Question

Cos (π/6-a) - sin (a-π/6). помогите пж :)

Answer 1

использую формулу cos(a-b),следственно во втором sin(a-b).

Удачи.

Answer 2

Sure, I'd be happy to help you simplify the expression:

We'll use the trigonometric identities for cosine and sine of the difference of two angles:

1. Cosine of the difference of two angles: cos(x - y) = cos(x) * cos(y) + sin(x) * sin(y)

2. Sine of the difference of two angles: sin(x - y) = sin(x) * cos(y) - cos(x) * sin(y)

Now, let's apply these identities to your expression:

cos(π/6 - a) - sin(a - π/6)

Step 1: Identify the angles: x = π/6 y = a

Step 2: Plug the angles into the identities:

cos(π/6 - a) = cos(π/6) * cos(a) + sin(π/6) * sin(a) sin(a - π/6) = sin(a) * cos(π/6) - cos(a) * sin(π/6)

Step 3: Simplify the expressions for cos(π/6) and sin(π/6):

cos(π/6) = √3/2 sin(π/6) = 1/2

Step 4: Substitute the simplified expressions back into the identities:

cos(π/6 - a) = (√3/2) * cos(a) + (1/2) * sin(a) sin(a - π/6) = sin(a) * (√3/2) - cos(a) * (1/2)

Step 5: Rearrange the terms:

cos(π/6 - a) = (√3/2) * cos(a) + (1/2) * sin(a) sin(a - π/6) = (1/2) * sin(a) - (√3/2) * cos(a)

Finally, the simplified expression is:

cos(π/6 - a) - sin(a - π/6) = (√3/2) * cos(a) + (1/2) * sin(a) - ((1/2) * sin(a) - (√3/2) * cos(a))

Now, we can see that the (1/2) * sin(a) terms cancel each other out, leaving us with:

cos(π/6 - a) - sin(a - π/6) = (√3/2) * cos(a) - (√3/2) * cos(a)

The two terms with cos(a) also cancel each other out, resulting in a simplified expression of:

cos(π/6 - a) - sin(a - π/6) = 0

So the final answer is 0.