TBQ - Prove the following I — task. Maths TNSB Mentoring, Class 10 Crash course.

\cos p + \sin p = \sqrt{2} \cos p

, then check that

\sqrt{2} \sin p = \cos p - \sin p

Proof:

Consider

\cos p + \sin p = \sqrt{2} \cos p

Squaring on both sides, we get:

Hence, we proved.

Answer variants:

2 \cos p \sin p = 2 \cos^{2} p - \cos^{2} p - \sin^{2} p

\cos^{2} p + \sin^{2} p + 2 \cos p \sin p = 2 \cos^{2} p

\sqrt{2} \sin p = \cos p - \sin p

{(\cos p + \sin p)}^{2} = {(\sqrt{2} \cos p)}^{2}

2 \cos p \sin p = (\cos p + \sin p) (\cos p - \sin p)

\frac{2 \cos p \sin p}{\sqrt{2} \cos p} = \cos p - \sin p

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

2 \cos p \sin p = \cos^{2} p - \sin^{2} p

Did you find an error?

4. TBQ - Prove the following I