Trigonometry · real student question

Simplify sin(9x)/cot(12x).

Question

Simplify

sin(9x)cot(12x).\frac{\sin(9x)}{\cot(12x)}.

Step-by-step solution

  1. Replace the cotangent by its definition. The quotient identity is

    cotθ=cosθsinθ,\cot\theta=\frac{\cos\theta}{\sin\theta},

    so with θ=12x\theta=12x the expression becomes

    sin(9x)cos(12x)sin(12x).\frac{\sin(9x)}{\dfrac{\cos(12x)}{\sin(12x)}}.

    Rewriting in terms of sine and cosine is almost always the right first move when a reciprocal trig function sits in an awkward place.

  2. Divide by a fraction by multiplying by its reciprocal. A compound fraction AB/C\dfrac{A}{B/C} equals ACBA\cdot\dfrac{C}{B}:

    sin(9x)cos(12x)sin(12x)=sin(9x)sin(12x)cos(12x).\frac{\sin(9x)}{\dfrac{\cos(12x)}{\sin(12x)}}=\sin(9x)\cdot\frac{\sin(12x)}{\cos(12x)}.

    The classic error is to write sin(9x)cos(12x)sin(12x)\sin(9x)\cdot\frac{\cos(12x)}{\sin(12x)} — dividing by cot\cot is the same as multiplying by tan\tan, not by cot\cot.

  3. Recognise the tangent. Since sinθcosθ=tanθ\dfrac{\sin\theta}{\cos\theta}=\tan\theta,

    sin(9x)cot(12x)=sin(9x)tan(12x).\frac{\sin(9x)}{\cot(12x)}=\sin(9x)\tan(12x).

    Equivalently the answer can be left as sin(9x)sin(12x)cos(12x)\dfrac{\sin(9x)\sin(12x)}{\cos(12x)}; both forms are correct, and the tan\tan version is the more compact one.

  4. State the domain restrictions. The original expression requires cot(12x)\cot(12x) to be defined and nonzero, so sin(12x)0\sin(12x)\ne0 and cos(12x)0\cos(12x)\ne0, i.e. 12xkπ212x\ne\frac{k\pi}{2}. The simplified form sin(9x)tan(12x)\sin(9x)\tan(12x) only needs cos(12x)0\cos(12x)\ne0, so it is defined at slightly more points than the original — the two agree wherever both make sense.

  5. Check numerically. At x=0.1x=0.1: sin(0.9)=0.783327\sin(0.9)=0.783327 and cot(1.2)=0.3887796\cot(1.2)=0.3887796, so the original quotient is 0.7833270.3887796=2.014836\frac{0.783327}{0.3887796}=2.014836; and sin(0.9)tan(1.2)=0.783327×2.572152=2.014836\sin(0.9)\tan(1.2)=0.783327\times2.572152=2.014836 ✓. Agreement at a value where neither function is near a pole confirms the algebra.

Answer

sin(9x)cot(12x)=sin(9x)tan(12x)=sin(9x)sin(12x)cos(12x)\frac{\sin(9x)}{\cot(12x)}=\sin(9x)\tan(12x)=\frac{\sin(9x)\sin(12x)}{\cos(12x)}

Need to solve a different problem like this? Open the solver →