Question

pls...find the value of A (trignometry) if;

2cos^2(A/2)=1

Answer 1

cos x =0 after that you can find all answers by trig equations

Answer 2

cos^2 (A/2) = (1 + cosA)/2 [An Identity] {Tell me if you want the proof}

So

\[\frac{ 1+cosA }{ 2 } = \frac{ 1 }{ 2 }\]

cos A = 0

A = 90'

Understood? :)

Answer 3

\[2\cos^2(\frac{A}{2})=1 \]
\[\cos^2(\frac{A}{2})=\frac{1 }{2} \rightarrow \cos(\frac{A}{2})=\sqrt{\frac{1 }{2} }\]
\[\rightarrow \cos(\frac{A}{2})=\frac{1 }{\sqrt 2}\]
\[\rightarrow \cos(\frac{A}{2})=\cos45^0 \rightarrow \frac{A}{2} =45^0\]
\[A=2 \times 45^0 \rightarrow A = 90^0\]

Answer 4

@basith

Answer 5

As you are not there
I'll just write the proof

\[\cos 2\theta= \cos(\theta+ \theta)= \cos \theta.\cos \theta - \sin \theta. \sin \theta = \cos^2 \theta - \sin^2 \theta = \cos^2 \theta - (1-\cos^2 \theta)\]

\[2\cos^2 \theta - 1 = \cos 2 \theta\]

NOW YOU CAN USE THIS!

OR

You can use @dpasingh 's method.

Answer 6

im here

Answer 7

thnx guys i understud.... :)

Answer 8

:)