Question

If a, b, c are interior angles of triangle abc , show that



cos² a/2 + cos ²( b + c/2) = 1

Answer 1

Hi...☺

Here is your answer...✌

GIVEN THAT,
a, b and c are interior angles of triangle

=> a + b + c = 180°
[ Angle sum property of triangle ]

Dividing both side by 2
We get,

$\frac{a + b + c}{2} = \frac{180 \degree}{2} \\ \\ \frac{a}{2} + \frac{b + c}{2} = 90 \degree\\ \\ \frac{a}{2} = 90\degree - \frac{b + c}{2} \: \: \: \: \: \: \: \: \: \: .....(1) \\ \\ Now ,\: \\ We \: know \:that, \\ \\ \cos {}^{2} ( \frac{a}{2} ) + \sin {}^{2} ( \frac{a}{2} ) = 1 \\ \\ \cos {}^{2} ( \frac{a}{2} ) + \sin {}^{2} (90\degree - \frac{b + c}{2} ) = 1 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \:\: \: \: \: \: [from \: (1)] \\ \cos {}^{2} ( \frac{a}{2} ) + \cos {}^{2} ( \frac{b + c}{2} ) = 1$

HENCE PROVED

Answer 2

Step-by-step explanation:

the above is the ans

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