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# Proving trigonometric identities

Summary:

Proving trigonometric identities involves showing that two different trigonometric expressions are equivalent. This is done by manipulating one or both sides of the identity using fundamental trigonometric identities, algebraic operations, and known values to transform them into the same expression.

Prove the identity cotx*sinx=cosx/(cos^2x+sin^2x)

We have to prove that cot x*sin x = cos x /((cos x)^2 + (sin x)^2)

Now we know that (cos x)^2 + (sin x)^2 = 1

Also, cot x = cos x / sin x

So cot x*sin x = (cos x / sin x)* sin x = cos x

cos x /((cos x)^2 + (sin x)^2) = cos x /1 = cos x

Therefore both the sides are equal to cos x.

We prove that cot x*sin x = cos x /((cos x)^2 + (sin x)^2).

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Prove the following identity: (1+ sin x + cos x) / (1+ sin x - cos x) = cot x/2

We have to prove that (1+ sin x + cos x) / (1+ sin x - cos x) = cot x/2

Now sin 2x = 2 sin x * cos x and cos 2x = 2 (cos x)^2 - 1

(1+ sin x + cos x) / (1+ sin x - cos x)

=> [1+2*(sin x/2)*(cos x/2) + 2*(cos x/2)^2 - 1]/ [1 + 2*(sin x/2)*(cos x/2)-1+2( sin x/2)^2]

eliminate 1 as we have -1 and  +1 in the numerator and denominator.

[2*(sin x/2)*(cos x/2) + 2*(cos x/2)^2]/ [2*(sin x/2)*(cos x/2)+2( sin x/2)^2]

cancel 2 from the numerator and denominator

=> [(sin x/2)*(cos x/2) + (cos x/2)^2 ] / [(sin x/2)*(cos x/2) + ( sin x/2)^2]

factorize

=> [cos x/2*( sin x/2 + cos x/2)]/[sin x/2*( cos x/2 + sin x/2)]

cancel (sin x/2 + cos x/2)

=> (cos x/2) / (sin x/2)

=> cot x/2

Therefore we have proved that (1+ sin x + cos x) / (1+ sin x - cos x) = cot x/2

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Prove the following identity: cos x + cos 2x + cos 3x = cos 2x(1 + 2cos x)

We have to prove : cos x + cos 2x + cos 3x = cos 2x(1 + 2cos x)

We know that cos A + cos B = 2 * cos (A + B)/2 * cos (A − B)/2

Now cos x + cos 2x + cos 3x

=> 2 * cos (4x / 2) * cos (2x / 2) + cos 2x

=> 2 cos 2x * cos x + cos 2x

=> cos 2x ( 1 + 2 cos x)

Therefore we prove that

cos x + cos 2x + cos 3x = cos 2x(1 + 2cos x)