Question

Sin альфа +2 sin2альфа+sin3aлфа=tg2 альфа/cosальфа+ 2cosa+cos3альфа

Answer 1

sin(-a)=-sina.

Тогда заданная разность преобразуется в сумму - и это формула косинуса разности двух углов: , т. е. получится в ответе cos(a-2a)=cos(-a)=cosa.

б) заданное выражение представляет из себя формулу синуса суммы двух углов: sin(2a+3a)=sin5a.

Answer 2

To prove the trigonometric identity sin α + 2sin^2α + sin 3α = tan^2α / (cos α + 2cosα + cos 3α), we'll first simplify both sides of the equation.

Step 1: Simplify the left-hand side (LHS):

sin α + 2sin^2α + sin 3α

Step 2: Apply the trigonometric identity sin 3α = 3sin α - 4sin^3α:

sin α + 2sin^2α + 3sin α - 4sin^3α

Step 3: Combine like terms:

4sin α + 2sin^2α - 4sin^3α

Step 4: Factor out common terms:

2sin α(2 - 2sin^2α)

Step 5: Use the identity 2sin^2α = 1 - cos(2α):

2sin α(1 - cos(2α))

Step 6: Now let's simplify the right-hand side (RHS):

tan^2α / (cos α + 2cosα + cos 3α)

Step 7: Apply the trigonometric identities: tan^2α = sec^2α - 1, and cos 3α = 4cos^3α - 3cos α:

(sec^2α - 1) / (cos α + 2cosα + 4cos^3α - 3cos α)

Step 8: Combine like terms:

(sec^2α - 1) / (4cos^3α + cos α + 2cos α)

Step 9: Use the identity sec^2α = 1 + tan^2α:

(1 + tan^2α - 1) / (4cos^3α + 3cos α)

Step 10: Simplify further:

tan^2α / (4cos^3α + 3cos α)

Now, we need to show that the LHS equals the RHS, so we can compare them:

2sin α(1 - cos(2α)) = tan^2α / (4cos^3α + 3cos α)

We know that 1 - cos(2α) = sin^2α and 2sin α cos α = sin(2α), so the left-hand side becomes:

2sin α(1 - cos(2α)) = 2sin α sin^2α = 2sin^3α

Now, the equation becomes:

2sin^3α = tan^2α / (4cos^3α + 3cos α)

Next, we'll convert tan^2α to its trigonometric equivalent using the identity tan^2α = sec^2α - 1:

(sec^2α - 1) = tan^2α sec^2α = tan^2α + 1

Now, let's rewrite the equation:

2sin^3α = (sec^2α + 1) / (4cos^3α + 3cos α)

Now, we'll use the identity sec^2α = 1 / cos^2α:

2sin^3α = (1 / cos^2α + 1) / (4cos^3α + 3cos α)

Next, multiply both sides by cos^2α:

2sin^3α cos^2α = (1 + cos^2α) / (4cos^3α + 3cos α)

Now, we'll express sin^2α in terms of cos^2α using the identity sin^2α = 1 - cos^2α:

2(1 - cos^2α) cos^2α = (1 + cos^2α) / (4cos^3α + 3cos α)

Expand the left-hand side:

2cos^2α - 2cos^4α = (1 + cos^2α) / (4cos^3α + 3cos α)

Now, let's combine the terms on the left-hand side:

2cos^2α - 2cos^4α = 1 + cos^2α / (4cos^3α + 3cos α)

Now, multiply both sides by (4cos^3α + 3cos α) to eliminate the denominator:

(2cos^2α - 2cos^4α) * (4cos^3α + 3cos α) = 1 + cos^2α

Expand the left-hand side:

8cos^5α + 6cos^3α - 8cos^7α - 6cos^5α = 1 + cos^2α

Simplify the equation:

6cos^3α - 8cos^7α = 1 + cos^2α

Now, apply the identity 1 + cos^2α = sin^2α:

6cos^3α - 8cos^7α = sin^2α

Finally, recall that sin^2α = 1 - cos^2α:

6cos^3α - 8cos^7α = 1 - cos^2α

Now, add cos^2α to both sides of the equation:

6cos^3α - 7cos^2α - 8cos^7α = 1

At this point, it seems we cannot reduce the equation further to get a direct identity. The original statement provided might contain a mistake or an error in transcription. Double-checking the original statement or reevaluating the steps might help clarify the correct identity.