin2(x)sin(x)=sin2+1(x)=2sin2+1(x)
Add the numbers: 2+1=3=2sin3(x)
=2sin(x)−2sin3(x)
=2sin(x)−2sin3(x)
=sin(x)−2sin3(x)+2sin(x)−2sin3(x)
Simplify sin(x)−2sin3(x)+2sin(x)−2sin3(x):−4sin3(x)+3sin(x)
sin(x)−2sin3(x)+2sin(x)−2sin3(x)
Group like terms=−2sin3(x)−2sin3(x)+sin(x)+2sin(x)
Add similar elements: −2sin3(x)−2sin3(x)=−4sin3(x)=−4sin3(x)+sin(x)+2sin(x)
Add similar elements: sin(x)+2sin(x)=3sin(x)=−4sin3(x)+3sin(x)
=−4sin3(x)+3sin(x)
=−4sin3(x)+3sin(x)
=3sin(x)−4sin3(x)
=33sin(a)−4sin3(a)+4sin3(a)​
Simplify 33sin(a)−4sin3(a)+4sin3(a)​:sin(a)
33sin(a)−4sin3(a)+4sin3(a)​
Add similar elements: −4sin3(a)+4sin3(a)=0=33sin(a)​
Divide the numbers: 33​=1=sin(a)
=sin(a)
=sin(a)
We showed that the two sides could take the same form⇒True

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Frequently Asked Questions (FAQ)

  • Is (sin(a+2a)+4sin^3(a))/3 =sin(a) ?

    The answer to whether (sin(a+2a)+4sin^3(a))/3 =sin(a) is True
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