Question

If (cos theta + sin theta) = root2 cos theta, then prove that cos theta - sin theta = root2 sin theta.....

Answer 1

Trigonometry,

Let 'e' instead of theta

We have
(cose + sine) = √2cose

Have to prove that,
cose - sine = √2sine

Now,
(cose + sine) = √2cose
or, sine = √2cose - cose
or, sine = cose(√2 - 1)
or, sine/(√2 - 1) = cose
or, √2sine + sine = cose. [multiplying √2 + 1 in the numerator and the denominator at the LHS]
or, √2sine = cose - sine [proved]

That's it
Hope it helped ^_^

Answer 2

-> cos θ + sin θ = √2 cos θ

--> sin θ = √2 cos θ - cos θ

=> sin θ = ( √2 - 1 ) cos θ

=> [ sin θ / ( √2 - 1 ) ] = cos θ

=> [ sin θ ( √2 + 1 ) / ( 2 - 1 ) ] = cos θ

0_0 --> We rationalized the denominator in the 2nd step ^_^ 

=> [ √2 sin θ + sin θ ] = cos θ

=> cos θ - sin θ = √2 sin θ