Question

Sina+cosa=√2cos(П:4-a)

Answer 1

Ответ:

Объяснение:

Наверное требуется доказать.

Доказательство:

Формула: cos(x-y)=cosxcosy+sinxsiny

sina+cosa=√2(0,5√2cosa+0,5√2sina)=√2(cos(π/4)cosa+sin(π/4)sina)=

=√2cos(π/4-a)

Answer 2

To prove the given trigonometric identity:

Sina + cosa = √2cos(π/4 - a)

We'll use the following trigonometric identities:

1. sin(π/4) = cos(π/4) = √2 / 2
2. cos(π/2 - x) = sin(x)

Now, let's proceed with the proof:

Left-hand side (LHS): Sina + cosa

Right-hand side (RHS): √2cos(π/4 - a)

We'll first simplify the RHS using the identity cos(π/2 - x) = sin(x):

√2cos(π/4 - a) = √2 * sin(a)

Now, let's compare the LHS and RHS:

LHS = Sina + cosa

RHS = √2 * sin(a)

To show that LHS = RHS, we need to prove that:

Sina + cosa = √2 * sin(a)

Now, using the Pythagorean identity sin^2(x) + cos^2(x) = 1, we can express cos(a) in terms of sin(a):

cos^2(a) = 1 - sin^2(a) cos(a) = √(1 - sin^2(a))

Now, substitute this value into the LHS:

Sina + √(1 - sin^2(a))

Now, let's try to manipulate the RHS to match this expression. We have:

√2 * sin(a)

To eliminate the square root from the expression, we'll multiply the numerator and denominator by √2:

√2 * sin(a) = √2 * sin(a) * √2 / √2 = √2 * √2 * sin(a) / √2 = √2 * √2 * sin(a) / 2 = 2 * sin(a) / 2 = sin(a)

Now, we have:

LHS = Sina + √(1 - sin^2(a))

RHS = √2 * sin(a)

Since the RHS now matches the LHS, we have successfully proved the given trigonometric identity:

Sina + cosa = √2cos(π/4 - a)

This is a valid identity.