Prove that 2cos3θ−cosθsinθ−2sin3θ=tanθ.
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Step-by-Step Solution
Step 1: Factor out common terms
To simplify the given expression, we first factor out the common terms from the numerator and the denominator. In the numerator, sinθ is common, and in the denominator, cosθ is common.
Step 2: Apply trigonometric identities
We know the identity sin2θ+cos2θ=1, which implies sin2θ=1−cos2θ. We substitute this into the numerator to express everything in terms of cosθ.
Step 3: Simplify the numerator
After substituting and simplifying the numerator, we get 1−2+2cos2θ=2cos2θ−1. Now, the term (2cos2θ−1) appears in both the numerator and the denominator.
Step 4: Cancel common terms and conclude
Since (2cos2θ−1) is a common factor in both the numerator and the denominator, we can cancel it out. This leaves us with cosθsinθ, which is equal to tanθ by definition.