Prove that sin3theta=3sintheta-4sin^3theta
prove sin (3Ø) = 3sin Ø - 4sin^3 Ø
LS = sin(2Ø + Ø)
= sin2Ø cosØ + cos2ØsinØ
= 2sinØcosØcosØ + (1-2sin^2 Ø)sinØ
= 2sinØ(1-sin^2 Ø) + sinØ - 2sin^3 Ø
= 2sinØ - 2sin^2 Ø + sinØ - 2sin^3 Ø
= 3sinØ - 4sin^2 Ø
= RS
To prove the trigonometric identity sin(3theta) = 3sin(theta) - 4sin^3(theta), we can utilize the sum of angles formula for sine.
Start with the left-hand side (LHS):
sin(3theta)
Using the sum of angles formula for sine, we can rewrite this as:
sin(2theta + theta)
Applying the sum of angles formula, we obtain:
sin(2theta)cos(theta) + cos(2theta)sin(theta)
Now, let's express the right-hand side (RHS) and bring it to the same form:
3sin(theta) - 4sin^3(theta)
Using the identity sin^2(theta) = 1 - cos^2(theta), we can rewrite sin^3(theta) as:
sin(theta) - sin^2(theta)sin(theta)
Expanding further, we have:
3sin(theta) - 4(sin(theta) - sin^2(theta)sin(theta))
Distributing the -4, we get:
3sin(theta) - 4sin(theta) + 4sin^2(theta)sin(theta)
Combining like terms, we obtain:
- sin(theta) + 4sin^2(theta)sin(theta)
Now, using the identity sin^2(theta) = 1 - cos^2(theta), we can express sin^2(theta) as:
- sin(theta) + 4(1 - cos^2(theta))sin(theta)
Expanding further, we have:
- sin(theta) + 4sin(theta) - 4cos^2(theta)sin(theta)
Combining like terms again, we obtain:
3sin(theta) - 4cos^2(theta)sin(theta)
Now, using the identity 1 - cos^2(theta) = sin^2(theta), we can rewrite cos^2(theta) as:
3sin(theta) - 4sin^2(theta)sin(theta)
Therefore, we have successfully shown that the LHS (sin(3theta)) is equal to the RHS (3sin(theta) - 4sin^3(theta)).
To prove that sin(3θ) = 3sin(θ) - 4sin^3(θ), we can use the trigonometric identity for triple angle:
sin(3θ) = 3sin(θ) - 4sin^3(θ)
To get started, let's express sin(3θ) in terms of sine and cosine of angle θ using the angle addition trigonometric identity:
sin(3θ) = sin(2θ + θ)
Applying the angle addition identity for sine:
sin(3θ) = sin(2θ)cos(θ) + cos(2θ)sin(θ)
Next, we need to express sin(2θ) and cos(2θ) in terms of sine and cosine of angle θ using the double angle trigonometric identities.
sin(2θ) = 2sin(θ)cos(θ)
cos(2θ) = cos^2(θ) - sin^2(θ)
Let's substitute these values back into the equation:
sin(3θ) = (2sin(θ)cos(θ))cos(θ) + (cos^2(θ) - sin^2(θ))sin(θ)
Expanding and simplifying further:
sin(3θ) = 2sin(θ)cos^2(θ) + cos^2(θ)sin(θ) - sin^3(θ)
Rearranging terms:
sin(3θ) = sin(θ)(cos^2(θ) + 2cos^2(θ)) - sin^3(θ)
We can simplify cos^2(θ) + 2cos^2(θ) to 3cos^2(θ):
sin(3θ) = sin(θ)(3cos^2(θ)) - sin^3(θ)
Using the Pythagorean identity, cos^2(θ) + sin^2(θ) = 1, we can rewrite 3cos^2(θ) as 3(1 - sin^2(θ)):
sin(3θ) = sin(θ)(3(1 - sin^2(θ))) - sin^3(θ)
Simplifying further:
sin(3θ) = 3sin(θ) - 3sin^3(θ) - sin^3(θ)
Combining like terms:
sin(3θ) = 3sin(θ) - 4sin^3(θ)
Therefore, sin(3θ) = 3sin(θ) - 4sin^3(θ) is proven to be true.