Question

1) \\ cos( \pi - \alpha )ctg( \pi /2- \alpha ) 2) \\ sin(270- \alpha )-sin(270+ \alpha ) 3) \\ sin(2t-21 \pi )

Answer 1

1) cos(P-a)=-cosa (II четверть)  ctg(P/2-a)=tga=sina/cosa (I четверть) -cosa * sina/cosa=-sina  2)sin(270-a)=-cosa (III ч.)  sin(270+a)=-cosa (IV ч.) -cosa - (-cosa)=0       3)sin(2t-21P)=-sin(21P-2t)=-sin(P-2t)=-sin2t=-2sintcost      4)ctg(P/2-a)=tga   sin('beta'-P/2)=-sin(P/2-'beta')=-cos'beta'
  cos(P-'beta')=cos'beta' tg(-a)=-tga tga * (-cos'beta')/cos/'beta' * (-tga)=1

Answer 2

1. Starting with the first expression:

cos(π - α)ctg(π/2 - α)

Using the fact that ctg(x) = 1/tan(x), we can rewrite the expression as:

cos(π - α) / tan(π/2 - α)

Using the trigonometric identities cos(π - x) = -cos(x) and tan(π/2 - x) = cot(x), we get:

-cos(α) / cot(α)

Using the fact that cot(x) = 1/tan(x), we can simplify to:

-cos(α)tan(α)

So the final expression is:

• cos(α) tan(α)

2. Moving on to the second expression:

sin(270 - α) - sin(270 + α)

Using the fact that sin(x + π) = -sin(x), we can rewrite the second term as:

sin(π + 270 + α) = -sin(α - π/2)

Using the fact that sin(x - π/2) = cos(x), we get:

-sin(π/2 - α)

Now, using the fact that sin(π/2 - x) = cos(x), we can rewrite the first term as:

cos(α)

So the final expression is:

cos(α) - sin(π/2 - α) = cos(α) - cos(α) = 0

3. For the third expression:

sin(2t - 21π)

Using the identity sin(x - π) = -sin(x), we can rewrite the expression as:

-sin(21π - 2t)

Using the identity sin(x + π) = -sin(x), we can further simplify to:

sin(2t - 21π) = -sin(2t + π)

Finally, using the identity sin(x + π) = -sin(x), we can rewrite the expression as:

-sin(2t)