(sin A/2 - cos A/2)² = 1 - sin A | Prove | Sub Multiple Angles | SciPiPupil

Question: Prove the following simple trigonometric identity: $\left ( sin \frac{A}{2} - cos \frac{A}{2} \right )^2 = 1 - sin A$

Solution:

To prove: $\left ( sin \frac{A}{2} - cos \frac{A}{2} \right )^2 = 1 - sin A$

Taking LHS,

$\left ( sin \frac{A}{2} - cos \frac{A}{2} \right )^2$

$= (sin \frac{A}{2} )^2 - 2 × sin \frac{A}{2} × cos \frac{A}{2} + (cos \frac{A}{2})^2$

$= sin^2 \frac{A}{2} + cos^2 \frac{A}{2} - 2 sin \frac{A}{2} cos \frac{A}{2}$

$= 1 - sin A$

= RHS

Some SUB Multiple Angles Formulae

$sin A = 2 sin\frac{A}{2}cos\frac{A}{2}$

$cos A = cos^2 \frac{A}{2} - sin^2 \frac{A}{2}$

$tan A = \dfrac{2 tan \frac{A}{2} }{1 - tan^2 \frac{A}{2}}$

$cot A = \dfrac{cot^2 \frac{A}{2} -1}{2 cot \frac{A}{2}}$

$sin A = 3 sin \frac{A}{3} - 4sin^3 \frac{A}{3}$

$cos A = 4cos^3 \frac{A}{3} - 3 cos \frac{A}{3}$

$tan A = \dfrac{3tan \frac{A}{3} - tan^3 \frac{A}{3}}{ 1 - 3tan^2 \frac{A}{3}}$

$cot A = \dfrac{3cot \frac{A}{3} - cot^3 \frac{A}{3}}{ 1 - 3cot^2 \frac{A}{3}}$

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