Question

3)cos(-π)+ctg(-π/2)-sin(-3/2π)+ctg(-π/4)= 4)3-sin^2(-π/3)-cos^2(-π/3)/2cos(-π/4)=

Answer 1

3)cos(-π)+ctg(-π/2)-sin(-3/2π)+ctg(-π/4)=   -1+0 -1 -1=-3

4)3-sin^2(-π/3)-cos^2(-π/3)/2cos(-π/4)=3 -3/4-1/4/1=2

Answer 2

Let's simplify each expression step by step:

1. cos(-π) + ctg(-π/2) - sin(-3/2π) + ctg(-π/4)

Remember the trigonometric identities:

• cos(-θ) = cos(θ) (cosine is an even function)
• sin(-θ) = -sin(θ) (sine is an odd function)
• cot(-θ) = -cot(θ) (cotangent is an odd function)

So, the expression becomes: cos(π) - cot(π/2) + sin(3/2π) + cot(π/4)

Now, let's find the values of each term: cos(π) = -1 (cosine of π is -1) cot(π/2) = 0 (cotangent of π/2 is not defined, but it approaches 0) sin(3/2π) = -1 (sine of 3/2π is -1) cot(π/4) = 1 (cotangent of π/4 is 1)

Now substitute the values into the expression: -1 - 0 - (-1) + 1 = -1 + 1 = 0

So, the first expression simplifies to 0.

2. 3 - sin^2(-π/3) - cos^2(-π/3) / 2cos(-π/4)

Using the same trigonometric identities as before, we have:

• sin(-θ) = -sin(θ) (sine is an odd function)
• cos(-θ) = cos(θ) (cosine is an even function)

So, the expression becomes: 3 - sin^2(π/3) - cos^2(π/3) / 2cos(π/4)

Now, let's find the values of each term: sin(π/3) = √(3)/2 (sine of π/3 is √(3)/2) cos(π/3) = 1/2 (cosine of π/3 is 1/2) cos(π/4) = √(2)/2 (cosine of π/4 is √(2)/2)

Now substitute the values into the expression: 3 - (√(3)/2)^2 - (1/2)^2 / 2(√(2)/2) 3 - (3/4) - (1/4) / (√(2)/2) 3 - 1 / (√(2)/2) 2 / (√(2)/2) 2 * (2/√(2)) 2 * √(2) 2√(2)

So, the second expression simplifies to 2√(2).

Therefore, the final simplified forms of the expressions are:

1. cos(-π) + ctg(-π/2) - sin(-3/2π) + ctg(-π/4) = 0
2. 3 - sin^2(-π/3) - cos^2(-π/3) / 2cos(-π/4) = 2√(2)