Question

Sin^2a+cos^2a.cos2b=cos^2b-sin^2b.cos2a

Answer 1

Answer on Question #79082 – Math – Trigonometry

$$\sin^2 a + \cos^2 a \cdot \cos^2 b = \cos^2 b - \sin^2 b \cdot \cos^2 a$$

We shall use the equality

$$\cos^2 x = \cos^2 x - \sin^2 x.$$

Then:

$$\begin{array}{l} \sin^2 a + \cos^2 a \cdot (\cos^2 b - \sin^2 b) = \cos^2 b - \sin^2 b \cdot (\cos^2 a - \sin^2 a), \\ \sin^2 a + \cos^2 a \cdot \cos^2 b - \cos^2 a \cdot \sin^2 b = \cos^2 b - \sin^2 b \cdot \cos^2 a + \sin^2 b \cdot \sin^2 a, \\ \sin^2 a + \cos^2 a \cdot \cos^2 b = \cos^2 b + \sin^2 b \cdot \sin^2 a, \\ \sin^2 a - \sin^2 b \cdot \sin^2 a = \cos^2 b - \cos^2 a \cdot \cos^2 b, \\ \sin^2 a \cdot (1 - \sin^2 b) = \cos^2 b \cdot (1 - \cos^2 a), \\ \sin^2 a \cdot \cos^2 b = \cos^2 b \cdot \sin^2 a. \end{array}$$

So,

$$\sin^2 a \cdot \cos^2 b = \sin^2 a \cdot \cos^2 b,$$

which is true.

If backward steps will be performed, then the initial formula (1) will be obtained.

It shows that the formula (1) is true.

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