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Exercise

$\frac{\sin\left(2^kx\right)}{2^k\sin\left(x\right)}+\:\cos\left(2^kx\right)$

Step-by-step Solution

1

Combine $\frac{\sin\left(2^kx\right)}{2^k\sin\left(x\right)}+\cos\left(2^kx\right)$ in a single fraction

$\frac{\sin\left(2^kx\right)+2^k\cos\left(2^kx\right)\sin\left(x\right)}{2^k\sin\left(x\right)}$

Final answer to the exercise

$\frac{\sin\left(2^kx\right)+2^k\cos\left(2^kx\right)\sin\left(x\right)}{2^k\sin\left(x\right)}$

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Main Topic: Equivalent Expressions

Two algebraic expressions are equivalent if they stay equal to each other, for any values their variables can take.

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