Math Doubts

Pythagorean identity of Sine and Cosine functions

The summation of the squares of sine and cosine functions at an angle equals to one, is called Pythagorean identity for sine and cosine functions.


$\sin^2{\theta} + \cos^2{\theta} = 1$


right angled triangle

$\Delta BAC$ is a right angled triangle and its angle is theta. $\overline{AB}$, $\overline{BC}$ and $\overline{CA}$ are adjacent side, opposite side and hypotenuse respectively. The lengths of them are $AB$, $BC$ and $CA$ respectively.

Expressing Relation between sides

According to the Pythagorean Theorem, write the relation between three sides in terms of lengths of all three sides of the triangle.

${AC}^2 = {BC}^2 + {AB}^2$

$\implies {BC}^2 + {AB}^2 = {AC}^2$

Adjusting the mathematical relation

The mathematical equation can be expressed in terms of sine and cosine functions by dividing both sides of the expressions by ${AC}^2$.

$\implies \dfrac{{BC}^2 + {AB}^2}{{AC}^2} = \dfrac{{AC}^2}{{AC}^2}$

$\require{cancel} \implies \dfrac{{BC}^2}{{AC}^2} + \dfrac{{AB}^2}{{AC}^2} = \dfrac{\cancel{{AC}^2}}{\cancel{{AC}^2}}$

$\implies \dfrac{{BC}^2}{{AC}^2} + \dfrac{{AB}^2}{{AC}^2} = 1$

$\implies {\Bigg( \dfrac{BC}{AC} \Bigg)}^2 + {\Bigg(\dfrac{AB}{AC}\Bigg)}^2 = 1$

Expressing the Equation in terms of Sine and Cosine

According to the right angled triangle $BAC$.

$\dfrac{BC}{AC} = \sin \theta$ and $\dfrac{AB}{AC} = \cos \theta$

Replace the ratios of two sides in terms of associated trigonometric functions.

$\implies (\sin{\theta})^2 + (\cos{\theta})^2 = 1$

$\,\,\, \therefore \,\,\,\,\,\,\, \sin^2{\theta} + \cos^2{\theta} = 1$

Therefore, it is proved that the sum of the squares of the sine and cosine functions at an angle is equal to one. The trigonometric identity is derived on the basis of Pythagoras Theorem. Hence, it is called as the Pythagorean identity for sine and cosine functions.

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