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QuestionProve that sin⁡3θ+sin⁡θ=4sin⁡θ−4sin⁡3θ\sin 3\theta + \sin \theta = 4 \sin \theta - 4 \sin^3 \theta.

Studdy Solution

STEP 1

Assumptions1. We are dealing with trigonometric identities. . The variable θ\theta represents an angle in radians or degrees.
3. The symbol ≡\equiv means "is identically equal to", which means the equation holds true for all values of θ\theta.

STEP 2

We know the trigonometric identity for sin⁡θ\sin\theta is given bysin⁡θ=sin⁡θ−4sin⁡θ\sin\theta =\sin\theta -4\sin^\theta

STEP 3

Substitute the identity of sin⁡3θ\sin3\theta into the given equation.
sin⁡3θ+sin⁡θ≡sin⁡θ−sin⁡3θ\sin3\theta + \sin\theta \equiv\sin\theta -\sin^3\thetabecomes3sin⁡θ−sin⁡3θ+sin⁡θ≡sin⁡θ−sin⁡3θ3\sin\theta -\sin^3\theta + \sin\theta \equiv\sin\theta -\sin^3\theta

STEP 4

implify the left-hand side of the equation by combining like terms.
4sin⁡θ−4sin⁡3θ≡4sin⁡θ−4sin⁡3θ4\sin\theta -4\sin^3\theta \equiv4\sin\theta -4\sin^3\theta

STEP 5

Now, we can see that the left-hand side of the equation is identical to the right-hand side of the equation, which verifies the given identity.
Therefore, we have shown thatsin⁡3θ+sin⁡θ≡4sin⁡θ−4sin⁡3θ\sin3\theta + \sin\theta \equiv4\sin\theta -4\sin^3\theta

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