Question

Cos theta +sin theta = root 2 cos theta then show cos theta- sin theta = root 2 sin theta

Answer 1

Answer:

Step-by-step explanation:

Given: cos theta + sin theta = $\sqrt{2}$ cos theta

Thus, sin theta = cos theta ( $\sqrt{2}$ -1)

cos theta = $\frac{sin theta}{\sqrt{2}-1 }$

               = $\frac{sin theta (\sqrt{2}+1) }{(\sqrt{2} +1)(\sqrt{2}-1) }$

               = $\frac{sin theta(\sqrt{2}+1) }{2-1}$

              = sin theta($\sqrt{2}$+1)

              = $\sqrt{2}$sin theta + sin theta

Thus, cos theta - sin theta = $\sqrt{2}$sin theta

Answer 2

cosx + sinx = √2 cosx

sinx = √2 cosx - cosx

       = cosx (√2 - 1)

cosx = sinx / (√2 - 1)

To prove:

LHS

cosx - sinx

sinx/(√2-1) - sinx

(sinx-√2sinx +sinx)/(√2-1)

(2sinx - √2sinx)/(√2-1)

[√2sinx(√2-1)]/(√2-1)

= √2sinx = RHS