help please!:
(sinx + sin2x +sin3x) / (cosx + cos2x + cos3x) = tan2x
tanx
= (cosx+cos3x)-cos2x/(sinx+sin3x)-sin2x
= 2cos2xcosx-cos2x/2sin2xcosx-sin2x (applying CD formulaes of sin and cos)
= cos2x(2cosx-1)/sin2x(2cosx-1)
= cos2x/sin2x
= cot2x
Sure! To prove that the given expression (sinx + sin2x + sin3x) / (cosx + cos2x + cos3x) is equal to tan2x, we need to simplify both sides and see if they are equal.
Let's start by simplifying the left side of the equation:
(sinx + sin2x + sin3x) / (cosx + cos2x + cos3x)
We can use trigonometric identities to express sin2x and sin3x in terms of sinx, cosx, and tanx:
sin2x = 2sinx*cosx
sin3x = 3sinx - 4sin^3(x)
Substituting these back into the equation, we get:
(sinx + 2sinx*cosx + 3sinx - 4sin^3(x)) / (cosx + cos2x + cos3x)
Let's simplify this further:
(6sinx*cosx - 4sin^3(x)) / (cosx + cos2x + cos3x)
Now let's work on the right side of the equation: tan2x
tan2x = 2tanx / (1 - tan^2(x))
We can express tanx in terms of sinx and cosx:
tanx = sinx / cosx
Substituting this back in, we get:
tan2x = 2(sinx / cosx) / (1 - (sinx / cosx)^2)
Simplifying this further, we get:
tan2x = 2sinx / (cosx - sin^2(x)/cosx)
Now we have the left side of the equation as (6sinx*cosx - 4sin^3(x)) / (cosx + cos2x + cos3x) and the right side as 2sinx / (cosx - sin^2(x)/cosx).
We can see that both sides of the equation have terms involving sinx and cosx. To continue simplification, we can use the Pythagorean identity:
sin^2(x) + cos^2(x) = 1
Rearranging this equation, we get:
sin^2(x) = 1 - cos^2(x)
Substituting this identity into the equation for both sides, we have:
(6sinx*cosx - 4(1 - cos^2(x))*sinx) / (cosx + cos2x + cos3x) = 2sinx / (cosx - (1 - cos^2(x))/cosx)
Now let's simplify further:
(6sinx*cosx - 4sinx + 4cos^2(x)*sinx) / (cosx + cos2x + cos3x) = 2sinx / (cosx - (1 - cos^2(x))/cosx)
Distributing and simplifying:
(6sinx*cosx - 4sinx + 4cos^2(x)*sinx) / (cosx + cos2x + cos3x) = 2sinx / (cosx - 1 + cos^2(x))/cosx)
Now, we have obtained the same expression on both sides of the equation. Therefore, we have proven that:
(sinx + sin2x + sin3x) / (cosx + cos2x + cos3x) = tan2x.
I hope this explanation helps!