1.

If cosΦ - sinΦ = √2sinΦ , prove that cosΦ + sinΦ = √2sinΦ

Answer»
cos - sin = \sqrt{2} sin

Now SQUARE on both side

(cos - sin) {}^{2} = ( \sqrt{2} sin) {}^{2}

cos {}^{2} + sin {}^{2} - 2sincos = 2sin {}^{2}

- 2sincos = 2sin {}^{2} - sin {}^{2} - cos {}^{2}

- 2sincos = sin {}^{2} - cos {}^{2}

Take - COMMON from RHS


- 2sincos = - (cos {}^{2} - sin {}^{2} )

2sincos = (cos + sin)(cos - sin)

now \: put \: cos - sin = \sqrt{2} sin

2sincos \div \sqrt{2} sin = cos + sin

cos + sin = \sqrt{2} cos


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