Question

Sin(π+a)+cos(2π+a)-sin(-a)-cos(-a) ​

Answer 1

По формулам приведения:

sin(π+a)=-sina

cos(2π+a)=cosa

Так как синус - нечетная функция, а косинус четная, то

sin(-a)=-sina

cos(-a)=cosa ​

sin(π+a)+cos(2π+a)-sin(-a)-cos(-a)=-sina+cosa+sina-cosa=0

Answer 2

Let's simplify the given expression step by step:

1. Sin(π + a): Using the trigonometric identity sin(π + θ) = -sin(θ), we have: Sin(π + a) = -sin(a)

2. Cos(2π + a): Using the trigonometric identity cos(2π + θ) = cos(θ), we have: Cos(2π + a) = cos(a)

3. Sin(-a): Using the trigonometric identity sin(-θ) = -sin(θ), we have: Sin(-a) = -(-sin(a)) = sin(a)

4. Cos(-a): Using the trigonometric identity cos(-θ) = cos(θ), we have: Cos(-a) = cos(a)

Now, let's substitute these results back into the original expression:

(-sin(a)) + cos(a) - sin(a) - cos(a)

Combining like terms:

-2sin(a)

So, the simplified expression is -2sin(a).