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Basic Trigonometric Identities

For an acute ...

Question

For an acute angle A, prove that
${sin}^{2} A + {cos}^{2} A = 1$

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Solution

$sin^{2} A + cos^{2} A = 1 L H S = sin^{2} A + cos^{2} A = {(\frac{o p p o s i t e}{h y p o t e n u s e})}^{2} + {(\frac{a d j a c e n t}{h y p o t e n u s e})}^{2} = \frac{{B C}^{2}}{{A C}^{2}} + \frac{{A B}^{2}}{{A C}^{2}} = \frac{{B C}^{2} + {A B}^{2}}{{A C}^{2}}$
$= \frac{{A C}^{2}}{{A C}^{2}}$ [By pythagoras theorem]
$= 1 = R H S$

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