Question

Prove the following trigonometric identity:
$\frac{c o s A - 2 c o s ^{3} A}{2 s i n ^{3} A - s i n A} = cot A$

Approach

Prove the trigonometric identity $\frac{c o s A - 2 c o s ^{3} A}{2 s i n ^{3} A - s i n A} = cot A$ . LHS: factor numerator as $cos A (1 - 2 cos^{2} A)$ and denominator as $sin A (2 sin^{2} A - 1)$ . Substitute $cos^{2} A = 1 - sin^{2} A$ into the numerator to get $cos A [1 - 2 (1 - sin^{2} A)] = cos A (- 1 + 2 sin^{2} A)$ . Denominator is $sin A (2 sin^{2} A - 1)$ . Then LHS = $\frac{c o s A ( - 1 + 2 s i n ^{2} A )}{s i n A ( 2 s i n ^{2} A - 1 )} = \frac{c o s A}{s i n A} = cot A$ .

Step-by-step working

1
LHS = $\frac{c o s A ( 1 - 2 c o s ^{2} A )}{s i n A ( 2 s i n ^{2} A - 1 )}$ .
0.5 marks
2
Substitute $cos^{2} A = 1 - sin^{2} A$ : numerator becomes $cos A [1 - 2 (1 - sin^{2} A)] = cos A (- 1 + 2 sin^{2} A)$ .
1 mark
3
LHS = $\frac{c o s A ( - 1 + 2 s i n ^{2} A )}{s i n A ( 2 s i n ^{2} A - 1 )} = \frac{c o s A ( 2 s i n ^{2} A - 1 )}{s i n A ( 2 s i n ^{2} A - 1 )} = \frac{c o s A}{s i n A} = cot A$ (noting sign cancellation).
1 mark
4
Therefore, LHS = RHS = $cot A$ .
0.5 marks

Concepts used

trigonometric identityprovecot asimplify

Approach

Step-by-step working

Concepts used

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