Cos70cos10+sin70sin10 e =2cos^2 2x-1
Cos70cos10+sin70sin10 e =2cos^2 2x-1
don't know what that "e" is supposed to be, seems to be superfluous.
recall : cos (A-B) = cosAcoB + sinAsinB
Cos70cos10+sin70sin10 =2cos^2 2x-1
cos(70-10) = 2cos^2 2x - 1
cos60 = 2cos^2 2x - 1
recall cos 2A = 2cos^2 A - 1
so 2cos^2 2x - 1 = cos 4x
cos 4x = cos60
4x = 60° or 4x = 300°
x = 15° or x = 75°
since cos 2x has a period of 180°
other solutions would be
x = 15+180 = 195°
or
x = 75+180 = 255°
To prove the equation, we will simplify the left-hand side (LHS) and the right-hand side (RHS) separately and show that they are equal.
Starting with LHS:
cos70cos10 + sin70sin10
Using the formula for the cosine of the sum of two angles:
cos(A + B) = cosAcosB - sinAsinB
We can rewrite the equation as:
cos(70 - 10) = cos(2x)
Now, let's simplify the LHS using the formula:
cos(70 - 10) = cos60
The angle of 60 degrees has a special property where cos60 = 0.5. Therefore, we have:
cos60 = 0.5
Now let's move on to the RHS:
2cos^2(2x) - 1
Using trigonometric identities:
cos(2x) = 2cos^2(x) - 1
We can rewrite the equation as:
2cos^2(x) - 1 = 2cos^2(2x) - 1
Comparing the RHS with the LHS, we can see that they are equal:
cos60 = 0.5 = 2cos^2(x) - 1 = 2cos^2(2x) - 1
Therefore, we have proven that:
cos70cos10 + sin70sin10 = 2cos^2(2x) - 1